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# Angles, Angular Conversions, and Trigonometric Functions

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE and Malta Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
State the reasons for every step as applicable
Show all work
Simplify completely
Simplified radicals should be as is
Simplified fractions should be as is

Unless otherwise stated:
When asked to convert to only degrees, round to two decimal places as needed.
When asked to convert to $D^\circ M' S''$, round to the nearest whole number of seconds.

(1.) Determine $3$ positive and $3$ negative coterminal angles of $20^\circ$

$\alpha = 20^\circ \\[3ex] Let\:\: \beta = coterminal\:\: angle \\[3ex] \beta = \alpha + 360k \\[3ex] \underline{Positive\:\: coterminal\:\: angles} \\[3ex] k = 1 \\[3ex] \beta = 20 + 360(1) \\[3ex] \beta = 20 + 360 \\[3ex] \beta = 380^\circ \\[3ex] k = 2 \\[3ex] \beta = 20 + 360(2) \\[3ex] \beta = 20 + 720 \\[3ex] \beta = 740^\circ \\[3ex] k = 3 \\[3ex] \beta = 20 + 360(3) \\[3ex] \beta = 20 + 1080 \\[3ex] \beta = 1100^\circ \\[3ex] \underline{Negative\:\: coterminal\:\: angles} \\[3ex] k = -1 \\[3ex] \beta = 20 + 360(-1) \\[3ex] \beta = 20 - 360 \\[3ex] \beta = -340^\circ \\[3ex] k = -2 \\[3ex] \beta = 20 + 360(-2) \\[3ex] \beta = 20 - 720 \\[3ex] \beta = -700^\circ \\[3ex] k = -3 \\[3ex] \beta = 20 + 360(-3) \\[3ex] \beta = 20 - 1080 \\[3ex] \beta = -1060^\circ$
(2.) Determine $3$ positive and $3$ negative coterminal angles of $-20^\circ$

$\alpha = -20^\circ \\[3ex] Let\:\: \beta = coterminal\:\: angle \\[3ex] \beta = \alpha + 360k \\[3ex] \underline{Positive\:\: coterminal\:\: angles} \\[3ex] k = 1 \\[3ex] \beta = -20 + 360(1) \\[3ex] \beta = -20 + 360 \\[3ex] \beta = 340^\circ \\[3ex] k = 2 \\[3ex] \beta = -20 + 360(2) \\[3ex] \beta = -20 + 720 \\[3ex] \beta = 700^\circ \\[3ex] k = 3 \\[3ex] \beta = -20 + 360(3) \\[3ex] \beta = -20 + 1080 \\[3ex] \beta = 1060^\circ \\[3ex] \underline{Negative\:\: coterminal\:\: angles} \\[3ex] k = -1 \\[3ex] \beta = -20 + 360(-1) \\[3ex] \beta = -20 - 360 \\[3ex] \beta = -380^\circ \\[3ex] k = -2 \\[3ex] \beta = -20 + 360(-2) \\[3ex] \beta = -20 - 720 \\[3ex] \beta = -740^\circ \\[3ex] k = -3 \\[3ex] \beta = -20 + 360(-3) \\[3ex] \beta = -20 - 1080 \\[3ex] \beta = -1100^\circ$
(3.) Do not use a calculator to solve.
Use only what you were given.
But, you may check/verify with a calculator.
Find the six function values of $22^\circ$ if

$\sin 68^\circ = 0.9272 \\[3ex] \cos 68^\circ = 0.3746 \\[3ex] \tan 68^\circ = 2.4751 \\[3ex] \cot 68^\circ = 0.4040 \\[3ex] \sec 68^\circ = 2.6695 \\[3ex] \csc 68^\circ = 1.0785$

$22^\circ \:\:and\:\: 68^\circ \:\:are\:\: complementary \\[3ex] \underline{Cofunction\:\: Identities} \\[3ex] \sin 22 = \cos 68 = 0.3746 \\[3ex] \cos 22 = \sin 68 = 0.9272 \\[3ex] \tan 22 = \cot 68 = 0.4040 \\[3ex] \cot 22 = \tan 68 = 2.4751 \\[3ex] \sec 22 = \csc 68 = 1.0785 \\[3ex] \csc 22 = \sec 68 = 2.6695$
(4.) Find the six function values of $39^\circ$ in terms of $x$, $y$, and $p$
Given that:

$\sin 51^\circ = x \\[3ex] \cos 51^\circ = y \\[3ex] \tan 51^\circ = p$

$\underline{Reciprocal\:\: Identities} \\[3ex] \csc 51 = \dfrac{1}{\sin 51} = \dfrac{1}{x} \\[5ex] \sec 51 = \dfrac{1}{\cos 51} = \dfrac{1}{y} \\[5ex] \cot 51 = \dfrac{1}{\tan 51} = \dfrac{1}{p} \\[5ex] 39^\circ \:\:and\:\: 51^\circ \:\:are\:\: complementary \\[3ex] \underline{Cofunction\:\: Identities} \\[3ex] \sin 39 = \cos 51 = y \\[3ex] \cos 39 = \sin 51 = x \\[3ex] \tan 39 = \cot 51 = \dfrac{1}{p} \\[5ex] \cot 39 = \tan 51 = p \\[3ex] \sec 39 = \csc 51 = \dfrac{1}{x} \\[3ex] \csc 39 = \sec 51 = \dfrac{1}{y}$
(5.) Given that $\sec \beta = 1.5183$, find $\sin(90 - \beta)$

$\beta \:\:and\:\: (90 - \beta) \:\:are\:\: complementary \\[3ex] \sec \beta = \csc (90 - \beta) ...Cofunction\:\: Identity \\[3ex] \implies \csc(90 - \beta) = 1.5183 \\[3ex] \csc(90 - \beta) = \dfrac{1}{\sin(90 - \beta)} ...Reciprocal\:\: Identity \\[5ex] \implies \dfrac{1}{\sin(90 - \beta)} = 1.5183 \\[5ex] 1 = 1.5183 * \sin(90 - \beta) \\[3ex] 1.5183 * \sin(90 - \beta) = 1 \\[3ex] \sin(90 - \beta) = \dfrac{1}{1.5183} \\[5ex] \sin(90 - \beta) = 0.6586$
(6.) Convert $24^\circ 54'$ to $^\circ$

$24^\circ 54' \:\:to\:\: ^\circ \\[3ex] 54' = \dfrac{54}{60} = 0.9^\circ \\[5ex] = 24^\circ + 0.9^\circ \\[3ex] = 24.9^\circ \\[3ex]$
(7.) Convert $83^\circ 30''$ to $^\circ$

$83^\circ 30'' \:\:to\:\: ^\circ \\[3ex] 30'' = \dfrac{30}{3600} = 0.0083333^\circ \\[5ex] = 83^\circ + 0.0083333^\circ \\[3ex] = 83.0083333^\circ \\[3ex] \approx 83.01^\circ$
(8.) Convert $52' 58''$ to $^\circ$

$52' 58'' \:\:to\:\: ^\circ \\[3ex] 52' = \dfrac{52}{60} = 0.866667^\circ \\[3ex] 58'' = \dfrac{58}{3600} = 0.016111^\circ \\[3ex] = 0.866667^\circ + 0.016111^\circ \\[3ex] = 0.88278^\circ \\[3ex] \approx 0.88^\circ$
(9.) Convert $42^\circ 48' 43''$ to $^\circ$

$42^\circ 48' 43'' \:\:to\:\: ^\circ \\[3ex] 48' = \dfrac{48}{60} = 0.8^\circ \\[3ex] 43'' = \dfrac{43}{3600} = 0.01194^\circ \\[3ex] = 42^\circ + 0.8^\circ + 0.01194^\circ \\[3ex] = 42.81194^\circ \\[3ex] \approx 42.81^\circ$
(10.) Convert $40.25^\circ$ to $D^\circ M' S''$

$40.25^\circ \:\:to\:\: D^\circ M'S'' \\[3ex] = 40^\circ + 0.25^\circ \\[3ex] 0.25^\circ \:\:to\:\: ' = 0.25 * 60 = 15' \\[3ex] No\:\: remainder \implies 0'' \\[3ex] = 40^\circ 15' 0''$
(11.) Convert $28.821^\circ$ to $D^\circ M' S''$

$28.821^\circ \:\:to\:\: D^\circ M'S'' \\[3ex] = 28^\circ + 0.821^\circ \\[3ex] 0.821^\circ \:\:to\:\: ' = 0.821 * 60 = 49.26' \\[3ex] 49.26' = 49' + 0.26' \\[3ex] 0.26' \:\:to\:\: '' = 0.26 * 60 = 15.6'' \\[3ex] = 28^\circ 49' 15.6'' \\[3ex] \approx 28^\circ 49' 16''$
(12.) Convert $345.5782^\circ$ to $D^\circ M' S''$

$345.5782^\circ \:\:to\:\: D^\circ M'S'' \\[3ex] = 345^\circ + 0.5782^\circ \\[3ex] 0.5782^\circ \:\:to\:\: ' = 0.5782 * 60 = 34.692' \\[3ex] 34.692' = 34' + 0.692' \\[3ex] 0.692' \:\:to\:\: '' = 0.692 * 60 = 41.52'' \\[3ex] = 345^\circ 34' 41.52'' \\[3ex] \approx 345^\circ 34' 42''$
(13.) Convert $0.9^\circ$ to $D^\circ M' S''$

$0.9^\circ \:\:to\:\: D^\circ M'S'' \\[3ex] = 0^\circ + 0.9^\circ \\[3ex] 0.9^\circ \:\:to\:\: ' = 0.9 * 60 = 54' \\[3ex] No\:\: remainder \implies 0'' \\[3ex] = 0^\circ 54' 0''$
(14.) Determine the reference angle and the exact value of $\sin 3240^\circ$

$3240^\circ \gt 360^\circ \\[3ex] \beta = coterminal\:\: \angle \:\:of\:\: 3240^\circ \\[3ex] \beta = \alpha + 360k \\[3ex] \alpha = 3240^\circ \\[3ex] Let\:\: k = -9 \\[3ex] \beta = 3240 + 360(-9) \\[3ex] \beta = 3240 - 3240 \\[3ex] \beta = 0^\circ \\[3ex] 0^\circ \:\:is\:\: quadrantal\:\: means\:\: No\:\: reference\:\: \angle \\[3ex] \therefore \sin 3240^\circ = \sin 0^\circ = 0$
(15.) ACT Angle $A$ measures $\dfrac{9}{2}\pi$ radians from its initial side to its terminal side.
Angle $B$ has the same initial side and terminal side as Angle $A$.
Which of the following measures could be that of Angle $B$?

$A.\:\: 5^\circ \\[3ex] B.\:\: 14^\circ \\[3ex] C.\:\: 25^\circ \\[3ex] D.\:\: 90^\circ \\[3ex] E.\:\: 180^\circ$

The answer options are in degrees. So, we need to convert Angle $A$ to degrees

$A = \dfrac{9}{2}\pi \\[5ex] \dfrac{9}{2}\pi \:\:to\:\: ^\circ = \dfrac{9}{2}\pi * \dfrac{180}{\pi} \\[5ex] A = 9 * 90 = 810^\circ \\[3ex] B = A + 360k ...Coterminal\:\: \angle s \\[3ex]$ Looking at the options, let $k = -2$

$B = 810 + (360)(-2) \\[3ex] B = 810 - 720 \\[3ex] B = 90^\circ$
(16.) Use a calculator but do not use the trigonometric functions on the calculator.
Apply the Trigonometric Identities.
Find the six function values of $24^\circ 23' 42''$ if

$\sin 65^\circ 36' 18'' \approx 0.9107 \\[3ex] \cos 65^\circ 36' 18'' \approx 0.4130 \\[3ex] \tan 65^\circ 36' 18'' \approx 2.2050 \\[3ex]$

$\underline{Reciprocal\:\: Identities} \\[3ex] \csc 65^\circ 36' 18'' = \dfrac{1}{\sin 65^\circ 36' 18''} = \dfrac{1}{0.9107} \approx 1.0981 \\[5ex] \sec 65^\circ 36' 18'' = \dfrac{1}{\cos 65^\circ 36' 18''} = \dfrac{1}{0.4130} \approx 2.4213 \\[5ex] \cot 65^\circ 36' 18'' = \dfrac{1}{\tan 65^\circ 36' 18''} = \dfrac{1}{2.2050} \approx 0.4535 \\[5ex] 24^\circ 23' 42'' \:\:and\:\: 65^\circ 36' 18'' \:\:are\:\: complementary \\[3ex] \underline{Cofunction\:\: Identities} \\[3ex] \sin 24^\circ 23' 42'' = \cos 65^\circ 36' 18'' \approx 0.4130 \\[3ex] \cos 24^\circ 23' 42'' = \sin 65^\circ 36' 18'' \approx 0.9107 \\[3ex] \tan 24^\circ 23' 42'' = \cot 65^\circ 36' 18'' \approx 0.4535 \\[3ex] \cot 24^\circ 23' 42'' = \tan 65^\circ 36' 18'' \approx 2.2050 \\[3ex] \sec 24^\circ 23' 42'' = \csc 65^\circ 36' 18'' \approx 1.0981 \\[3ex] \csc 24^\circ 23' 42'' = \sec 65^\circ 36' 18'' \approx 2.4213$
(17.) Determine the reference angle and the exact value of $\cos 675^\circ$

$675^\circ \gt 360^\circ \\[3ex] \beta = coterminal\:\: \angle \:\:of\:\: 675^\circ \\[3ex] \beta = \alpha + 360k \\[3ex] \alpha = 675^\circ \\[3ex] Let\:\: k = -1 \\[3ex] \beta = 675 + 360(-1) \\[3ex] \beta = 675 - 360 \\[3ex] \beta = 315^\circ \\[3ex] 315 \:\:is\:\: in\:\: the\:\: 4th\:\: quadrant \\[3ex] reference\:\: \angle = 360 - 315 = 45 \\[3ex] \cos 315 = \cos 45 = \dfrac{\sqrt{2}}{2} ...4th\:\: Quadrant\:\: Identity \\[5ex] \therefore \cos 675^\circ = \cos 315^\circ = \cos 45^\circ = \dfrac{\sqrt{2}}{2}$
(18.) Determine the reference angle and the exact value of $\cos 585^\circ$

$585^\circ \gt 360^\circ \\[3ex] \beta = coterminal\:\: \angle \:\:of\:\: 585^\circ \\[3ex] \beta = \alpha + 360k \\[3ex] \alpha = 585^\circ \\[3ex] Let\:\: k = -1 \\[3ex] \beta = 585 + 360(-1) \\[3ex] \beta = 585 - 360 \\[3ex] \beta = 225^\circ \\[3ex] 225 \:\:is\:\: in\:\: the\:\: 3rd\:\: quadrant \\[3ex] reference\:\: \angle = 225 - 180 = 45 \\[3ex] \cos 225 = -\cos 45 = -\dfrac{\sqrt{2}}{2} ...3rd\:\: Quadrant\:\: Identity \\[5ex] \therefore \cos 585^\circ = \cos 225^\circ = -\cos 45^\circ = -\dfrac{\sqrt{2}}{2}$
(19.) Determine the reference angle and the exact value of $\csc (-240)^\circ$

$-240^\circ \lt 0^\circ \\[3ex] \beta = coterminal\:\: \angle \:\:of\:\: -240^\circ \\[3ex] \beta = \alpha + 360k \\[3ex] \alpha = -240^\circ \\[3ex] Let\:\: k = 1 \\[3ex] \beta = -240 + 360(1) \\[3ex] \beta = -240 + 360 \\[3ex] \beta = 120^\circ \\[3ex] 120 \:\:is\:\: in\:\: the\:\: 2nd\:\: quadrant \\[3ex] reference\:\: \angle = 180 - 120 = 60 \\[3ex] \csc(-240) = \csc 60 = \dfrac{2\sqrt{3}}{3} ...2nd\:\: Quadrant\:\: Identity \\[5ex] \therefore \csc(-240)^\circ = \csc 60^\circ = \dfrac{2\sqrt{3}}{3}$
(20.) Determine the reference angle and the exact value of $\cot 390^\circ$

$390^\circ \gt 360^\circ \\[3ex] \beta = coterminal\:\: \angle \:\:of\:\: 390^\circ \\[3ex] \beta = \alpha + 360k \\[3ex] \alpha = 390^\circ \\[3ex] Let\:\: k = -1 \\[3ex] \beta = 390 + 360(-1) \\[3ex] \beta = 390 - 360 \\[3ex] \beta = 30^\circ \\[3ex] 30 \:\:is\:\: in\:\: the\:\: 1st\:\: quadrant \\[3ex] reference\:\: \angle = 30 \\[3ex] \cot 390 = \cot 30 = \sqrt{3} ...1st\:\: Quadrant\:\: Identity \\[3ex] \therefore \cot 390^\circ = \cot 30^\circ = \sqrt{3}$

(21.) Determine the reference angle and the exact value of $\tan (315)^\circ$

$315 \:\:is\:\: in\:\: the\:\: 4th\:\: quadrant \\[3ex] reference\:\: \angle = 360 - 315 = 45 \\[3ex] \tan 315 = -\tan 45 = -1 ...4th\:\: Quadrant\:\: Identity \\[3ex] \therefore \tan 315^\circ = -\tan 45^\circ = -1$
(22.) Determine the reference angle and the exact value of $\tan (-315)^\circ$

$-315^\circ \lt 0^\circ \\[3ex] \beta = coterminal\:\: \angle \:\:of\:\: -315^\circ \\[3ex] \beta = \alpha + 360k \\[3ex] \alpha = -315^\circ \\[3ex] Let\:\: k = 1 \\[3ex] \beta = -315 + 360(1) \\[3ex] \beta = -315 + 360 \\[3ex] \beta = 45^\circ \\[3ex] 45 \:\:is\:\: in\:\: the\:\: 1st\:\: quadrant \\[3ex] reference\:\: \angle = 45 \\[3ex] \tan (-315) = \tan 45 = 1 ...1st\:\: Quadrant\:\: Identity \\[3ex] \therefore \tan (-315)^\circ = \tan 45^\circ = 1$
(23.) Determine the other exact trigonometric values of $\phi$ for
$\sin \phi = \dfrac{1}{8}$, $\phi$ is the second quadrant

$\sin \phi = \dfrac{1}{8} = \dfrac{opp}{hyp} ... SOHCAHTOA \\[5ex] opp = 1 \\[3ex] hyp = 8 \\[3ex] hyp^2 = opp^2 + adj^2 ... Pythagorean\:\: Theorem \\[3ex] adj^2 = hyp^2 - opp^2 \\[3ex] adj^2 = 8^2 - 1^2 \\[3ex] adj^2 = 64 - 1 \\[3ex] adj^2 = 63 \\[3ex] adj = \sqrt{63} \\[3ex] adj = \sqrt{9 * 7} \\[3ex] adj = \sqrt{9} * \sqrt{7} \\[3ex] adj = 3\sqrt{7} \\[3ex] \cos \phi = -\dfrac{adj}{hyp} ... 2nd\:\: Quadrant\:\: Identity\:\: and\:\: SOHCAHTOA \\[5ex] \cos \phi = -\dfrac{3\sqrt{7}}{8} \\[5ex] \tan \phi = -\dfrac{opp}{adj} ... 2nd\:\: Quadrant\:\: Identity\:\: and\:\: SOHCAHTOA \\[5ex] \tan \phi = -\dfrac{1}{3\sqrt{7}} \\[5ex] \tan \phi = -\dfrac{1}{3\sqrt{7}} * \dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac{-1 * \sqrt{7}}{3\sqrt{7} * \sqrt{7}} = -\dfrac{\sqrt{7}}{3 * 7} \\[5ex] \tan \phi = -\dfrac{\sqrt{7}}{21} \\[5ex] \csc \phi = \dfrac{1}{\sin \phi} ... Reciprocal\:\: Identity \\[5ex] \csc \phi = 1 \div \sin \phi \\[3ex] \csc \phi = 1 \div \dfrac{1}{8} \\[5ex] \csc \phi = 1 * \dfrac{8}{1} \\[5ex] \csc \phi = 8 \\[3ex] \sec \phi = \dfrac{1}{\cos \phi} ... Reciprocal\:\: Identity \\[5ex] \sec \phi = 1 \div \cos \phi \\[3ex] \sec \phi = 1 \div -\dfrac{3\sqrt{7}}{8} \\[5ex] \sec \phi = 1 * -\dfrac{8}{3\sqrt{7}} \\[5ex] \sec \phi = -\dfrac{8}{3\sqrt{7}} * \dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac{-8 * \sqrt{7}}{3\sqrt{7} * \sqrt{7}} = -\dfrac{8\sqrt{7}}{3 * 7} \\[5ex] \sec \phi = -\dfrac{8\sqrt{7}}{21} \\[5ex] \cot \phi = \dfrac{1}{\tan \phi} ... Reciprocal\:\: Identity \\[5ex] \cot \phi = 1 \div \tan \phi \\[3ex] \cot \phi = 1 \div -\dfrac{1}{3\sqrt{7}} \\[5ex] \cot \phi = 1 * -\dfrac{3\sqrt{7}}{1} \\[5ex] \cot \phi = -3\sqrt{7}$
(24.) Determine the other exact trigonometric values of $\beta$ for
$\cot\beta = 3$, $\beta$ is the first quadrant

$\cot\beta = \dfrac{1}{\tan\beta} ... Reciprocal\:\: Identity \\[5ex] \tan\beta = \dfrac{opp}{adj} ... Quotient\:\: Identity \\[5ex] \cot\beta = 1 \div \tan\beta = 1 \div \dfrac{opp}{adj} = 1 * \dfrac{adj}{opp} = \dfrac{adj}{opp} \\[5ex] \cot\beta = \dfrac{adj}{opp} = 3 = \dfrac{3}{1} \\[5ex] adj = 3 \\[3ex] opp = 1 \\[3ex] \tan\beta = \dfrac{opp}{adj} = \dfrac{1}{3} \\[5ex] hyp^2 = opp^2 + adj^2 ... Pythagorean\:\: Theorem \\[3ex] hyp^2 = 1^2 + 3^2 \\[3ex] hyp^2 = 1 + 9 \\[3ex] hyp^2 = 10 \\[3ex] hyp = \sqrt{10} \\[3ex] \sin\beta = \dfrac{opp}{hyp} ... 1st\:\: Quad.\:\: Identity\:\: and\:\: SOHCAHTOA \\[5ex] \sin\beta = \dfrac{1}{\sqrt{10}} \\[5ex] \sin\beta = \dfrac{1}{\sqrt{10}} * \dfrac{\sqrt{10}}{\sqrt{10}} \\[5ex] \sin\beta = \dfrac{1 * \sqrt{10}}{\sqrt{10} * \sqrt{10}} \\[5ex] \sin\beta = \dfrac{\sqrt{10}}{10} \\[5ex] \csc\beta = \dfrac{1}{\sin\beta} ... Reciprocal\:\: Identity \\[5ex] \csc\beta = 1 \div \sin\beta \\[3ex] \csc\beta = 1 \div \dfrac{1}{\sqrt{10}} \\[5ex] \csc\beta = 1 * \dfrac{\sqrt{10}}{1} \\[5ex] \csc\beta = \sqrt{10} \\[3ex] \cos\beta = \dfrac{adj}{hyp} ... 1st\:\: Quad.\:\: Identity\:\: and\:\: SOHCAHTOA \\[5ex] \cos\beta = \dfrac{3}{\sqrt{10}} \\[5ex] \cos\beta = \dfrac{3}{\sqrt{10}} * \dfrac{\sqrt{10}}{\sqrt{10}} \\[5ex] \cos\beta = \dfrac{3 * \sqrt{10}}{\sqrt{10} * \sqrt{10}} \\[5ex] \cos\beta = \dfrac{3\sqrt{10}}{10} \\[5ex] \sec\beta = \dfrac{1}{\cos\beta} ... Reciprocal\:\: Identity \\[5ex] \sec\beta = 1 \div \cos\beta \\[3ex] \sec\beta = 1 \div \dfrac{3}{\sqrt{10}} \\[5ex] \sec\beta = 1 * \dfrac{\sqrt{10}}{3} \\[5ex] \sec\beta = \dfrac{\sqrt{10}}{3}$
(25.)

(26.) ACT If $90^\circ \lt \theta \lt 180^\circ$ and $\sin \theta = \dfrac{8}{17}$,
then $\cos \theta = ?$

$A.\;\; \dfrac{8}{15} \\[5ex] B.\;\; \dfrac{17}{8} \\[5ex] C.\;\; -\dfrac{17}{8} \\[5ex] D.\;\; -\dfrac{15}{17} \\[5ex] E.\;\; -\dfrac{17}{15} \\[5ex]$

$SOH:CAH:TOA \\[3ex] \sin A = \dfrac{opp}{hyp} = \dfrac{8}{17} \\[5ex] hyp = 17 \\[3ex] opp = leg = 8 \\[3ex] adj = leg \\[3ex] hyp^2 = leg^2 + leg^2...Pythagorean\:\: Theorem \\[3ex] hyp^2 = opp^2 + adj^2 adj^2 = hyp^2 - opp^2 \\[3ex] adj^2 = 17^2 - 8^2 \\[3ex] adj^2 = 289 - 64 = 225 \\[3ex] adj = \sqrt{225} = 15 \\[3ex] \implies \cos \theta = \dfrac{adj}{hyp} = \dfrac{15}{17} \\[5ex] However; \\[3ex] 0 \lt \theta \lt 90 \:\:(\theta \:\:is\:\: acute) \\[3ex] Based\:\: on\:\: the\:\: question \\[3ex] 90 \lt \theta \lt 180 \:\:(\theta \:\:is\:\: obtuse) \rightarrow Second\:\: Quadrant \\[3ex] Second\:\: Quadrant \\[3ex] cosine \:\:is\:\: negative \\[3ex] \therefore \cos \theta = -\dfrac{15}{17}$
(27.)

(28.) ACT If $\sin x = \dfrac{\sqrt{3}}{2}$ and $\cos x = -\dfrac{1}{2}$, then $\sec x =$?

$A.\:\: -2 \\[3ex] B.\:\: -\sqrt{3} \\[3ex] C.\:\: -\dfrac{2}{\sqrt{3}} \\[5ex] D.\:\: \dfrac{2}{\sqrt{3}} \\[5ex] E.\:\: 2 \\[3ex]$

$\sec x = \dfrac{1}{\cos x} ...Reciprocal\:\: Identity \\[5ex] \sec x = 1 \div \cos x \\[3ex] \sec x = 1 \div -\dfrac{1}{2} \\[5ex] \sec x = 1 * -\dfrac{2}{1} \\[5ex] \sec x = -2$
(29.) Determine the other exact trigonometric values of $\gamma$ for
$\cos \gamma = \dfrac{3}{5}$, $\gamma$ is the fourth quadrant

$\cos\gamma = \dfrac{3}{5} = \dfrac{adj}{hyp} ... SOHCAHTOA \\[5ex] adj = 3 \\[3ex] hyp = 5 \\[3ex] hyp^2 = opp^2 + adj^2 ... Pythagorean\:\: Theorem \\[3ex] opp^2 = hyp^2 - adj^2 \\[3ex] opp^2 = 5^2 - 3^2 \\[3ex] opp^2 = 25 - 9 \\[3ex] opp^2 = 16 \\[3ex] opp = \sqrt{16} \\[3ex] opp = 4 \\[3ex] \sec\gamma = \dfrac{1}{\cos\gamma} ... Reciprocal\:\: Identity \\[5ex] \sec\gamma = 1 \div \cos\gamma \\[3ex] \sec\gamma = 1 \div \dfrac{3}{5} \\[5ex] \sec\gamma = 1 * \dfrac{5}{3} \\[5ex] \sec\gamma = \dfrac{5}{3} \\[5ex] \sin\gamma = -\dfrac{opp}{hyp} ... 4th\:\: Quadrant\:\: Identity\:\: and\:\: SOHCAHTOA \\[5ex] \sin \gamma = -\dfrac{4}{5} \\[5ex] \csc\gamma = \dfrac{1}{\sin\gamma} ... Reciprocal\:\: Identity \\[5ex] \csc\gamma = 1 \div \sin\gamma \\[3ex] \csc\gamma = 1 \div -\dfrac{4}{5} \\[5ex] \sec\gamma = 1 * -\dfrac{5}{4} \\[5ex] \sec\gamma = -\dfrac{5}{4} \\[5ex] \tan\gamma = -\dfrac{opp}{adj} ... 4th\:\: Quadrant\:\: Identity\:\: and\:\: SOHCAHTOA \\[5ex] \tan\gamma = -\dfrac{4}{3} \\[5ex] \cot\gamma = -\dfrac{adj}{opp} ... 4th\:\: Quadrant\:\: Identity\:\: and\:\: SOHCAHTOA \\[5ex] \cot\gamma = -\dfrac{3}{4}$
(30.) Determine the reference angle and the exact value of $\cos (-900)^\circ$

$-900^\circ \lt 0^\circ \\[3ex] \beta = coterminal\:\: \angle \:\:of\:\: -900^\circ \\[3ex] \beta = \alpha + 360k \\[3ex] \alpha = -900^\circ \\[3ex] Let\:\: k = 2 \\[3ex] \beta = -900 + 360(2) \\[3ex] \beta = -900 + 720 \\[3ex] \beta = 180^\circ \\[3ex] 180^\circ \:\:is\:\: quadrantal\:\: means\:\: No\:\: reference\:\: \angle \\[3ex] \therefore \cos (-900)^\circ = \cos 180^\circ = -1$
(31.) Determine the exact trigonometric values of the angle $\phi$ if the terminal side of $\phi$ in standard position lies on the line $5x + 3y = 0$ in the fourth quadrant

Let us find a point on that line in that quadrant
Draw the sketch of the quadrant
We want to avoid decimals.

$5x + 3y = 0 \\[3ex] 3y = -5x \\[3ex] y = -\dfrac{5}{3}x \\[5ex] 4th\:\: quadrant\:\: are\:\: positive-x \:\:values \\[3ex] Let\:\: x = 3 \\[3ex] y = -\dfrac{5}{3} * 3 = -5 \\[5ex] (3, -5) \:\:lies \:\:on\:\: 5x + 3y = 0 \\[3ex] (3, -5) \:\:is \:\:in\:\: 4th\:\: quadrant \\[3ex] x = adj = 3 \\[3ex] y = opp = -5 \\[3ex] \tan\phi = \dfrac{opp}{adj} = -\dfrac{5}{3} ... SOHCAHTOA \\[5ex] \cot\phi = \dfrac{1}{\tan\phi} ... Reciprocal\:\: Identity \\[5ex] \cot\phi = 1 \div \tan\phi \\[3ex] \cot\phi = 1 \div -\dfrac{5}{3} \\[5ex] \cot\phi = 1 * -\dfrac{3}{5} \\[5ex] \cot\phi = -\dfrac{3}{5} \\[5ex] hyp^2 = opp^2 + adj^2 ... Pythagorean\:\: Theorem \\[3ex] hyp^2 = (-5)^2 + 3^2 \\[3ex] hyp^2 = 25 + 9 = 34 \\[3ex] hyp = \sqrt{34} \\[3ex] \sin\phi = \dfrac{opp}{hyp} = -\dfrac{5}{\sqrt{34}} ... SOHCAHTOA \\[5ex] \sin\phi = -\dfrac{5}{\sqrt{34}} * \dfrac{\sqrt{34}}{\sqrt{34}} \\[5ex] \sin\phi = -\dfrac{5 * \sqrt{34}}{\sqrt{34} * \sqrt{34}} \\[5ex] \sin\phi = -\dfrac{5\sqrt{34}}{34} \\[5ex] \csc\phi = \dfrac{1}{\sin\phi} ... Reciprocal\:\: Identity \\[5ex] \csc\phi = 1 \div \sin\phi \\[3ex] \csc\phi = 1 \div -\dfrac{5}{\sqrt{34}} \\[5ex] \csc\phi = 1 * -\dfrac{\sqrt{34}}{5} \\[5ex] \csc\phi = -\dfrac{\sqrt{34}}{5} \\[5ex] \cos\phi = \dfrac{adj}{hyp} = \dfrac{3}{\sqrt{34}} ... SOHCAHTOA \\[5ex] \cos\phi = \dfrac{3}{\sqrt{34}} * \dfrac{\sqrt{34}}{\sqrt{34}} \\[5ex] \cos\phi = \dfrac{3 * \sqrt{34}}{\sqrt{34} * \sqrt{34}} \\[5ex] \cos\phi = \dfrac{3\sqrt{34}}{34} \\[5ex] \sec\phi = \dfrac{1}{\cos\phi} ... Reciprocal\:\: Identity \\[5ex] \sec\phi = 1 \div \cos\phi \\[3ex] \sec\phi = 1 \div \dfrac{3}{\sqrt{34}} \\[5ex] \sec\phi = 1 * \dfrac{\sqrt{34}}{3} \\[5ex] \sec\phi = \dfrac{\sqrt{34}}{3}$
(32.) Determine the exact trigonometric values of the angle $\phi$ if the terminal side of $\phi$ in standard position lies on the line $5x + 3y = 0$ in the second quadrant

Let us find a point on that line in that quadrant
Draw the sketch of the quadrant
We want to avoid decimals.

$5x + 3y = 0 \\[3ex] 3y = -5x \\[3ex] y = -\dfrac{5}{3}x \\[5ex] 2nd\:\: quadrant\:\: are\:\: negative-x \:\:values \\[3ex] Let\:\: x = -3 \\[3ex] y = -\dfrac{5}{3} * -3 = 5 \\[5ex] (-3, 5) \:\:lies \:\:on\:\: 5x + 3y = 0 \\[3ex] (-3, 5) \:\:is \:\:in\:\: 2nd\:\: quadrant \\[3ex] x = adj = -3 \\[3ex] y = opp = 5 \\[3ex] \tan\phi = \dfrac{opp}{adj} = \dfrac{5}{-3} = -\dfrac{5}{3} ... SOHCAHTOA \\[5ex] \cot\phi = \dfrac{1}{\tan\phi} ... Reciprocal\:\: Identity \\[5ex] \cot\phi = 1 \div \tan\phi \\[3ex] \cot\phi = 1 \div -\dfrac{5}{3} \\[5ex] \cot\phi = 1 * -\dfrac{3}{5} \\[5ex] \cot\phi = -\dfrac{3}{5} \\[5ex] hyp^2 = opp^2 + adj^2 ... Pythagorean\:\: Theorem \\[3ex] hyp^2 = 5^2 + (-3)^2 \\[3ex] hyp^2 = 25 + 9 = 34 \\[3ex] hyp = \sqrt{34} \\[3ex] \sin\phi = \dfrac{opp}{hyp} = \dfrac{5}{\sqrt{34}} ... SOHCAHTOA \\[5ex] \sin\phi = \dfrac{5}{\sqrt{34}} * \dfrac{\sqrt{34}}{\sqrt{34}} \\[5ex] \sin\phi = \dfrac{5 * \sqrt{34}}{\sqrt{34} * \sqrt{34}} \\[5ex] \sin\phi = \dfrac{5\sqrt{34}}{34} \\[5ex] \csc\phi = \dfrac{1}{\sin\phi} ... Reciprocal\:\: Identity \\[5ex] \csc\phi = 1 \div \sin\phi \\[3ex] \csc\phi = 1 \div \dfrac{5}{\sqrt{34}} \\[5ex] \csc\phi = 1 * \dfrac{\sqrt{34}}{5} \\[5ex] \csc\phi = \dfrac{\sqrt{34}}{5} \\[5ex] \cos\phi = \dfrac{adj}{hyp} = -\dfrac{3}{\sqrt{34}} ... SOHCAHTOA \\[5ex] \cos\phi = -\dfrac{3}{\sqrt{34}} * \dfrac{\sqrt{34}}{\sqrt{34}} \\[5ex] \cos\phi = -\dfrac{3 * \sqrt{34}}{\sqrt{34} * \sqrt{34}} \\[5ex] \cos\phi = -\dfrac{3\sqrt{34}}{34} \\[5ex] \sec\phi = \dfrac{1}{\cos\phi} ... Reciprocal\:\: Identity \\[5ex] \sec\phi = 1 \div \cos\phi \\[3ex] \sec\phi = 1 \div -\dfrac{3}{\sqrt{34}} \\[5ex] \sec\phi = 1 * -\dfrac{\sqrt{34}}{3} \\[5ex] \sec\phi = -\dfrac{\sqrt{34}}{3}$
(33.) Determine the exact trigonometric values of the angle $\beta$ if the terminal side of $\beta$ in standard position lies on the line $2x - 7y = 0$ in the third quadrant

Let us find a point on that line in that quadrant
Draw the sketch of the quadrant
We want to avoid decimals.

$2x - 7y = 0 \\[3ex] 2x = 7y \\[3ex] 7y = 2x \\[3ex] y = \dfrac{2}{7}x \\[5ex] 3rd\:\: quadrant\:\: are\:\: negative-x \:\:values \\[3ex] Let\:\: x = -7 \\[3ex] y = \dfrac{2}{7} * -7 = -2 \\[5ex] (-7, -2) \:\:lies \:\:on\:\: 2x - 7y = 0 \\[3ex] (-7, -2) \:\:is \:\:in\:\: 3rd\:\: quadrant \\[3ex] x = adj = -7 \\[3ex] y = opp = -2 \\[3ex] \tan\beta = \dfrac{opp}{adj} = \dfrac{-2}{-7} = \dfrac{2}{7} ... SOHCAHTOA \\[5ex] \cot\beta = \dfrac{1}{\tan\beta} ... Reciprocal\:\: Identity \\[5ex] \cot\beta = 1 \div \tan\beta \\[3ex] \cot\beta = 1 \div \dfrac{2}{7} \\[5ex] \cot\beta = 1 * \dfrac{7}{2} \\[5ex] \cot\beta = \dfrac{7}{2} \\[5ex] hyp^2 = opp^2 + adj^2 ... Pythagorean\:\: Theorem \\[3ex] hyp^2 = (-2)^2 + (-7)^2 \\[3ex] hyp^2 = 4 + 49 = 53 \\[3ex] hyp = \sqrt{53} \\[3ex] \sin\beta = \dfrac{opp}{hyp} = -\dfrac{2}{\sqrt{53}} ... SOHCAHTOA \\[5ex] \sin\beta = -\dfrac{2}{\sqrt{53}} * \dfrac{\sqrt{53}}{\sqrt{53}} \\[5ex] \sin\beta = -\dfrac{2 * \sqrt{53}}{\sqrt{53} * \sqrt{53}} \\[5ex] \sin\beta = -\dfrac{2\sqrt{53}}{53} \\[5ex] \csc\beta = \dfrac{1}{\sin\beta} ... Reciprocal\:\: Identity \\[5ex] \csc\beta = 1 \div \sin\beta \\[3ex] \csc\beta = 1 \div -\dfrac{2}{\sqrt{53}} \\[5ex] \csc\beta = 1 * -\dfrac{\sqrt{53}}{2} \\[5ex] \csc\beta = -\dfrac{\sqrt{53}}{2} \\[5ex] \cos\beta = \dfrac{adj}{hyp} = -\dfrac{7}{\sqrt{53}} ... SOHCAHTOA \\[5ex] \cos\beta = -\dfrac{7}{\sqrt{53}} * \dfrac{\sqrt{53}}{\sqrt{53}} \\[5ex] \cos\beta = \dfrac{-7 * \sqrt{53}}{\sqrt{53} * \sqrt{53}} \\[5ex] \cos\beta = -\dfrac{7\sqrt{53}}{53} \\[5ex] \sec\beta = \dfrac{1}{\cos\beta} ... Reciprocal\:\: Identity \\[5ex] \sec\beta = 1 \div \cos\beta \\[3ex] \sec\beta = 1 \div -\dfrac{7}{\sqrt{53}} \\[5ex] \sec\beta = 1 * -\dfrac{\sqrt{53}}{7} \\[5ex] \sec\beta = -\dfrac{\sqrt{53}}{7}$
(34.) Determine the exact trigonometric values of the angle $\beta$ if the terminal side of $\beta$ in standard position lies on the line $2x - 7y = 0$ in the first quadrant

Let us find a point on that line in that quadrant
Draw the sketch of the quadrant
We want to avoid decimals.

$2x - 7y = 0 \\[3ex] 2x = 7y \\[3ex] 7y = 2x \\[3ex] y = \dfrac{2}{7}x \\[5ex] 1st\:\: quadrant\:\: are\:\: positive-x \:\:values \\[3ex] Let\:\: x = 7 \\[3ex] y = \dfrac{2}{7} * 7 = 2 \\[5ex] (7, 2) \:\:lies \:\:on\:\: 2x - 7y = 0 \\[3ex] (7, 2) \:\:is \:\:in\:\: 3rd\:\: quadrant \\[3ex] x = adj = 7 \\[3ex] y = opp = 2 \\[3ex] \tan\beta = \dfrac{opp}{adj} = \dfrac{2}{7} ... SOHCAHTOA \\[5ex] \cot\beta = \dfrac{1}{\tan\beta} ... Reciprocal\:\: Identity \\[5ex] \cot\beta = 1 \div \tan\beta \\[3ex] \cot\beta = 1 \div \dfrac{2}{7} \\[5ex] \cot\beta = 1 * \dfrac{7}{2} \\[5ex] \cot\beta = \dfrac{7}{2} \\[5ex] hyp^2 = opp^2 + adj^2 ... Pythagorean\:\: Theorem \\[3ex] hyp^2 = (-2)^2 + (-7)^2 \\[3ex] hyp^2 = 4 + 49 = 53 \\[3ex] hyp = \sqrt{53} \\[3ex] \sin\beta = \dfrac{opp}{hyp} = \dfrac{2}{\sqrt{53}} ... SOHCAHTOA \\[5ex] \sin\beta = \dfrac{2}{\sqrt{53}} * \dfrac{\sqrt{53}}{\sqrt{53}} \\[5ex] \sin\beta = \dfrac{2 * \sqrt{53}}{\sqrt{53} * \sqrt{53}} \\[5ex] \sin\beta = \dfrac{2\sqrt{53}}{53} \\[5ex] \csc\beta = \dfrac{1}{\sin\beta} ... Reciprocal\:\: Identity \\[5ex] \csc\beta = 1 \div \sin\beta \\[3ex] \csc\beta = 1 \div \dfrac{2}{\sqrt{53}} \\[5ex] \csc\beta = 1 * \dfrac{\sqrt{53}}{2} \\[5ex] \csc\beta = \dfrac{\sqrt{53}}{2} \\[5ex] \cos\beta = \dfrac{adj}{hyp} = \dfrac{7}{\sqrt{53}} ... SOHCAHTOA \\[5ex] \cos\beta = \dfrac{7}{\sqrt{53}} * \dfrac{\sqrt{53}}{\sqrt{53}} \\[5ex] \cos\beta = \dfrac{7 * \sqrt{53}}{\sqrt{53} * \sqrt{53}} \\[5ex] \cos\beta = \dfrac{7\sqrt{53}}{53} \\[5ex] \sec\beta = \dfrac{1}{\cos\beta} ... Reciprocal\:\: Identity \\[5ex] \sec\beta = 1 \div \cos\beta \\[3ex] \sec\beta = 1 \div \dfrac{7}{\sqrt{53}} \\[5ex] \sec\beta = 1 * \dfrac{\sqrt{53}}{7} \\[5ex] \sec\beta = \dfrac{\sqrt{53}}{7}$
(35.) Determine the exact value of $\cos 75^\circ$

Find two special angles whose sum or difference is $75^\circ$

$\cos 75^\circ \\[3ex] = \cos(45 + 30) \\[3ex] = \cos 45 \cos 30 - \sin 45 \sin 30 ... Sum\:\: Formula \\[3ex] = \dfrac{\sqrt{2}}{2} * \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{2}}{2} * \dfrac{1}{2} \\[5ex] = \dfrac{\sqrt{6}}{4} - \dfrac{\sqrt{2}}{4} \\[5ex] = \dfrac{\sqrt{6} - \sqrt{2}}{4}$
(36.) Determine the exact value of $\sin \dfrac{7\pi}{12}$

Find two special angles whose sum or difference is $\dfrac{7\pi}{12}$

$\dfrac{7\pi}{12} = \dfrac{7 * 180}{12} = 105 \\[5ex] 105 = 45 + 60 \\[3ex] \dfrac{7\pi}{12} = \dfrac{3\pi}{12} + \dfrac{4\pi}{12} \\[5ex] \dfrac{3\pi}{12} = \dfrac{\pi}{4} = \dfrac{180}{4} = 45 \\[5ex] \dfrac{4\pi}{12} = \dfrac{\pi}{3} = \dfrac{180}{3} = 60 \\[5ex] \sin \dfrac{7\pi}{12} \\[3ex] = \sin\left(\dfrac{3\pi}{12} + \dfrac{4\pi}{12}\right) \\[5ex] = \sin\left(\dfrac{\pi}{4} + \dfrac{\pi}{3}\right) \\[5ex] = \sin\dfrac{\pi}{4} \cos\dfrac{\pi}{3} + \cos\dfrac{\pi}{4} \sin\dfrac{\pi}{3} ... Sum\:\: Formula \\[5ex] = \dfrac{\sqrt{2}}{2} * \dfrac{1}{2} + \dfrac{\sqrt{2}}{2} * \dfrac{\sqrt{3}}{2} \\[5ex] = \dfrac{\sqrt{2}}{4} + \dfrac{\sqrt{6}}{4} \\[5ex] = \dfrac{\sqrt{2} + \sqrt{6}}{4}$
(37.) Determine the exact value of $\tan 15^\circ$

Find two special angles whose sum or difference is $15^\circ$

$\tan 15^\circ \\[3ex] = \tan(45 - 30) \\[3ex] = \dfrac{\tan45 - \tan30}{1 + \tan45\tan30} ... Difference\:\: Formula \\[5ex] \tan45 - \tan30 = 1 - \dfrac{\sqrt{3}}{3} = \dfrac{3}{3} - \dfrac{\sqrt{3}}{3} = \dfrac{3 - \sqrt{3}}{3} \\[5ex] \tan45\tan30 = 1 * \dfrac{\sqrt{3}}{3} = \dfrac{\sqrt{3}}{3} \\[5ex] 1 + \tan45\tan30 = 1 + \dfrac{\sqrt{3}}{3} = \dfrac{3}{3} + \dfrac{\sqrt{3}}{3} = \dfrac{3 + \sqrt{3}}{3} \\[5ex] = \dfrac{3 - \sqrt{3}}{3} \div \dfrac{3 + \sqrt{3}}{3} \\[5ex] = \dfrac{3 - \sqrt{3}}{3} * \dfrac{3}{3 + \sqrt{3}} \\[5ex] = \dfrac{3 - \sqrt{3}}{3 + \sqrt{3}} \\[5ex] = \dfrac{3 - \sqrt{3}}{3 + \sqrt{3}} * \dfrac{3 + \sqrt{3}}{3 + \sqrt{3}} \\[5ex] = \dfrac{9 - 3\sqrt{3} - 3\sqrt{3} + (\sqrt{3})^2}{3^2 - (\sqrt{3})^2} \\[5ex] = \dfrac{12 - 6\sqrt{3}}{9 - 3} \\[5ex] = \dfrac{12 - 6\sqrt{3}}{6} \\[5ex] = \dfrac{12}{6} - \dfrac{6\sqrt{3}}{6} \\[5ex] = 2 - \sqrt{3}$
(38.) Determine the exact value of $\cos \dfrac{\pi}{12}$

Find two special angles whose sum or difference is $\dfrac{\pi}{12}$

$\dfrac{\pi}{12} = \dfrac{180}{12} = 15 \\[5ex] 15 = 45 - 30 \\[3ex] \dfrac{\pi}{12} = \dfrac{3\pi}{12} - \dfrac{2\pi}{12} \\[5ex] \dfrac{3\pi}{12} = \dfrac{\pi}{4} = \dfrac{180}{4} = 45 \\[5ex] \dfrac{2\pi}{12} = \dfrac{\pi}{6} = \dfrac{180}{6} = 30 \\[5ex] \cos\dfrac{\pi}{12} \\[3ex] = \cos\left(\dfrac{3\pi}{12} - \dfrac{2\pi}{12}\right) \\[5ex] = \cos\left(\dfrac{\pi}{4} - \dfrac{\pi}{6}\right) \\[5ex] = \cos\dfrac{\pi}{4} \cos\dfrac{\pi}{6} + \sin\dfrac{\pi}{4} \sin\dfrac{\pi}{6} ... Difference\:\: Formula \\[5ex] = \dfrac{\sqrt{2}}{2} * \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} * \dfrac{1}{2} \\[5ex] = \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4} \\[5ex] = \dfrac{\sqrt{6} + \sqrt{2}}{4}$
(39.)

(40.) ACT If $\sin \alpha = \dfrac{12}{13}$, and $\cos \alpha = \dfrac{5}{13}$, then $\tan \alpha = ?$

$F.\;\; \dfrac{5}{12} \\[5ex] G.\;\; \dfrac{7}{13} \\[5ex] H.\;\; \dfrac{12}{5} \\[5ex] J.\;\; \dfrac{17}{13} \\[5ex] K.\;\; \dfrac{60}{13} \\[5ex]$

$\sin \alpha = \dfrac{12}{13} \\[5ex] \cos \alpha = \dfrac{5}{13} \\[5ex] \tan \alpha = \dfrac{\sin \alpha}{\cos \alpha}...Quotient\;\;Identities \\[5ex] = \sin \alpha \div \cos \alpha \\[3ex] = \dfrac{12}{13} \div \dfrac{5}{13} \\[5ex] = \dfrac{12}{13} * \dfrac{13}{5} \\[5ex] = \dfrac{12}{5}$

(41.) Determine the exact value of $\cos(\theta + \gamma)$ if $\sin\theta = -\dfrac{3}{5}$ and $\sin\gamma = -\dfrac{5}{13}$, where $\theta$ and $\gamma$ are in the third quadrant.

$\cos(\theta + \gamma) = \cos\theta\cos\gamma - \sin\theta\sin\gamma ... Sum\:\: Formula \\[3ex] \underline{For\:\: \theta} \\[3ex] \sin\theta = -\dfrac{3}{5} = \dfrac{opp}{hyp} ... SOHCAHTOA \\[5ex] opp = -3 \\[3ex] hyp = 5 \\[3ex] hyp^2 = opp^2 + adj^2 ... Pythagorean\:\: Theorem \\[3ex] adj^2 = hyp^2 - opp^2 \\[3ex] adj^2 = 5^2 - (-3)^2 \\[3ex] adj^2 = 25 - 9 = 16 \\[3ex] adj = \sqrt{16} \\[3ex] adj = 4 \\[3ex] \cos\theta = \dfrac{adj}{hyp} = \dfrac{4}{5} ... SOHCAHTOA \\[5ex] \underline{For\:\: \gamma} \\[3ex] \sin\gamma = -\dfrac{5}{13} = \dfrac{opp}{hyp} ... SOHCAHTOA \\[5ex] opp = -5 \\[3ex] hyp = 13 \\[3ex] hyp^2 = opp^2 + adj^2 ... Pythagorean\:\: Theorem \\[3ex] adj^2 = hyp^2 - opp^2 \\[3ex] adj^2 = 13^2 - (-5)^2 \\[3ex] adj^2 = 169 - 25 = 144 \\[3ex] adj = \sqrt{144} \\[3ex] adj = 12 \\[3ex] \cos\gamma = \dfrac{adj}{hyp} = \dfrac{12}{13} ... SOHCAHTOA \\[5ex] = \dfrac{4}{5} * \dfrac{12}{13} - \left(-\dfrac{3}{5} * -\dfrac{5}{13}\right) \\[5ex] = \dfrac{4 * 12}{5 * 13} - \dfrac{-3 * -5}{5 * 13} \\[5ex] = \dfrac{48}{65} - \dfrac{15}{65} \\[5ex] = \dfrac{48 - 15}{65} \\[5ex] = \dfrac{33}{65}$
(42.) Determine the exact value of $\sin \dfrac{13\pi}{12}$

Find two special angles whose sum or difference is $\dfrac{13\pi}{12}$

$\dfrac{13\pi}{12} = \dfrac{13 * 180}{12} = 195 \\[5ex] 195 = 60 + 135 \\[3ex] \dfrac{13\pi}{12} = \dfrac{4\pi}{12} + \dfrac{9\pi}{12} \\[5ex] \dfrac{4\pi}{12} = \dfrac{\pi}{3} = \dfrac{180}{3} = 60 \\[5ex] \dfrac{9\pi}{12} = \dfrac{3\pi}{4} = \dfrac{3 * 180}{4} = 135 \\[5ex] \sin \dfrac{13\pi}{12} \\[3ex] = \sin\left(\dfrac{4\pi}{12} + \dfrac{9\pi}{12}\right) \\[5ex] = \sin\left(\dfrac{\pi}{3} + \dfrac{3\pi}{4}\right) \\[5ex] = \sin\dfrac{\pi}{3} \cos\dfrac{3\pi}{4} + \cos\dfrac{\pi}{3} \sin\dfrac{3\pi}{4} ... Sum\:\: Formula \\[5ex] = \dfrac{\sqrt{3}}{2} * -\dfrac{\sqrt{2}}{2} + \dfrac{1}{2} * \dfrac{\sqrt{2}}{2} \\[5ex] = \dfrac{\sqrt{3} * -\sqrt{2}}{2 * 2} + \dfrac{1 * \sqrt{2}}{2 * 2} \\[5ex] = -\dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4} \\[5ex] = \dfrac{\sqrt{2}}{4} - \dfrac{\sqrt{6}}{4} \\[5ex] = \dfrac{\sqrt{2} - \sqrt{6}}{4}$
(43.) Determine the exact value of $\sin(\theta - \gamma)$ if $\sin\theta = \dfrac{5}{13}$ and $\sin\gamma = \dfrac{15}{17}$, where $\theta$ and $\gamma$ are between $0$ and $\dfrac{\pi}{2}$

$\theta$ and $\gamma$ are between $0$ and $\dfrac{\pi}{2}$

This means that $\theta$ and $\gamma$ are in the first quadrant

$\sin(\theta - \gamma) = \sin\theta\cos\gamma - \cos\theta\sin\gamma ... Difference\:\: Formula \\[3ex] \underline{For\:\: \theta} \\[3ex] \sin\theta = \dfrac{5}{13} = \dfrac{opp}{hyp} ... SOHCAHTOA \\[5ex] opp = 5 \\[3ex] hyp = 13 \\[3ex] hyp^2 = opp^2 + adj^2 ... Pythagorean\:\: Theorem \\[3ex] adj^2 = hyp^2 - opp^2 \\[3ex] adj^2 = 13^2 - 5^2 \\[3ex] adj^2 = 169 - 25 = 144 \\[3ex] adj = \sqrt{144} \\[3ex] adj = 12 \\[3ex] \cos\theta = \dfrac{adj}{hyp} = \dfrac{12}{13} ... SOHCAHTOA \\[5ex] \underline{For\:\: \gamma} \\[3ex] \sin\gamma = \dfrac{15}{17} = \dfrac{opp}{hyp} ... SOHCAHTOA \\[5ex] opp = 15 \\[3ex] hyp = 17 \\[3ex] hyp^2 = opp^2 + adj^2 ... Pythagorean\:\: Theorem \\[3ex] adj^2 = hyp^2 - opp^2 \\[3ex] adj^2 = 17^2 - 15^2 \\[3ex] adj^2 = 289 - 225 = 64 \\[3ex] adj = \sqrt{64} \\[3ex] adj = 8 \\[3ex] \cos\gamma = \dfrac{adj}{hyp} = \dfrac{8}{17} ... SOHCAHTOA \\[5ex] = \dfrac{5}{13} * \dfrac{8}{17} - \left(\dfrac{12}{13} * \dfrac{15}{17}\right) \\[5ex] = \dfrac{5 * 8}{13 * 17} - \dfrac{12 * 15}{13 * 17} \\[5ex] = \dfrac{40}{221} - \dfrac{180}{221} \\[5ex] = \dfrac{40 - 180}{221} \\[5ex] = \dfrac{-140}{221} \\[5ex] = -\dfrac{140}{221}$
(44.) WASSCE If $\cos \theta = \dfrac{15}{17}$, find the value of $\dfrac{\tan \theta}{1 + 2\tan \theta}$

$17^2 = opp^2 + 15^2 ... Pythagorean\:\: Theorem \\[3ex] 289 = opp^2 + 225 \\[3ex] 289 - 225 = opp^2 \\[3ex] 64 = opp^2 \\[3ex] opp^2 = 64 \\[3ex] opp = \sqrt{64} \\[3ex] opp = 8 \\[3ex] SOH:CAH:TOA \\[3ex] \tan \theta = \dfrac{opp}{adj} = \dfrac{8}{15} \\[5ex] 2\tan \theta = 2\left(\dfrac{8}{15}\right) = \dfrac{16}{15} \\[5ex] 1 + 2\tan \theta = 1 + \dfrac{16}{15} = \dfrac{15}{15} + \dfrac{16}{15} = \dfrac{31}{15} \\[5ex] \dfrac{\tan \theta}{1 + 2\tan \theta} \\[5ex] = \dfrac{8}{15} \div \dfrac{31}{15} \\[5ex] = \dfrac{8}{15} * \dfrac{15}{31} \\[5ex] = \dfrac{8}{31}$
(45.) Determine the exact value of $\cos(\alpha + \omega)$ if $\cos\alpha = -\dfrac{3}{7}$ and $\cos\omega = \dfrac{5}{11}$, where $\dfrac{\pi}{2} \le \alpha \lt \pi$ and $\dfrac{3\pi}{2} \le \omega \lt 2\pi$

$\cos(\alpha + \omega) = \cos\alpha\cos\omega - \sin\alpha\sin\omega ... Sum\:\: Formula \\[3ex] \underline{For\:\: \alpha} \\[3ex] \dfrac{\pi}{2} \le \alpha \lt \pi \rightarrow \alpha \:\:is\:\:in\:\:second\:\:quadrant \\[5ex] \cos\alpha = -\dfrac{3}{7} = \dfrac{adj}{hyp} ... SOHCAHTOA \\[5ex] adj = -3 \\[3ex] hyp = 7 \\[3ex] hyp^2 = opp^2 + adj^2 ... Pythagorean\:\: Theorem \\[3ex] opp^2 = hyp^2 - adj^2 \\[3ex] opp^2 = 7^2 - (-3)^2 \\[3ex] opp^2 = 49 - 9 = 40 \\[3ex] opp = \sqrt{40} = \sqrt{4 * 10} = \sqrt{4} * \sqrt{10} = 2 * \sqrt{10} \\[3ex] opp = 2\sqrt{10} \\[3ex] \sin\alpha = \dfrac{opp}{hyp} = \dfrac{2\sqrt{10}}{7} ... SOHCAHTOA \\[5ex] \underline{For\:\: \omega} \\[3ex] \dfrac{3\pi}{2} \le \alpha \lt 2\pi \rightarrow \omega \:\:is\:\:in\:\:fourth\:\:quadrant \\[5ex] \cos\omega = \dfrac{5}{11} = \dfrac{adj}{hyp} ... SOHCAHTOA \\[5ex] adj = 5 \\[3ex] hyp = 11 \\[3ex] hyp^2 = opp^2 + adj^2 ... Pythagorean\:\: Theorem \\[3ex] opp^2 = hyp^2 - adj^2 \\[3ex] opp^2 = 11^2 - 5^2 \\[3ex] opp^2 = 121 - 25 = 96 \\[3ex] opp = \sqrt{96} = \sqrt{16 * 6} = \sqrt{16} * \sqrt{6} = 4 * \sqrt{6} \\[3ex] opp = 4\sqrt{6} \\[3ex] \sin\omega = \dfrac{opp}{hyp} = \dfrac{4\sqrt{6}}{11} ... SOHCAHTOA \\[5ex] \omega \:\:is\:\:in\:\:fourth\:\:quadrant \\[3ex] \sin \:\:is\:\:negative\:\:in\:\:fourth\:\:quadrant \\[3ex] \therefore \sin\omega = -\dfrac{4\sqrt{6}}{11} \\[5ex] = -\dfrac{3}{7} * \dfrac{5}{11} - \left(\dfrac{2\sqrt{10}}{7} * -\dfrac{4\sqrt{6}}{11}\right) \\[5ex] = \dfrac{-3 * 5}{7 * 11} - \dfrac{2\sqrt{10} * -4\sqrt{6}}{7 * 11} \\[5ex] = -\dfrac{15}{77} - -\dfrac{8\sqrt{10 * 6}}{77} \\[5ex] = -\dfrac{15}{77} + \dfrac{8\sqrt{10 * 6}}{77} \\[5ex] = -\dfrac{15}{77} + \dfrac{8\sqrt{60}}{77} \\[5ex] \sqrt{60} = \sqrt{4 * 15} = \sqrt{4} * \sqrt{15} = 2 * \sqrt{15} = 2\sqrt{15} \\[3ex] = -\dfrac{15}{77} + \dfrac{8 * 2\sqrt{15}}{77} \\[5ex] = -\dfrac{15}{77} + \dfrac{16\sqrt{15}}{77} \\[5ex] = \dfrac{-15 + 16\sqrt{15}}{77}$
(46.) Determine $2$ positive and $2$ negative coterminal angles of $73^\circ$ that are greater than $-720^\circ$ but less than $1080^\circ$

$\gt -720^\circ \:\:but\:\: \lt 1080^\circ \\[3ex] \alpha = 73^\circ \\[3ex] Let\:\: \beta = coterminal\:\: angle \\[3ex] \beta = \alpha + 360k \\[3ex] \underline{Positive\:\: coterminal\:\: angles} \\[3ex] k = 1 \\[3ex] \beta = 73 + 360(1) \\[3ex] \beta = 73 + 360 \\[3ex] \beta = 433^\circ \\[3ex] k = 2 \\[3ex] \beta = 73 + 360(2) \\[3ex] \beta = 73 + 720 \\[3ex] \beta = 793^\circ \\[3ex] \underline{Negative\:\: coterminal\:\: angles} \\[3ex] k = -1 \\[3ex] \beta = 73 + 360(-1) \\[3ex] \beta = 73 - 360 \\[3ex] \beta = -287^\circ \\[3ex] k = -2 \\[3ex] \beta = 73 + 360(-2) \\[3ex] \beta = 73 - 720 \\[3ex] \beta = -647^\circ$
(47.) Determine the slope of a line, Line $1$ given that the slope of another line, Line $2$ = $\dfrac{5}{8}$ and the smallest positive angle from Line $1$ to Line $2$ is $30^\circ$

Let the:
slope of Line $1$ = $m_1$
slope of Line $2$ = $m_2$
smallest positive angle from Line $1$ to Line $2$ = $\alpha$

$m_1 = ? \\[3ex] m_2 = \dfrac{5}{8} \\[5ex] \alpha = 30^\circ \\[3ex] \tan\alpha = \dfrac{m_2 - m_1}{1 + m_1m_2} ... Difference\:\: Formula\:\: For\:\: Tangent \\[5ex] \tan\alpha = \tan30 = \dfrac{\sqrt{3}}{3} ... Unit\:\: Circle\:\: Trigonometry \\[5ex] \rightarrow \dfrac{\sqrt{3}}{3} = \dfrac{\dfrac{5}{8} - m_1}{1 + m_1 * \dfrac{5}{8}} \\[7ex] Cross\:\: Multiply \\[3ex] \dfrac{\sqrt{3}}{3}\left(1 + m_1 * \dfrac{5}{8}\right) = \dfrac{5}{8} - m_1 \\[5ex] \dfrac{\sqrt{3}}{3}\left(1 + \dfrac{5}{8}m_1\right) = \dfrac{5}{8} - m_1 \\[5ex] \dfrac{\sqrt{3}}{3} + \dfrac{5\sqrt{3}}{24}m_1 = \dfrac{5}{8} - m_1 \\[5ex] \dfrac{5\sqrt{3}}{24}m_1 + m_1 = \dfrac{5}{8} - \dfrac{\sqrt{3}}{3} \\[5ex] m_1\left(\dfrac{5\sqrt{3}}{24} + 1\right) = \dfrac{5(3) - \sqrt{3}(8)}{24} \\[5ex] m_1\left(\dfrac{5\sqrt{3}}{24} + \dfrac{24}{24}\right) = \dfrac{15 - 8\sqrt{3}}{24} \\[5ex] m_1\left(\dfrac{5\sqrt{3} + 24}{24}\right) = \dfrac{15 - 8\sqrt{3}}{24} \\[5ex] m_1 = \dfrac{15 - 8\sqrt{3}}{24} \div \dfrac{5\sqrt{3} + 24}{24} \\[5ex] = \dfrac{15 - 8\sqrt{3}}{24} * \dfrac{24}{5\sqrt{3} + 24} \\[5ex] = \dfrac{15 - 8\sqrt{3}}{5\sqrt{3} + 24} \\[5ex] = \dfrac{15 - 8\sqrt{3}}{5\sqrt{3} + 24} * \dfrac{5\sqrt{3} - 24}{5\sqrt{3} - 24} \\[5ex] = \dfrac{(15 - 8\sqrt{3})(5\sqrt{3} - 24)}{(5\sqrt{3})^2 - 24^2} \\[5ex] = \dfrac{75\sqrt{3} - 360 - 40(3) + 192\sqrt{3}}{25(3) - 576} \\[5ex] = \dfrac{267\sqrt{3} - 360 - 120}{75 - 576} \\[5ex] = \dfrac{267\sqrt{3} - 480}{-501} \\[5ex] = \dfrac{-(267\sqrt{3} - 480)}{501} \\[5ex] = \dfrac{-267\sqrt{3} + 480}{501} \\[5ex] \therefore m_1 = \dfrac{480 - 267\sqrt{3}}{501} \\[5ex]$
(48.) Determine the complement and supplement of $57.21^\circ$

$Complement\:\:of\:\: 57.21 = 90 - 57.21 = 32.79^\circ \\[3ex] Supplement\:\:of\:\: 57.11 = 180 = 57.21 = 122.79^\circ$
(49.) Determine the reference angle and the exact value of $\csc (-240)^\circ$

$-240^\circ \lt 0^\circ \\[3ex] \beta = coterminal\:\: \angle \:\:of\:\: -240^\circ \\[3ex] \beta = \alpha + 360k \\[3ex] \alpha = -240^\circ \\[3ex] Let\:\: k = 1 \\[3ex] \beta = -240 + 360(1) \\[3ex] \beta = -240 + 360 \\[3ex] \beta = 120^\circ \\[3ex] 120 \:\:is\:\: in\:\: the\:\: 2nd\:\: quadrant \\[3ex] reference\:\: \angle = 180 - 120 = 60 \\[3ex] \csc(-240) = \csc 60 = \dfrac{2\sqrt{3}}{3} ...2nd\:\: Quadrant\:\: Identity \\[5ex] \therefore \csc(-240)^\circ = \csc 60^\circ = \dfrac{2\sqrt{3}}{3}$
(50.) Determine the complement and supplement of $45^\circ 41'50''$

$Complement\:\:of\:\: 45^\circ41'50'' = 90 - 45^\circ 41'50'' \\[3ex] 90^\circ 0' 0'' - 45^\circ 41'50'' \\[3ex] Take\:\: 1^\circ \:\:from\:\: 90^\circ \\[3ex] Remaining\:\: 89^\circ \\[3ex] 1^\circ = 60' \\[3ex] Take\:\: 1' \:\:from\:\: 60' \\[3ex] Remaining\:\: 59' \\[3ex] 1' = 60'' \\[3ex] \therefore 60'' - 50'' = 10'' \\[3ex] \therefore 59' - 41' = 18' \\[3ex] \therefore 89^\circ - 45^\circ = 44^\circ \\[3ex] \therefore \:\:the\:\:complement\:\:of\:\: 45^\circ41'50'' = 44^\circ 18'10'' \\[5ex] Supplement\:\:of\:\: 45^\circ41'50'' = 180 - 45^\circ 41'50'' \\[3ex] 180^\circ 0' 0'' - 45^\circ 41'50'' \\[3ex] Take\:\: 1^\circ \:\:from\:\: 180^\circ \\[3ex] Remaining\:\: 179^\circ \\[3ex] 1^\circ = 60' \\[3ex] Take\:\: 1' \:\:from\:\: 60' \\[3ex] Remaining\:\: 59' \\[3ex] 1' = 60'' \\[3ex] \therefore 60'' - 50'' = 10'' \\[3ex] \therefore 59' - 41' = 18' \\[3ex] \therefore 179^\circ - 45^\circ = 134^\circ \\[3ex] \therefore \:\:the\:\:supplement\:\:of\:\: 45^\circ41'50'' = 134^\circ 18'10''$
(51.) Determine the six trigonometric function values of the angle between the point $(3, 4)$ and the horizontal axis.

Sketch the graph and locate the point.
Point $(3, 4)$ is in the $1st$ quadrant.
The horizontal axis is the $x-axis$
Draw a right triangle joining the origin, the point, and the horizonal axis.
Let the angle between the point and the horizontal axis = $\theta$

$adj = 3 \\[3ex] opp = 4 \\[3ex] hyp = ? \\[3ex] hyp^2 = opp^2 + adj^2 \\[3ex] hyp^2 = 4^2 + 3^2 \\[3ex] hyp^2 = 16 + 9 \\[3ex] hyp^2 = 25 \\[3ex] hyp = \sqrt{25} = 5 \\[3ex] \sin\theta = \dfrac{opp}{hyp} ...1st\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \sin\theta = \dfrac{4}{5} \\[5ex] \csc\theta = \dfrac{1}{\sin\theta} ...Reciprocal\:\:Identity \\[5ex] \csc\theta = \dfrac{5}{4} \\[5ex] \cos\theta = \dfrac{adj}{hyp} ...1st\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \cos\theta = \dfrac{3}{5} \\[5ex] \sec\theta = \dfrac{1}{\cos\theta} ...Reciprocal\:\:Identity \\[5ex] \sec\theta = \dfrac{5}{3} \\[5ex] \tan\theta = \dfrac{opp}{adj} ...1st\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \tan\theta = \dfrac{4}{3} \\[5ex] \cot\theta = \dfrac{1}{\tan\theta} ...Reciprocal\:\:Identity \\[5ex] \cot\theta = \dfrac{3}{4}$
(52.) Determine the six trigonometric function values of the angle between the point $(3, 4)$ and the vertical axis.

Sketch the graph and locate the point.
Point $(3, 4)$ is in the $1st$ quadrant.
The vertical axis is the $y-axis$
Draw a right triangle joining the origin, the point, and the vertical axis.
Let the angle between the point and the vertical axis = $\theta$

$adj = 4 \\[3ex] opp = 3 \\[3ex] hyp = ? \\[3ex] hyp^2 = opp^2 + adj^2 \\[3ex] hyp^2 = 3^2 + 4^2 \\[3ex] hyp^2 = 9 + 16 \\[3ex] hyp^2 = 25 \\[3ex] hyp = \sqrt{25} = 5 \\[3ex] \sin\theta = \dfrac{opp}{hyp} ...1st\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \sin\theta = \dfrac{3}{5} \\[5ex] \csc\theta = \dfrac{1}{\sin\theta} ...Reciprocal\:\:Identity \\[5ex] \csc\theta = \dfrac{5}{3} \\[5ex] \cos\theta = \dfrac{adj}{hyp} ...1st\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \cos\theta = \dfrac{4}{5} \\[5ex] \sec\theta = \dfrac{1}{\cos\theta} ...Reciprocal\:\:Identity \\[5ex] \sec\theta = \dfrac{5}{4} \\[5ex] \tan\theta = \dfrac{opp}{adj} ...1st\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \tan\theta = \dfrac{3}{4} \\[5ex] \cot\theta = \dfrac{1}{\tan\theta} ...Reciprocal\:\:Identity \\[5ex] \cot\theta = \dfrac{4}{3}$
(53.) Determine the six trigonometric function values of the angle between the point $(7, -2)$ and the horizontal axis.

Sketch the graph and locate the point.
Point $(7, -2)$ is in the $4th$ quadrant.
The horizontal axis is the $x-axis$
Draw a right triangle joining the origin, the point, and the horizonal axis.
Let the angle between the point and the horizontal axis = $\theta$

$adj = 7 \\[3ex] opp = 2 \\[3ex] hyp = ? \\[3ex] hyp^2 = opp^2 + adj^2 \\[3ex] hyp^2 = 2^2 + 7^2 \\[3ex] hyp^2 = 4 + 49 \\[3ex] hyp^2 = 53 \\[3ex] hyp = \sqrt{53} \\[3ex] \sin\theta = -\dfrac{opp}{hyp} ...4th\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \sin\theta = -\dfrac{2}{\sqrt{53}} \\[5ex] \sin\theta = -\dfrac{2}{\sqrt{53}} * \dfrac{\sqrt{53}}{\sqrt{53}} \\[5ex] \sin\theta = \dfrac{-2 * \sqrt{53}}{\sqrt{53} * \sqrt{53}} \\[5ex] \sin\theta = -\dfrac{2\sqrt{53}}{53} \\[5ex] \csc\theta = \dfrac{1}{\sin\theta} ...Reciprocal\:\:Identity \\[5ex] \csc\theta = -\dfrac{\sqrt{53}}{2} \\[5ex] \cos\theta = \dfrac{adj}{hyp} ...4th\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \cos\theta = \dfrac{7}{\sqrt{53}} \\[5ex] \cos\theta = \dfrac{7}{\sqrt{53}} * \dfrac{\sqrt{53}}{\sqrt{53}} \\[5ex] \cos\theta = \dfrac{7 * \sqrt{53}}{\sqrt{53} * \sqrt{53}} \\[5ex] \cos\theta = \dfrac{7\sqrt{53}}{53} \\[5ex] \sec\theta = \dfrac{1}{\cos\theta} ...Reciprocal\:\:Identity \\[5ex] \sec\theta = \dfrac{\sqrt{53}}{7} \\[5ex] \tan\theta = -\dfrac{opp}{adj} ...4th\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \tan\theta = -\dfrac{2}{7} \\[5ex] \cot\theta = \dfrac{1}{\tan\theta} ...Reciprocal\:\:Identity \\[5ex] \cot\theta = -\dfrac{7}{2}$
(54.) Determine the six trigonometric function values of the angle between the point $(-8, 15)$ and the vertical axis.

Sketch the graph and locate the point.
Point $(-8, 15)$ is in the $2nd$ quadrant.
The vertical axis is the $y-axis$
Draw a right triangle joining the origin, the point, and the vertical axis.
Let the angle between the point and the vertical axis = $\theta$

$adj = 15 \\[3ex] opp = 8 \\[3ex] hyp = ? \\[3ex] hyp^2 = opp^2 + adj^2 \\[3ex] hyp^2 = 8^2 + 15^2 \\[3ex] hyp^2 = 64 + 225 \\[3ex] hyp^2 = 289 \\[3ex] hyp = \sqrt{289} = 17 \\[3ex] \sin\theta = \dfrac{opp}{hyp} ...2nd\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \sin\theta = \dfrac{8}{17} \\[5ex] \csc\theta = \dfrac{1}{\sin\theta} ...Reciprocal\:\:Identity \\[5ex] \csc\theta = \dfrac{17}{8} \\[5ex] \cos\theta = -\dfrac{adj}{hyp} ...2nd\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \cos\theta = -\dfrac{15}{17} \\[5ex] \sec\theta = \dfrac{1}{\cos\theta} ...Reciprocal\:\:Identity \\[5ex] \sec\theta = -\dfrac{17}{15} \\[5ex] \tan\theta = -\dfrac{opp}{adj} ...2nd\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \tan\theta = -\dfrac{8}{15} \\[5ex] \cot\theta = \dfrac{1}{\tan\theta} ...Reciprocal\:\:Identity \\[5ex] \cot\theta = -\dfrac{15}{8}$
(55.) Determine the six trigonometric function values of the angle between the point $(-3\sqrt{2}, -5)$ and the horizontal axis.

Sketch the graph and locate the point.
Point $(-3\sqrt{2}, -5)$ is in the $3rd$ quadrant.
The horizontal axis is the $x-axis$
Draw a right triangle joining the origin, the point, and the horizonal axis.
Let the angle between the point and the horizontal axis = $\theta$

$adj = 3\sqrt{2} \\[3ex] opp = 5 \\[3ex] hyp = ? \\[3ex] hyp^2 = opp^2 + adj^2 \\[3ex] hyp^2 = (3\sqrt{2})^2 + 5^2 \\[3ex] hyp^2 = (3)^2 * (\sqrt{2})^2 + 25 \\[3ex] hyp^2 = 9(2) + 25 \\[3ex] hyp^2 = 18 + 25 \\[3ex] hyp^2 = 43 \\[3ex] hyp = \sqrt{43} \\[3ex] \sin\theta = -\dfrac{opp}{hyp} ...3rd\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \sin\theta = -\dfrac{5}{\sqrt{43}} \\[5ex] \sin\theta = -\dfrac{5}{\sqrt{43}} * \dfrac{\sqrt{43}}{\sqrt{43}} \\[5ex] \sin\theta = \dfrac{-5 * \sqrt{43}}{\sqrt{43} * \sqrt{43}} \\[5ex] \sin\theta = -\dfrac{5\sqrt{43}}{43} \\[5ex] \csc\theta = \dfrac{1}{\sin\theta} ...Reciprocal\:\:Identity \\[5ex] \csc\theta = -\dfrac{\sqrt{43}}{5} \\[5ex] \cos\theta = -\dfrac{adj}{hyp} ...3rd\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \cos\theta = -\dfrac{3\sqrt{2}}{\sqrt{43}} \\[5ex] \cos\theta = -\dfrac{3\sqrt{2}}{\sqrt{43}} * \dfrac{\sqrt{43}}{\sqrt{43}} \\[5ex] \cos\theta = -\dfrac{3 * \sqrt{2} * \sqrt{43}}{\sqrt{43} * \sqrt{43}} \\[5ex] cos\theta = -\dfrac{3 * \sqrt{2 * 43}}{43} \\[5ex] \cos\theta = -\dfrac{3\sqrt{86}}{46} \\[5ex] \sec\theta = \dfrac{1}{\cos\theta} ...Reciprocal\:\:Identity \\[5ex] \sec\theta = -\dfrac{\sqrt{43}}{3\sqrt{2}} \\[5ex] \sec\theta = -\dfrac{\sqrt{43}}{3\sqrt{2}} * \dfrac{\sqrt{2}}{\sqrt{2}} \\[5ex] \sec\theta = -\dfrac{\sqrt{43} * \sqrt{2}}{3 * \sqrt{2} * \sqrt{2}} \\[5ex] \sec\theta = -\dfrac{\sqrt{43 * 2}}{3 * 2} \\[5ex] \sec\theta = -\dfrac{\sqrt{86}}{6} \\[5ex] \tan\theta = \dfrac{opp}{adj} ...3rd\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \tan\theta = \dfrac{5}{3\sqrt{2}} \\[5ex] \tan\theta = \dfrac{5}{3\sqrt{2}} * \dfrac{\sqrt{2}}{\sqrt{2}} \\[5ex] \tan\theta = \dfrac{5 * \sqrt{2}}{3 * \sqrt{2} * \sqrt{2}} \\[5ex] \tan\theta = \dfrac{5\sqrt{2}}{3 * 2} \\[5ex] \tan\theta = \dfrac{5\sqrt{2}}{6} \\[5ex] \cot\theta = \dfrac{1}{\tan\theta} ...Reciprocal\:\:Identity \\[5ex] \cot\theta = \dfrac{3\sqrt{2}}{5}$
(56.) Determine the six trigonometric function values of the angle between the point $(-12, 5)$ and the horizontal axis.

Sketch the graph and locate the point.
Point $(-12, 5)$ is in the $2nd$ quadrant.
The horizontal axis is the $x-axis$
Draw a right triangle joining the origin, the point, and the horizontal axis.
Let the angle between the point and the horizontal axis = $\theta$

$adj = 12 \\[3ex] opp = 5 \\[3ex] hyp = ? \\[3ex] hyp^2 = opp^2 + adj^2 \\[3ex] hyp^2 = 5^2 + 12^2 \\[3ex] hyp^2 = 25 + 144 \\[3ex] hyp^2 = 169 \\[3ex] hyp = \sqrt{169} = 13 \\[3ex] \sin\theta = \dfrac{opp}{hyp} ...2nd\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \sin\theta = \dfrac{5}{13} \\[5ex] \csc\theta = \dfrac{1}{\sin\theta} ...Reciprocal\:\:Identity \\[5ex] \csc\theta = \dfrac{13}{5} \\[5ex] \cos\theta = -\dfrac{adj}{hyp} ...2nd\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \cos\theta = -\dfrac{12}{13} \\[5ex] \sec\theta = \dfrac{1}{\cos\theta} ...Reciprocal\:\:Identity \\[5ex] \sec\theta = -\dfrac{13}{12} \\[5ex] \tan\theta = -\dfrac{opp}{adj} ...2nd\:\:Quadrant\:\:Identity\:\:and\:\:SOHCAHTOA \\[5ex] \tan\theta = -\dfrac{5}{12} \\[5ex] \cot\theta = \dfrac{1}{\tan\theta} ...Reciprocal\:\:Identity \\[5ex] \cot\theta = -\dfrac{12}{5}$
(57.) Use a calculator but do not use the trigonometric functions on the calculator.
Apply the Trigonometric Identities.
Find the trigonometric function values of $352^\circ$ if

$\sin 8^\circ \approx 0.1392 \\[3ex] \cos 8^\circ \approx 0.9903 \\[3ex] \tan 8^\circ \approx 0.1405 \\[3ex]$

$8 + 352 = 360 \\[3ex] \rightarrow 352 = 360 - 8 \\[3ex] 352^\circ \:\:is\:\:in\:\:the\:\:4th\:\:Quadrant \\[3ex] \sin 352^\circ = -\sin(360 - 352) ...4th\:\:Quadrant\:\:Identity \\[3ex] \sin 352^\circ = -\sin 8^\circ\\[3ex] \sin 325^\circ \approx -0.1392 \\[3ex] \csc 325^\circ = \dfrac{1}{\sin 352^\circ} ...Reciprocal\:\:Identity \\[5ex] \csc 325^\circ = \dfrac{1}{-0.1392} \\[5ex] \csc 325^\circ \approx -7.1839 \\[3ex] \cos 352^\circ = \cos(360 - 352) ...4th\:\:Quadrant\:\:Identity \\[3ex] \cos 352^\circ = \cos 8^\circ\\[3ex] \cos 325^\circ \approx 0.9903 \\[3ex] \sec 325^\circ = \dfrac{1}{\cos 352^\circ} ...Reciprocal\:\:Identity \\[5ex] \sec 325^\circ = \dfrac{1}{0.9903} \\[5ex] \sec 325^\circ \approx 1.0098 \\[3ex] \tan 352^\circ = -\tan(360 - 352) ...4th\:\:Quadrant\:\:Identity \\[3ex] \tan 352^\circ = -\tan 8^\circ\\[3ex] \tan 325^\circ \approx -0.1405 \\[3ex] \cot 325^\circ = \dfrac{1}{\tan 352^\circ} ...Reciprocal\:\:Identity \\[5ex] \cot 325^\circ = \dfrac{1}{-0.1405} \\[5ex] \cot 325^\circ \approx -7.1174$
(58.) Use a calculator but do not use the trigonometric functions on the calculator.
Apply the Trigonometric Identities.
Find the trigonometric function values of $159^\circ$ if

$\sin 21^\circ \approx 0.3584 \\[3ex] \cos 21^\circ \approx 0.9336 \\[3ex] \tan 21^\circ \approx 0.3839 \\[3ex]$

$21 + 159 = 180 \\[3ex] \rightarrow 159 = 180 - 21 \\[3ex] 159^\circ \:\:is\:\:in\:\:the\:\:2nd\:\:Quadrant \\[3ex] \sin 159^\circ = \sin(180 - 159) ...2nd\:\:Quadrant\:\:Identity \\[3ex] \sin 159^\circ = \sin 21^\circ\\[3ex] \sin 159^\circ \approx 0.3584 \\[3ex] \csc 159^\circ = \dfrac{1}{\sin 159^\circ} ...Reciprocal\:\:Identity \\[5ex] \csc 159^\circ = \dfrac{1}{0.3584} \\[5ex] \csc 159^\circ \approx 2.7902 \\[3ex] \cos 159^\circ = -\cos(180 - 159) ...2nd\:\:Quadrant\:\:Identity \\[3ex] \cos 159^\circ = -\cos 21^\circ\\[3ex] \cos 159^\circ \approx -0.9336 \\[3ex] \sec 159^\circ = \dfrac{1}{\cos 159^\circ} ...Reciprocal\:\:Identity \\[5ex] \sec 159^\circ = \dfrac{1}{-0.9336} \\[5ex] \sec 159^\circ \approx -1.0711 \\[3ex] \tan 159^\circ = -\tan(180 - 159) ...2nd\:\:Quadrant\:\:Identity \\[3ex] \tan 159^\circ = -\tan 21^\circ\\[3ex] \tan 159^\circ \approx -0.3839 \\[3ex] \cot 159^\circ = \dfrac{1}{\tan 159^\circ} ...Reciprocal\:\:Identity \\[5ex] \cot 159^\circ = \dfrac{1}{-0.3839} \\[5ex] \cot 159^\circ \approx -2.6048$
(59.) Determine the value of $\alpha$ in the interval $(270^\circ, 360^\circ)$ if $\sin \alpha = -0.7771$

$\sin \alpha = -0.7771 \\[3ex] \alpha = \sin^{-1} (-0.7771) \\[3ex] \alpha = -50.9958157 \\[3ex] Let\:\: \beta = coterminal\:\: angle \\[3ex] \beta = \alpha + 360k \\[3ex] \underline{Positive\:\: coterminal\:\: angle} \\[3ex] k = 1 \\[3ex] \beta = -50.9958157 + 360(1) \\[3ex] \beta = -50.9958157 + 360 \\[3ex] \beta = 309.004184^\circ \\[3ex] \implies \alpha = 309.004184^\circ \\[3ex] \underline{Check} \\[3ex] \sin 309.004184^\circ \approx -0.7771$
(60.) Determine the value of $\alpha$ in the interval $(180^\circ, 270^\circ)$ if $\cos \alpha = -0.9848$
Round to the nearest integer.

$\cos \alpha = -0.9848 \\[3ex] \alpha = \cos^{-1} (-0.9848) \\[3ex] \alpha = 169.997442 \approx 170^\circ \\[3ex] Let\:\: \beta = coterminal\:\: angle \\[3ex] \beta = \alpha + 360k \\[3ex] \underline{Negative\:\: coterminal\:\: angle} \\[3ex] k = -1 \\[3ex] \beta = 170 + 360(-1) \\[3ex] \beta = 170 - 360 \\[3ex] \beta = -190^\circ \\[3ex] But\:\: \cos -190^\circ = \cos 190^\circ ...Even Identity \\[3ex] \therefore \beta = 190^\circ \\[3ex] \implies \alpha = 190^\circ \\[3ex] \underline{Check} \\[3ex] \cos 190^\circ \approx -0.9848$

(61.) Determine the value of $\alpha$ in the interval $(90^\circ, 180^\circ)$ if $\tan \alpha = -0.3640$
Round to the nearest integer.

$\tan \alpha = -0.3640 \\[3ex] \alpha = \tan^{-1} (-0.3640) \\[3ex] \alpha = -20.0015059 \approx -20^\circ \\[3ex] Let\:\: \beta = coterminal\:\: angle \\[3ex] \beta = \alpha + 360k \\[3ex] \underline{Positive\:\: coterminal\:\: angle} \\[3ex] k = 1 \\[3ex] \beta = -20 + 360(1) \\[3ex] \beta = -20 + 360 \\[3ex] \beta = 340^\circ \\[3ex] (90^\circ, 180^\circ) \rightarrow 2nd\:\:Quadrant \\[3ex] \tan 340 = -\tan(180 - 340) ... 2nd\:\:Quadrant\:\:Identity \\[3ex] \tan 340 = -\tan (-160) \\[3ex] But\:\: \tan(-160) = -\tan 160 ... Odd\:\:Identity \\[3ex] \rightarrow \tan 340 = -(-\tan 160) \\[3ex] \tan 340 = \tan 160 \\[3ex] \therefore \alpha = 160^\circ \\[3ex] \underline{Check} \\[3ex] \tan 160^\circ \approx -0.3640$
(62.) Determine the value of $\alpha$ in the interval $(180^\circ, 270^\circ)$ if $\sec \alpha = -2.3662$
Round to the nearest integer.

$\sec \alpha = -2.3662 \\[3ex] \sec \alpha = \dfrac{1}{\cos \alpha} ... Reciprocal\:\:Identity \\[5ex] \rightarrow \dfrac{1}{\cos \alpha} = -2.3662 \\[5ex] \cos \alpha = \dfrac{1}{-2.3662} \\[5ex] \cos \alpha = -0.422618545 \\[3ex] \alpha = \cos^{-1} (-0.422618545) \\[3ex] \alpha = 115.000018 \approx 115^\circ \\[3ex] Let\:\: \beta = coterminal\:\: angle \\[3ex] \beta = \alpha + 360k \\[3ex] \underline{Negative\:\: coterminal\:\: angle} \\[3ex] k = -1 \\[3ex] \beta = 115 + 360(-1) \\[3ex] \beta = 115 - 360 \\[3ex] \beta = -245^\circ \\[3ex] But\:\: \cos -245^\circ = \cos 245^\circ ...Even Identity \\[3ex] Similarly\:\: \sec -245^\circ = \sec 245^\circ ...Even Identity \\[3ex] \therefore \beta = 245^\circ \\[3ex] \underline{Check} \\[3ex] \cos 245^\circ \approx -0.4226 \\[3ex] \sec 245^\circ \approx -2.3662$
(63.) Determine the value of $\alpha$ in the interval $(90^\circ, 180^\circ)$ if $\csc \alpha = 1.0154$
Round to the nearest integer.

$\csc \alpha = 1.0154 \\[3ex] \csc \alpha = \dfrac{1}{\sin \alpha} ... Reciprocal\:\:Identity \\[5ex] \rightarrow \dfrac{1}{\sin \alpha} = 0.9848 \\[5ex] \sin \alpha = \dfrac{1}{1.0154} \\[5ex] \sin \alpha = 0.984833563 \\[3ex] \alpha = \sin^{-1} (0.984833563) \\[3ex] \alpha = 80.0085197 \approx 80^\circ \\[3ex] (90^\circ, 180^\circ) \rightarrow 2nd\:\:Quadrant \\[3ex] \sin 80 = \sin(180 - 80) ... 2nd\:\:Quadrant\:\:Identity \\[3ex] \sin 80^\circ = \sin 100^\circ \\[3ex] Similarly\:\: \csc 80^\circ = \csc 100^\circ \\[3ex] \implies \alpha = 100^\circ \\[3ex] \underline{Check} \\[3ex] \sin 100^\circ \approx 0.9848 \\[3ex] \csc 100^\circ \approx 1.0154$
(64.) Determine the value of $\alpha$ in the interval $(270^\circ, 360^\circ)$ if $\cot \alpha = -0.8391$
Round to the nearest integer.

$\cot \alpha = -0.8391 \\[3ex] \cot \alpha = \dfrac{1}{\tan \alpha} ... Reciprocal\:\:Identity \\[5ex] \rightarrow \dfrac{1}{\tan \alpha} = -0.8391 \\[5ex] \tan \alpha = \dfrac{1}{-0.8391} \\[5ex] \tan \alpha = -1.19175307 \\[3ex] \alpha = \tan^{-1} (-1.19175307) \\[3ex] \alpha = -49.9999876 \approx -50^\circ \\[3ex] Let\:\: \beta = coterminal\:\: angle \\[3ex] \beta = \alpha + 360k \\[3ex] \underline{Positive\:\: coterminal\:\: angle} \\[3ex] k = 1 \\[3ex] \beta = -50 + 360(1) \\[3ex] \beta = -50 + 360 \\[3ex] \beta = 310^\circ \\[3ex] \implies \alpha = 310^\circ \\[3ex] \underline{Check} \\[3ex] \tan 310^\circ \approx -1.1918 \\[3ex] \cot 310^\circ \approx -0.8391$
(65.) Determine $3$ positive and $3$ negative coterminal angles of $\dfrac{11\pi}{9}$

$\alpha = \dfrac{11\pi}{9} \\[5ex] Let\:\: \beta = coterminal\:\: angle \\[3ex] \beta = \alpha + 2\pi k \\[3ex] \underline{Positive\:\: coterminal\:\: angles} \\[3ex] k = 1 \\[3ex] \beta = \dfrac{11\pi}{9} + 2\pi(1) \\[5ex] \beta = \dfrac{11\pi}{9} + 2\pi \\[5ex] \beta = \dfrac{11\pi}{9} + \dfrac{18\pi}{9} \\[5ex] \beta = \dfrac{11\pi + 18\pi}{9} \\[5ex] \beta = \dfrac{29\pi}{9} \\[5ex] k = 2 \\[3ex] \beta = \dfrac{11\pi}{9} + 2\pi(2) \\[5ex] \beta = \dfrac{11\pi}{9} + 4\pi \\[5ex] \beta = \dfrac{11\pi}{9} + \dfrac{36\pi}{9} \\[5ex] \beta = \dfrac{11\pi + 36\pi}{9} \\[5ex] \beta = \dfrac{47\pi}{9} \\[5ex] k = 3 \\[3ex] \beta = \dfrac{11\pi}{9} + 2\pi(3) \\[5ex] \beta = \dfrac{11\pi}{9} + 6\pi \\[5ex] \beta = \dfrac{11\pi}{9} + \dfrac{54\pi}{9} \\[5ex] \beta = \dfrac{11\pi + 54\pi}{9} \\[5ex] \beta = \dfrac{65\pi}{9} \\[5ex] \underline{Negative\:\: coterminal\:\: angles} \\[3ex] k = -1 \\[3ex] \beta = \dfrac{11\pi}{9} + 2\pi(-1) \\[5ex] \beta = \dfrac{11\pi}{9} - 2\pi \\[5ex] \beta = \dfrac{11\pi}{9} - \dfrac{18\pi}{9} \\[5ex] \beta = \dfrac{11\pi - 18\pi}{9} \\[5ex] \beta = -\dfrac{7\pi}{9} \\[5ex] k = -2 \\[3ex] \beta = \dfrac{11\pi}{9} + 2\pi(-2) \\[5ex] \beta = \dfrac{11\pi}{9} - 4\pi \\[5ex] \beta = \dfrac{11\pi}{9} - \dfrac{36\pi}{9} \\[5ex] \beta = \dfrac{11\pi - 36\pi}{9} \\[5ex] \beta = -\dfrac{25\pi}{9} \\[5ex] k = -3 \\[3ex] \beta = \dfrac{11\pi}{9} + 2\pi(-3) \\[5ex] \beta = \dfrac{11\pi}{9} - 6\pi \\[5ex] \beta = \dfrac{11\pi}{9} - \dfrac{54\pi}{9} \\[5ex] \beta = \dfrac{11\pi - 54\pi}{9} \\[5ex] \beta = -\dfrac{43\pi}{9}$
(66.) Determine $3$ positive and $3$ negative coterminal angles of $-\dfrac{5\pi}{6}$

$\alpha = -\dfrac{5\pi}{6} \\[5ex] Let\:\: \beta = coterminal\:\: angle \\[3ex] \beta = \alpha + 2\pi k \\[3ex] \underline{Positive\:\: coterminal\:\: angles} \\[3ex] k = 1 \\[3ex] \beta = -\dfrac{5\pi}{6} + 2\pi(1) \\[5ex] \beta = -\dfrac{5\pi}{6} + 2\pi \\[5ex] \beta = -\dfrac{5\pi}{6} + \dfrac{12\pi}{6} \\[5ex] \beta = \dfrac{-5\pi + 12\pi}{6} \\[5ex] \beta = \dfrac{7\pi}{6} \\[5ex] k = 2 \\[3ex] \beta = -\dfrac{5\pi}{6} + 2\pi(2) \\[5ex] \beta = -\dfrac{5\pi}{6} + 4\pi \\[5ex] \beta = -\dfrac{5\pi}{6} + \dfrac{24\pi}{6} \\[5ex] \beta = \dfrac{-5\pi + 24\pi}{6} \\[5ex] \beta = \dfrac{19\pi}{6} \\[5ex] k = 3 \\[3ex] \beta = -\dfrac{5\pi}{6} + 2\pi(3) \\[5ex] \beta = -\dfrac{5\pi}{6} + 6\pi \\[5ex] \beta = -\dfrac{5\pi}{6} + \dfrac{36\pi}{6} \\[5ex] \beta = \dfrac{-5\pi + 36\pi}{6} \\[5ex] \beta = \dfrac{31\pi}{6} \\[5ex] \underline{Negative\:\: coterminal\:\: angles} \\[3ex] k = -1 \\[3ex] \beta = -\dfrac{5\pi}{6} + 2\pi(-1) \\[5ex] \beta = -\dfrac{5\pi}{6} - 2\pi \\[5ex] \beta = -\dfrac{5\pi}{6} - \dfrac{12\pi}{6} \\[5ex] \beta = \dfrac{-5\pi - 12\pi}{6} \\[5ex] \beta = -\dfrac{17\pi}{6} \\[5ex] k = -2 \\[3ex] \beta = -\dfrac{5\pi}{6} + 2\pi(-2) \\[5ex] \beta = -\dfrac{5\pi}{6} - 4\pi \\[5ex] \beta = -\dfrac{5\pi}{6} - \dfrac{24\pi}{6} \\[5ex] \beta = \dfrac{-5\pi - 24\pi}{6} \\[5ex] \beta = -\dfrac{29\pi}{6} \\[5ex] k = -3 \\[3ex] \beta = -\dfrac{5\pi}{6} + 2\pi(-3) \\[5ex] \beta = -\dfrac{5\pi}{6} - 6\pi \\[5ex] \beta = -\dfrac{5\pi}{6} - \dfrac{36\pi}{6} \\[5ex] \beta = \dfrac{-5\pi - 36\pi}{6} \\[5ex] \beta = -\dfrac{41\pi}{6}$
(67.)

(68.) ACT If $A$ is the measure of an acute angle (that is, $0^\circ \lt A \lt 90^\circ$) and $\sin A = \dfrac{12}{13}$, what are the possible values of $\tan A$?

$A.\:\: \dfrac{5}{12} \:\:and\:\: -\dfrac{5}{12} \\[5ex] B.\:\: \dfrac{12}{5} \:\:and\:\: -\dfrac{12}{5} \\[5ex] C.\:\: \dfrac{5}{12} \:\:only \\[5ex] D.\:\: \dfrac{12}{5} \:\:only \\[5ex] E.\:\: \dfrac{13}{12} \:\:only \\[5ex]$

$SOHCAHTOA \\[3ex] \sin A = \dfrac{opp}{hyp} = \dfrac{12}{13} \\[5ex] hyp = 13 \\[3ex] opp = leg = 12 \\[3ex] adj = leg \\[3ex] hyp^2 = leg^2 + leg^2...Pythagorean\:\: Theorem \\[3ex] hyp^2 = opp^2 + adj^2 adj^2 = hyp^2 - opp^2 \\[3ex] adj^2 = 13^2 - 12^2 \\[3ex] adj^2 = 169 - 144 = 25 \\[3ex] adj = \sqrt{25} = 5 \\[3ex] 0^\circ \lt A \lt 90^\circ - First\:\: Quadrant \\[3ex] \tan A = \dfrac{opp}{adj} \\[5ex] \tan A = \dfrac{12}{5}$
(69.) WASSCE:FM If $\sin X = \dfrac{p - q}{p + q}$, where $0^\circ \le x \le 90^\circ$, find $1 - \tan^2X$

$0^\circ \lt x \lt 90^\circ \implies \theta \;\;is\;\;acute \implies First\;\;Quadrant \\[3ex] SOH:CAH:TOA \\[3ex] \sin X = \dfrac{opp}{hyp} = \dfrac{p - q}{p + q} \\[5ex] opp = p - q \\[3ex] hyp = p + q \\[3ex] hyp^2 = opp^2 + adj^2...Pythagorean\:\: Theorem \\[3ex] (p + q)^2 = (p - q)^2 + adj^2 \\[3ex] (p - q)^2 + adj^2 = (p + q)^2 \\[3ex] adj^2 = (p + q)^2 - (p - q)^2 \\[3ex] = [(p + q) + (p - q)][(p + q) - (p - q)]...Difference\;\;of\;\;Two\;\;Squares \\[3ex] = [p + q + p - q][p + q - p + q] \\[3ex] = (2p)(2q) \\[3ex] = 4pq \\[3ex] \tan X = \dfrac{opp}{adj} \\[5ex] \tan^2 X = \dfrac{opp^2}{adj^2} \\[5ex] \tan^2 X = \dfrac{(p - q)^2}{4pq} \\[5ex] 1 - \tan^2 X \\[3ex] = 1 - \dfrac{(p - q)^2}{4pq} \\[5ex] = \dfrac{4pq}{4pq} - \dfrac{(p - q)^2}{4pq} \\[5ex] = \dfrac{4pq - (p - q)^2}{4pq} \\[5ex] = \dfrac{4pq - [(p - q)(p - q)]}{4pq} \\[5ex] = \dfrac{4pq - (p^2 - pq - pq + q^2)}{4pq} \\[5ex] = \dfrac{4pq - (p^2 - 2pq + q^2)}{4pq} \\[5ex] = \dfrac{4pq - p^2 + 2pq - q^2}{4pq} \\[5ex] = \dfrac{6pq - p^2 - q^2}{4pq}$
(70.) WASSCE Given that $\sin \theta = \dfrac{5}{6 + x}$ and $\tan \theta = \dfrac{5}{12}$, $0^\circ \lt x \lt 90^\circ$, find the value of $x$

$0^\circ \lt x \lt 90^\circ \implies x \;\;is\;\;acute \implies First\;\;Quadrant \\[3ex] SOH:CAH:TOA \\[3ex] \tan \theta = \dfrac{opp}{adj} = \dfrac{5}{12} \\[5ex] opp = 5 \\[3ex] adj = 12 \\[3ex] hyp^2 = opp^2 + adj^2...Pythagorean\:\: Theorem \\[3ex] hyp^2 = 5^2 + 12^2 \\[3ex] hyp^2 = 25 + 144 \\[3ex] hyp^2 = 169 \\[3ex] hyp = \sqrt{169} \\[3ex] hyp = 13 \\[3ex] \sin \theta = \dfrac{opp}{hyp} = \dfrac{5}{6 + x} \\[5ex] \implies \dfrac{5}{6 + x} = \dfrac{5}{13} \\[5ex] The\;\;two\;\;fractions\;\;are\;\;equal \\[3ex] Numerators\;\;are\;\;the\;\;same \\[3ex] So,\;\;equate\;\;the\;\;denominators \\[3ex] 6 + x = 13 \\[3ex] x = 13 - 6 \\[3ex] x = 7$
(71.) WASSCE:FM If $\sin p = \dfrac{1}{2}$ and $\cos q = \dfrac{1}{3}$ evaluate $\sin(p - q)$, where $0^\circ \le p \le 90^\circ$ and $90^\circ \le q \le 180^\circ$

$\sin(p - q) = \sin p \cos q - \cos p \sin q...Difference\;\;Formula \\[3ex] 0^\circ \lt p \lt 90^\circ \implies p \;\;is\;\;acute \implies First\;\;Quadrant \\[3ex] First\;\;Quadrant \implies \sin p \;\;is\;\;positive \\[3ex] First\;\;Quadrant \implies \cos p \;\;is\;\;positive \\[3ex] \sin p = \dfrac{1}{2} \\[5ex] p = \sin^{-1}\left(\dfrac{1}{2}\right) \\[5ex] p = 30^\circ \\[3ex] \cos p = \cos 30^\circ = \dfrac{\sqrt{3}}{2} \\[5ex] 90^\circ \lt q \lt 180^\circ \implies q \;\;is\;\;obtuse \implies Second\;\;Quadrant \\[3ex] Second\;\;Quadrant \implies \sin q \;\;is\;\;positive \\[3ex] Second\;\;Quadrant \implies \cos q \;\;is\;\;negative \\[3ex] \implies \cos q = -\dfrac{1}{3}...even\;\;though\;\;the\;\;question\;\;states\;\;it\;\;as\;\;positive \\[5ex] SOH:CAH:TOA \\[3ex] \cos q = -\dfrac{1}{3} = \dfrac{adj}{hyp} \\[5ex]$ Because q is not a special angle, and we do want to find the exact forms the trigonometric functions; it is better to draw the diagram.

$hyp^2 = opp^2 + adj^2...Pythagorean\:\: Theorem \\[3ex] 3^2 = opp^2 + 1^2 \\[3ex] 9 = opp^2 + 1 \\[3ex] opp^2 + 1 = 9 \\[3ex] opp^2 = 9 - 1 \\[3ex] opp^2 = 8 \\[3ex] opp = \sqrt{8} = \sqrt{4 * 2} = \sqrt{4} * \sqrt{2} \\[3ex] opp = 2\sqrt{2} \\[3ex] \sin q = \dfrac{opp}{hyp} = \dfrac{2\sqrt{2}}{3} \\[5ex] \therefore \sin(p - q) \\[3ex] = \sin p \cos q - \cos p \sin q \\[3ex] = \left(\dfrac{1}{2} * -\dfrac{1}{3}\right) - \left(\dfrac{\sqrt{3}}{2} * \dfrac{2\sqrt{2}}{3}\right) \\[5ex] = -\dfrac{1}{6} - \dfrac{\sqrt{6}}{3} \\[5ex] = \dfrac{-1 - 2\sqrt{6}}{6} \\[5ex] = \dfrac{-(1 + 2\sqrt{6})}{6}$
(72.) Determine the exact value of $\cos \dfrac{5\pi}{12}$

Find two special angles whose sum or difference is $\dfrac{5\pi}{12}$

$\dfrac{5\pi}{12} = \dfrac{5 * 180}{12} = \dfrac{900}{12} = 75 \\[5ex] 75 = 30 + 45 \\[3ex] \dfrac{5\pi}{12} = \dfrac{2\pi}{12} + \dfrac{3\pi}{12} \\[5ex] \dfrac{2\pi}{12} = \dfrac{\pi}{6} = \dfrac{180}{6} = 30 \\[5ex] \dfrac{3\pi}{12} = \dfrac{\pi}{4} = \dfrac{180}{4} = 45 \\[5ex] \cos\dfrac{5\pi}{12} \\[3ex] = \cos\left(\dfrac{2\pi}{12} + \dfrac{3\pi}{12}\right) \\[5ex] = \cos\left(\dfrac{\pi}{6} + \dfrac{\pi}{4}\right) \\[5ex] = \cos\dfrac{\pi}{6} \cos\dfrac{\pi}{4} - \sin\dfrac{\pi}{6} \sin\dfrac{\pi}{4} ... Sum\:\: Formula \\[5ex] = \dfrac{\sqrt{3}}{2} * \dfrac{\sqrt{2}}{2} - \dfrac{1}{2} * \dfrac{\sqrt{2}}{2} \\[5ex] = \dfrac{\sqrt{6}}{4} - \dfrac{\sqrt{2}}{4} \\[5ex] = \dfrac{\sqrt{6} - \sqrt{2}}{4}$
(73.) If $\sin t = \dfrac{2}{3}$, and $t$ is in quadrant I, find the exact value of $\sin (2t)$, $\cos (2t)$, and $\tan (2t)$ algebraically without solving for $t$

Let us draw the diagram.

$SOH:CAH:TOA \\[3ex] \sin t = \dfrac{2}{3} = \dfrac{opp}{hyp} \\[5ex] hyp^2 = opp^2 + adj^2...Pythagorean\:\: Theorem \\[3ex] 3^2 = 2^2 + adj^2 \\[3ex] 9 = 4 + adj^2 \\[3ex] 4 + adj^2 = 9 \\[3ex] adj^2 = 9 - 4 \\[3ex] adj^2 = 5 \\[3ex] adj = \sqrt{5} \\[3ex] \cos t = \dfrac{adj}{hyp} = \dfrac{\sqrt{5}}{3} \\[5ex] \tan t = \dfrac{opp}{adj} = \dfrac{2}{\sqrt{5}} \\[5ex] (a.) \\[3ex] \sin 2t = 2\sin t \cos t...Double-Angle\;\;Formula \\[3ex] = 2 * \dfrac{2}{3} * \dfrac{\sqrt{5}}{3} \\[5ex] = \dfrac{4\sqrt{5}}{9} \\[5ex] (b.) \\[3ex] \cos 2t = \cos^2t - \sin^2t...Double-Angle\;\;Formula \\[3ex] = \left(\dfrac{\sqrt{5}}{3}\right)^2 - \left(\dfrac{2}{3}\right)^2 \\[5ex] = \dfrac{5}{9} - \dfrac{4}{9} \\[5ex] = \dfrac{1}{9} \\[5ex] (c.) \\[3ex] \tan 2t = \dfrac{2\tan t}{1 - \tan^2t}...Double-Angle\;\;Formula \\[5ex] = (2\tan t) \div (1 - \tan^2t) \\[5ex] = \left(2 * \dfrac{2}{\sqrt{5}}\right) \div \left[1 - \left(\dfrac{2}{\sqrt{5}}\right)^2\right] \\[5ex] = \dfrac{4}{\sqrt{5}} \div \left(1 - \dfrac{4}{5}\right) \\[5ex] = \dfrac{4}{\sqrt{5}} \div \left(\dfrac{5}{5} - \dfrac{4}{5}\right) \\[5ex] = \dfrac{4}{\sqrt{5}} \div \dfrac{1}{5} \\[5ex] = \dfrac{4}{\sqrt{5}} * \dfrac{5}{1} \\[5ex] = \dfrac{20}{\sqrt{5}} \\[5ex] = \dfrac{20}{\sqrt{5}} * \dfrac{\sqrt{5}}{\sqrt{5}} \\[5ex] = \dfrac{20\sqrt{5}}{5} \\[5ex] = 4\sqrt{5}$
(74.) WASSCE:FM If $\dfrac{\tan \theta + \tan 30^\circ}{1 - \tan \theta \tan 30^\circ} + 1 = 0$, find $\tan \theta$, leaving the answer in surd form.

$\dfrac{\tan \theta + \tan 30^\circ}{1 - \tan \theta \tan 30^\circ} + 1 = 0 \\[5ex] \dfrac{\tan \theta + \tan 30^\circ}{1 - \tan \theta \tan 30^\circ} = -1 \\[5ex] \tan \theta + \tan 30 = -1(1 - \tan \theta \tan 30) \\[3ex] \tan \theta + \tan 30 = -1 + \tan\theta \tan 30 \\[3ex] \tan 30 + 1 = \tan\theta \tan 30 - \tan\theta \\[3ex] \tan 30 + 1 = \tan\theta(\tan 30 - 1) \\[3ex] \tan\theta(\tan 30 - 1) = \tan 30 + 1 \\[3ex] \tan\theta = \dfrac{\tan 30 + 1}{\tan 30 - 1} \\[5ex] \tan\theta = (\tan 30 + 1) \div (\tan 30 - 1) \\[3ex] \tan 30 = \dfrac{\sqrt{3}}{3} \\[5ex] \implies \\[3ex] \tan\theta \\[3ex] = \left(\dfrac{\sqrt{3}}{3} + 1\right) \div \left(\dfrac{\sqrt{3}}{3} - 1\right) \\[5ex] = \left(\dfrac{\sqrt{3}}{3} + \dfrac{3}{3}\right) \div \left(\dfrac{\sqrt{3}}{3} - \dfrac{3}{3}\right) \\[5ex] = \dfrac{\sqrt{3} + 3}{3} \div \dfrac{\sqrt{3} - 3}{3} \\[5ex] = \dfrac{\sqrt{3} + 3}{3} * \dfrac{3}{\sqrt{3} - 3} \\[5ex] = \dfrac{\sqrt{3} + 3}{\sqrt{3} - 3} \\[5ex] = \dfrac{\sqrt{3} + 3}{\sqrt{3} - 3} * \dfrac{\sqrt{3} + 3}{\sqrt{3} + 3} \\[5ex] = \dfrac{3 + 3\sqrt{3} + 3\sqrt{3} + 9}{\left(\sqrt{3}\right)^2 - 3^2} \\[7ex] = \dfrac{12 + 6\sqrt{3}}{3 - 9} \\[5ex] = \dfrac{6(2 + \sqrt{3})}{-6} \\[5ex] = \dfrac{2 + \sqrt{3}}{-1} \\[5ex] = -1(2 + \sqrt{3})$
(75.)

(76.) JAMB If $\tan \theta = \dfrac{4}{3}$, calculate $\sin^2\theta - \cos^2\theta$

$A.\;\; \dfrac{24}{25} \\[5ex] B.\;\; \dfrac{16}{25} \\[5ex] C.\;\; \dfrac{9}{25} \\[5ex] D.\;\; \dfrac{7}{25} \\[5ex]$

$SOH:CAH:TOA \\[3ex] \tan \theta = \dfrac{4}{3} = \dfrac{opp}{adj} \\[5ex] opp = 4 \\[3ex] adj = 3 \\[3ex] hyp^2 = opp^2 + adj^2...Pythagorean\:\: Theorem \\[3ex] hyp^2 = 4^2 + 3^2 \\[3ex] hyp^2 = 16 + 9 \\[3ex] hyp^2 = 25 \\[3ex] hyp = \sqrt{25} = 5 \\[3ex] \sin \theta = \dfrac{opp}{hyp} = \dfrac{4}{5} \\[5ex] \sin^2\theta = \left(\dfrac{4}{5}\right)^2 = \dfrac{16}{25} \\[5ex] \cos \theta = \dfrac{adj}{hyp} = \dfrac{3}{5} \\[5ex] \cos^2\theta = \left(\dfrac{3}{5}\right)^2 = \dfrac{9}{25} \\[5ex] \sin^2\theta - \cos^2\theta \\[3ex] = \dfrac{16}{25} - \dfrac{9}{25} \\[5ex] = \dfrac{16 - 9}{25} \\[5ex] = \dfrac{7}{25}$
(77.)

(78.) JAMB If $\tan \theta = \dfrac{3}{4}$, find the value of $\sin\theta + \cos\theta$

$A.\;\; 1\dfrac{2}{5} \\[5ex] B.\;\; 1\dfrac{1}{3} \\[5ex] C.\;\; 1\dfrac{2}{3} \\[5ex] D.\;\; 1\dfrac{3}{5} \\[5ex]$

$SOH:CAH:TOA \\[3ex] \tan \theta = \dfrac{3}{4} = \dfrac{opp}{adj} \\[5ex] opp = 3 \\[3ex] adj = 4 \\[3ex] hyp^2 = opp^2 + adj^2...Pythagorean\:\: Theorem \\[3ex] hyp^2 = 3^2 + 4^2 \\[3ex] hyp^2 = 9 + 16 \\[3ex] hyp^2 = 25 \\[3ex] hyp = \sqrt{25} = 5 \\[3ex] \sin \theta = \dfrac{opp}{hyp} = \dfrac{3}{5} \\[5ex] \cos \theta = \dfrac{adj}{hyp} = \dfrac{4}{5} \\[5ex] \sin\theta + \cos\theta \\[3ex] = \dfrac{3}{5} + \dfrac{4}{5} \\[5ex] = \dfrac{7}{5} \\[5ex] = 1\dfrac{2}{5}$
(79.)


(80.) ACT If $90^\circ \lt \theta \lt 180^\circ$ and $\sin \theta = \dfrac{20}{29}$,
then $\cos \theta = ?$

$SOH:CAH:TOA \\[3ex] \cos \theta = \dfrac{adj}{hyp} \\[5ex] \sin \theta = \dfrac{opp}{hyp} = \dfrac{20}{29} \\[5ex] hyp = 29 \\[3ex] opp = leg = 20 \\[3ex] adj = leg \\[3ex] hyp^2 = leg^2 + leg^2...Pythagorean\:\: Theorem \\[3ex] hyp^2 = opp^2 + adj^2 adj^2 = hyp^2 - opp^2 \\[3ex] adj^2 = 29^2 - 20^2 \\[3ex] adj^2 = 841 - 400 = 441 \\[3ex] adj = \sqrt{441} = 21 \\[3ex] \implies \cos \theta = \dfrac{adj}{hyp} = \dfrac{21}{29} \\[5ex] However; \\[3ex] 0 \lt \theta \lt 90 \:\:(\theta \:\:is\:\: acute) \\[3ex] Based\:\: on\:\: the\:\: question \\[3ex] 90 \lt \theta \lt 180 \:\:(\theta \:\:is\:\: obtuse) \rightarrow Second\:\: Quadrant \\[3ex] Second\:\: Quadrant \\[3ex] cosine \:\:is\:\: negative \\[3ex] \therefore \cos \theta = -\dfrac{21}{29}$

(81.) WASSCE:FM Simplify: $\dfrac{1}{1 - \cos\theta} + \dfrac{1}{1 + \cos\theta}$ and leave the answer in terms of $\sin$

$\dfrac{1}{1 - \cos\theta} + \dfrac{1}{1 + \cos\theta} \\[5ex] = \dfrac{1 + \cos\theta + 1 - \cos\theta}{(1 - \cos\theta)(1 + \cos\theta)} \\[5ex] (1 - \cos\theta)(1 + \cos\theta) = 1^2 - \cos^2\theta...Difference\;\;of\;\;Two\;\;Squares \\[3ex] 1^2 - \cos^2\theta = 1 - \cos^2\theta \\[3ex] \sin^2\theta + \cos^2\theta = 1 ...Pythagorean\;\;Identities \\[3ex] \sin^2\theta = 1 - \cos^2\theta \\[3ex] \implies 1 - \cos^2\theta = \sin^2\theta \\[3ex] \implies \\[3ex] \dfrac{1 + \cos\theta + 1 - \cos\theta}{(1 - \cos\theta)(1 + \cos\theta)} \\[5ex] = \dfrac{2}{1 - \cos^2\theta} \\[5ex] = \dfrac{2}{\sin^2\theta}$
(82.) WASSCE If $\sin x = \dfrac{5}{13}$ and $0^\circ \le x \le 90^\circ$,

find without using tables or calculator, $\dfrac{\cos x - 2\sin x}{2\tan x}$

$0^\circ \le x \le 90^\circ ...First\:\: Quadrant \\[3ex] \sin x = \dfrac{5}{13} = \dfrac{opp}{hyp} ... SOHCAHTOA \\[5ex] opp = 5 \\[3ex] hyp = 13 \\[3ex] hyp^2 = opp^2 + adj^2 ... Pythagorean\:\: Theorem \\[3ex] adj^2 = hyp^2 - opp^2 \\[3ex] adj^2 = 13^2 - 5^2 \\[3ex] adj^2 = 169 - 25 \\[3ex] adj^2 = 144 \\[3ex] adj = \sqrt{144} \\[3ex] adj = 12 \\[3ex] \cos x = \dfrac{adj}{hyp} = \dfrac{12}{13} ... SOHCAHTOA \\[5ex] \tan x = \dfrac{opp}{adj} = \dfrac{5}{12} ... SOHCAHTOA \\[5ex] 2\sin x = 2 * \dfrac{5}{13} = \dfrac{10}{13} \\[5ex] \cos x - 2\sin x = \dfrac{12}{13} - \dfrac{10}{13} = \dfrac{12 - 10}{13} = \dfrac{2}{13} \\[5ex] 2\tan x = 2 * \dfrac{5}{12} = \dfrac{5}{6} \\[5ex] \dfrac{\cos x - 2\sin x}{2\tan x} = (\cos x - 2\sin x) \div 2\tan x \\[5ex] = \dfrac{2}{13} \div \dfrac{5}{6} \\[5ex] = \dfrac{2}{13} * \dfrac{6}{5} \\[5ex] = \dfrac{2 * 6}{13 * 5} \\[5ex] = \dfrac{12}{65}$
(83.) Determine the exact value of $\sec\left(-\dfrac{5\pi}{12}\right)$

Find two special angles whose sum or difference is $-\dfrac{5\pi}{12}$

$\sec\left(-\dfrac{5\pi}{12}\right) = \sec\left(\dfrac{5\pi}{12}\right) ... Even\:\: Identity \\[5ex] \dfrac{5\pi}{12} = \dfrac{5 * 180}{12} = 75 \\[5ex] 75 = 30 + 45 \\[3ex] \dfrac{5\pi}{12} = \dfrac{3\pi}{12} + \dfrac{2\pi}{12} \\[5ex] \dfrac{2\pi}{12} = \dfrac{\pi}{6} = \dfrac{180}{6} = 30 \\[5ex] \dfrac{3\pi}{12} = \dfrac{\pi}{4} = \dfrac{180}{4} = 45 \\[5ex] \sec\left(\dfrac{5\pi}{12}\right) = \dfrac{1}{\cos\left(\dfrac{5\pi}{12}\right)} ... Reciprocal\:\: Identity \\[7ex] \cos\left(\dfrac{5\pi}{12}\right) = \cos\left(\dfrac{\pi}{6} + \dfrac{\pi}{4}\right) \\[5ex] \cos\dfrac{5\pi}{12} = \cos\dfrac{\pi}{6}\cos\dfrac{\pi}{4} - \sin\dfrac{\pi}{6}\sin\dfrac{\pi}{4} ... Sum\:\: Formula \\[3ex] \cos\dfrac{5\pi}{12} = \dfrac{\sqrt{3}}{2} * \dfrac{\sqrt{2}}{2} - \dfrac{1}{2} * \dfrac{\sqrt{2}}{2} \\[5ex] \cos\dfrac{5\pi}{12} = \dfrac{\sqrt{6}}{4} - \dfrac{\sqrt{2}}{4} \\[5ex] \cos\dfrac{5\pi}{12} = \dfrac{\sqrt{6} - \sqrt{2}}{4} \\[5ex] = 1 \div \cos\dfrac{5\pi}{12} \\[5ex] = 1 \div \dfrac{\sqrt{6} - \sqrt{2}}{4} \\[5ex] = 1 * \dfrac{4}{\sqrt{6} - \sqrt{2}} \\[5ex] = \dfrac{4}{\sqrt{6} - \sqrt{2}} * \dfrac{\sqrt{6} + \sqrt{2}}{\sqrt{6} + \sqrt{2}} \\[5ex] (\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2}) = (\sqrt{6})^2 - (\sqrt{2})^2 ... Difference \:\:of\:\: Two\:\: Squares \\[3ex] = \dfrac{4(\sqrt{6} + \sqrt{2})}{(\sqrt{6})^2 - (\sqrt{2})^2} \\[5ex] = \dfrac{4(\sqrt{6} + \sqrt{2})}{6 - 2} \\[5ex] = \dfrac{4(\sqrt{6} + \sqrt{2})}{4} \\[5ex] = \sqrt{6} + \sqrt{2}$
(84.)


(85.)


(86.) JAMB If $\cot\theta = \dfrac{8}{15}$, where $0\lt\theta\lt90^\circ$, find $\sin\theta$

$A.\;\; \dfrac{8}{17} \\[5ex] B.\;\; \dfrac{15}{17} \\[5ex] C.\;\; \dfrac{16}{17} \\[5ex] D.\;\; \dfrac{13}{15} \\[5ex]$

$\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{8}{15} ... Reciprocal\:\: Identity \\[5ex] \dfrac{\tan\theta}{1} = \dfrac{15}{8} \\[5ex] \tan\theta = \dfrac{opp}{adj} = \dfrac{15}{8} ... SOHCAHTOA \\[5ex] opp = 15 \\[3ex] adj = 8 \\[3ex] hyp^2 = opp^2 + adj^2 ... Pythagorean\:\: Theorem \\[3ex] hyp^2 = 15^2 + 8^2 \\[3ex] hyp^2 = 225 + 64 = 289 \\[3ex] hyp = \sqrt{289} \\[3ex] hyp = 17 \\[3ex] \sin\theta = \dfrac{opp}{hyp} = \dfrac{15}{17} ... SOHCAHTOA$
(87.)


(88.) ACT What angle measure, in radians, is equal to 30°?

$F.\;\; \dfrac{\pi}{6} \\[5ex] G.\;\; \dfrac{\pi}{5} \\[5ex] H.\;\; \dfrac{\pi}{3} \\[5ex] J.\;\; \dfrac{\pi}{2} \\[5ex] K.\;\; \dfrac{5\pi}{6} \\[5ex]$

$180^\circ = \pi\;radians \\[3ex] 30^\circ \\[3ex] = \dfrac{180^\circ}{6} \\[5ex] = \dfrac{\pi}{6}\;radians$
(89.)


(90.) WASSCE:FM If $\cos \theta = x,\;\;\;0^\circ \lt \theta \lt 90^\circ$, find the value of $\tan^2\theta$

$A.\;\; \dfrac{\sqrt{1 - x^2}}{x} \\[5ex] B.\;\; \sqrt{1 - x^2} \\[3ex] C.\;\; \dfrac{1}{x^2} - 1 \\[5ex] D.\;\; \dfrac{1- x^2}{x} \\[5ex]$

$\tan^2\theta + 1 = \sec^2\theta ...Pythagorean\;\;Identities \\[3ex] But: \\[3ex] \sec\theta = \dfrac{1}{\cos\theta} ... Reciprocal\;\;Identities \\[3ex] So: \\[3ex] \sec\theta = \dfrac{1}{x} \\[5ex] \sec^2\theta = \left(\dfrac{1}{x}\right)^2 = \dfrac{1}{x^2} \\[5ex] \implies \\[3ex] \tan^2\theta + 1 = \dfrac{1}{x^2} \\[5ex] \tan^2\theta = \dfrac{1}{x^2} - 1$
(91.)


(92.) ACT Which of the following expressions is equal to $(\sin 60^\circ)(\cos 30^\circ) + (\cos 60^\circ)(\sin 30^\circ)$?

$F.\;\; \cos(60^\circ - 30^\circ) \\[3ex] G.\;\; \cos(60^\circ + 30^\circ) \\[3ex] H.\;\; \sin(60^\circ - 30^\circ) \\[3ex] J.\;\; \sin(60^\circ + 30^\circ) \\[3ex] K.\;\; \sin\left(\dfrac{60^\circ + 30^\circ}{2}\right) \\[5ex]$

$(\sin 60^\circ)(\cos 30^\circ) + (\cos 60^\circ)(\sin 30^\circ) \\[3ex] = \sin(60^\circ + 30^\circ) ... Sum\;\;Formulas$
(93.)


(94.) WASSCE:FM Given that $\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta$, evaluate $\sin(30^\circ + \theta) + \sin(30^\circ - \theta)$

$A.\;\; \sin\theta \\[3ex] B.\;\; \cos\theta \\[3ex] C.\;\; \sqrt{3}\sin\theta \\[3ex] D.\;\; \sqrt{3}\cos\theta \\[3ex]$

$\sin 30 = \dfrac{1}{2}...Special\;\;\angle \\[5ex] \cos 30 = \dfrac{\sqrt{3}}{2} ...Special\;\;\angle \\[5ex] \sin(30 + \theta) \\[3ex] = \sin 30 \cos\theta + \cos 30 \sin\theta ...Given \\[3ex] = \dfrac{1}{2} * \cos\theta + \dfrac{\sqrt{3}}{2} * \sin\theta \\[5ex] = \dfrac{1}{2}\cos\theta + \dfrac{\sqrt{3}}{2}\sin\theta \\[5ex] \sin(30 - \theta) \\[3ex] = \sin 30 \cos\theta - \cos 30 \sin\theta ...Given \\[3ex] = \dfrac{1}{2} * \cos\theta - \dfrac{\sqrt{3}}{2} * \sin\theta \\[5ex] = \dfrac{1}{2}\cos\theta - \dfrac{\sqrt{3}}{2}\sin\theta \\[5ex] \sin(30^\circ + \theta) + \sin(30^\circ - \theta) \\[3ex] = \dfrac{1}{2}\cos\theta + \dfrac{\sqrt{3}}{2}\sin\theta + \left(\dfrac{1}{2}\cos\theta - \dfrac{\sqrt{3}}{2}\sin\theta\right) \\[5ex] = \dfrac{1}{2}\cos\theta + \dfrac{\sqrt{3}}{2}\sin\theta + \dfrac{1}{2}\cos\theta - \dfrac{\sqrt{3}}{2}\sin\theta \\[5ex] = \dfrac{1}{2}\cos\theta + \dfrac{1}{2}\cos\theta + \dfrac{\sqrt{3}}{2}\sin\theta - \dfrac{\sqrt{3}}{2}\sin\theta \\[5ex] = \cos\theta$
(95.)


(96.)


(97.)


(98.) WASSCE Given that $\cos 30^\circ = \sin 60^\circ = \dfrac{\sqrt{3}}{2}$ and

$\sin 30^\circ = \cos 60^\circ = \dfrac{1}{2}$, evaluate $\dfrac{\tan 60^\circ - 1}{1 - \tan 30^\circ}$

$A.\;\; \sqrt{3} - 2 \\[3ex] B.\;\; 2 - \sqrt{3} \\[3ex] C.\;\; \sqrt{3} \\[3ex] D.\;\; -2 \\[3ex]$

$\tan 60 \\[3ex] = \dfrac{\sin 60}{\cos 60} ...Quotient\;\;Identities \\[5ex] = \sin 60 \div \cos 60 \\[3ex] = \dfrac{\sqrt{3}}{2} \div \dfrac{1}{2} \\[5ex] = \dfrac{\sqrt{3}}{2} * \dfrac{2}{1} \\[5ex] = \sqrt{3} \\[3ex] \tan 30 \\[3ex] = \dfrac{\sin 30}{\cos 30} ...Quotient\;\;Identities \\[5ex] = \sin 30 \div \cos 30 \\[3ex] = \dfrac{1}{2} \div \dfrac{\sqrt{3}}{2} \\[5ex] = \dfrac{1}{2} * \dfrac{2}{\sqrt{3}} \\[5ex] = \dfrac{1}{\sqrt{3}} \\[5ex] \underline{Numerator} \\[3ex] \tan 60 - 1 \\[3ex] = \sqrt{3} - 1 \\[3ex] \underline{Denominator} \\[3ex] 1 - \tan 30 \\[3ex] = 1 - \dfrac{1}{\sqrt{3}} \\[5ex] = \dfrac{\sqrt{3}}{\sqrt{3}} - \dfrac{1}{\sqrt{3}} \\[5ex] = \dfrac{\sqrt{3} - 1}{\sqrt{3}} \\[5ex] \implies \\[3ex] \dfrac{\tan 60^\circ - 1}{1 - \tan 30^\circ} \\[5ex] = (\tan 60^\circ - 1) \div (1 - \tan 30^\circ) \\[3ex] = (\sqrt{3} - 1) \div \left(\dfrac{\sqrt{3} - 1}{\sqrt{3}}\right) \\[5ex] = \dfrac{\sqrt{3} - 1}{1} * \dfrac{\sqrt{3}}{\sqrt{3} - 1} \\[5ex] = \sqrt{3}$
(99.)


(100.)