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# Solved Examples on Trigonometric Functions and Graphs

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 and CMAT 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 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
Show all work

(1.) ACT What is the period of the function $f(x) = \csc(4x)$?

$f(x) = A \:\:\boldsymbol{function}\:\: [B(x \pm C)] \pm D...First\:\: General\:\: Form \\[3ex] f(x) = A \csc [B(x \pm C)] \pm D \\[3ex] f(x) = \csc(4x) \\[3ex] f(x) = \csc[4(x)] \\[3ex] B = 4 \\[3ex] \omega = \dfrac{2\pi}{|B|} \\[5ex] \omega = \dfrac{2\pi}{|4|} \\[5ex] \omega = \dfrac{2\pi}{4} \\[5ex] \omega = \dfrac{\pi}{2}$
(2.) ACT What is the amplitude of the graph of the equation $y + 2 = 3\sin(4\theta)$?
(Note: The amplitude is $\dfrac{1}{2}$ the difference between the maximum and the minimum values of $y$.)

$f(x) = A \:\:\boldsymbol{function}\:\: [B(x \pm C)] \pm D...General\:\: Form \\[3ex] f(x) = A \sin [B(x \pm C)] \pm D \\[3ex] y + 2 = 3\sin(4\theta) \\[3ex] y = 3\sin(4\theta) - 2 \\[3ex] y = 3\sin[4(\theta)] - 2 \\[3ex] A = 3$
(3.) (a.) Graph two full periods of the function $f(x) = \cos (6x)$

State the:
(b.) Amplitude
(c.) Period

$f(x) = A \:\:\boldsymbol{function}\:\: [B(x \pm C)] \pm D...First\:\: General\:\: Form \\[3ex] f(x) = A \cos [B(x \pm C)] \pm D \\[3ex] f(x) = \cos(6x) \\[3ex] f(x) = 1 * \cos[6(x)] \\[3ex] (b.) \\[3ex] Amplitude = |A| = |1| = 1 \\[3ex] (c.) \\[3ex] B = 6 \\[3ex] \omega = \dfrac{2\pi}{|B|} \\[5ex] \omega = \dfrac{2\pi}{|6|} \\[5ex] \omega = \dfrac{2\pi}{6} \\[5ex] \omega = \dfrac{\pi}{3} \\[3ex] f(x) = \cos(6x) \\[3ex]$ (a.)
(i.) Amplitude = 1

(ii.) Period = $\dfrac{\pi}{3}$

(iii.) C = 0...implies that there is no phase shift

(iv.) D = 0...implies that the midline is $y = 0$ and there is no vertical shift

(v.) The local minimum will occur 1 unit below the midline at $x = \dfrac{\pi}{6}$

(vi.) The local maximum will occur 1 unit above the midline at $x = \dfrac{\pi}{3}$

Based on these information, the correct graph is:
(4.) ACT The domain of the function $y(x) = 3\cos(5x - 4) + 1$ is all real numbers.
Which of the following is the range of the function $y(x)$?

$A.\:\: -3 \le y(x) \le 3 \\[3ex] B.\:\: -4 \le y(x) \le 3 \\[3ex] C.\:\: -4 \le y(x) \le 2 \\[3ex] D.\:\: -2 \le y(x) \le 4 \\[3ex] E.\:\: All\:\: real\:\: numbers \\[3ex]$

The cosine function is an even function
This means that $\cos (-x) = \cos(x)$

$y(x) = 3\cos(5x - 4) + 1 \\[3ex] For\:\: \cos(5x - 4) \\[3ex] Minimum = \cos 180 = -1 \\[3ex] Maximum = \cos 360 = 1 \\[3ex] For\:\: y(x) = 3\cos(5x - 4) + 1 \\[3ex] Minimum\:\: y(x) = 3(-1) + 1 = -3 + 1 = -2 \\[3ex] Maximum\:\: y(x) = 3(1) + 1 = 3 + 1 = 4 \\[3ex] \therefore -2 \le y(x) \le 4$
(5.) $\dfrac{1}{1 + \cos \theta} + \dfrac{1}{1 - \cos \theta}$

$\dfrac{1}{1 + \cos \theta} + \dfrac{1}{1 - \cos \theta} \\[5ex] LCD = (1 + \cos \theta)(1 - \cos \theta) \\[5ex] = \dfrac{1(1 - \cos \theta) + 1(1 + \cos \theta)}{(1 + \cos \theta)(1 - \cos \theta)} \\[5ex] = \dfrac{1 - \cos \theta + 1 + \cos \theta}{(1 + \cos \theta)(1 - \cos \theta)} \\[5ex] = \dfrac{2}{(1 + \cos \theta)(1 - \cos \theta)} \\[5ex] (1 + \cos \theta)(1 - \cos \theta) = 1^2 - \cos^2 \theta = 1 - \cos^2 \theta ...\: Difference\: of\: Two\: Squares \\[3ex] = \dfrac{2}{1 - \cos^2 \theta} \\[5ex] 1 - \cos^2 \theta = \sin^2 \theta ...\:Pythagorean\: Identity \\[3ex] = \dfrac{2}{\sin^2 \theta} \\[5ex] \dfrac{1}{\sin \theta} = \csc \theta ...\:Reciprocal\: Identity \\[5ex] = 2 * \dfrac{1}{\sin \theta} * \dfrac{1}{\sin \theta} \\[5ex] = 2 * \csc \theta * \csc \theta \\[3ex] = 2\csc^2 \theta$
(6.) $\dfrac{3\sin^2 \theta + \sin \theta - 10}{\sin \theta + 2}$

$Let\: p = \sin \theta \\[3ex] \dfrac{3\sin^2 \theta + \sin \theta - 10}{\sin \theta + 2} \\[5ex] = \dfrac{3p^2 + p - 10}{p + 2} \\[5ex] = \dfrac{(p + 2)(3p - 5)}{p + 2} \\[5ex] = 3p - 5 \\[3ex] = 3\sin \theta - 5$
(7.) $\dfrac{\cos \theta}{1 + \csc \theta} * \dfrac{1 - \csc \theta}{1 - \csc \theta}$

$\dfrac{\cos \theta}{1 + \csc \theta} * \dfrac{1 - \csc \theta}{1 - \csc \theta} \\[5ex] = \dfrac{\cos \theta(1 - \csc \theta)}{(1 + \csc \theta)(1 - \csc \theta)} \\[5ex] (1 + \csc \theta)(1 - \csc \theta) = 1^2 - \csc^2 \theta ... \:Difference\: of \:two \:Squares \\[3ex] = \dfrac{\cos \theta - \cos \theta \csc \theta}{1^2 - \csc^2 \theta} \\[5ex] 1^2 - \csc^2 \theta = 1 - \csc^2 \theta \\[3ex] 1 - \csc^2 \theta = -\cot^2 \theta ... \:Pythagorean\: Identity \\[3ex] = \dfrac{\cos \theta - \cos \theta \csc \theta}{-\cot^2 \theta} \\[5ex] \csc \theta = \dfrac{1}{\sin \theta} ... \:Reciprocal\: Identity \\[5ex] \tan \theta = \dfrac{1}{\cot \theta} ... \:From \:Reciprocal\: Identity \\[5ex] \cot \theta = \dfrac{\cos \theta}{\sin \theta} ... \:Quotient\: Identity \\[5ex] = \dfrac{\cos \theta - \cos \theta * \dfrac{1}{\sin \theta}}{-\cot^2 \theta} \\[5ex] = \dfrac{\cos \theta - {\dfrac{\cos \theta}{\sin \theta}}}{-\cot^2 \theta} \\[5ex] = \dfrac{\cos \theta - \cot \theta}{-\cot^2 \theta} \\[5ex] = \dfrac{\cos \theta}{-\cot^2 \theta} - \dfrac{\cot \theta}{-\cot^2 \theta} \\[5ex] = \dfrac{-\cos \theta}{\cot^2 \theta} - \dfrac{\cot \theta}{-\cot^2 \theta} \\[5ex] = (-\cos \theta \div \cot^2 \theta) + \dfrac{1}{\cot \theta} \\[5ex] = (-\cos \theta \div \dfrac{\cos^2 \theta}{\sin^2 \theta}) + \tan \theta \\[5ex] = (-\cos \theta * \dfrac{\sin^2 \theta}{\cos^2 \theta}) + \tan \theta \\[5ex] = \dfrac{-\sin^2 \theta}{\cos \theta} + \tan \theta \\[5ex] = -\sin \theta * \dfrac{\sin \theta}{\cos \theta} + \tan \theta \\[5ex] = -\sin \theta \tan \theta + \tan \theta \\[3ex] = \tan \theta - \sin \theta \tan \theta \\[3ex] = \tan \theta(1 - \sin \theta)$
(8.) ACT For all real values of $x$, which of the following equations is true?

$F.\:\: \sin(7x) + \cos(7x) = 7 \\[3ex] G.\:\: \sin(7x) + \cos(7x) = 1 \\[3ex] H.\:\: 7\sin(7x) + 7\cos(7x) = 14 \\[3ex] J.\:\: \sin^2(7x) + \cos^2(7x) = 7 \\[3ex] K.\:\: \sin^2(7x) + \cos^2(7x) = 1$

$F.\:\: -\sin(-x^\circ) = -1 * \sin(-x) \\[3ex] \sin(-x) = -\sin x ...Odd\:\: Identity \\[3ex] = -1 * -\sin x = \sin x ...YES \\[5ex] G.\:\: \sin(-x^\circ) \\[3ex] \sin(-x) = -\sin x ...Odd\:\: Identity \\[3ex] -\sin x \ne \sin x ...NO \\[5ex] H.\:\: \cos(90 - x)^\circ \\[3ex] \cos(90 - x) = \sin x ...Cofunction\:\: Identity \\[3ex] YES \\[5ex] J.\:\: \cos(x - 90)^\circ \\[3ex] -(x - 90) = -x + 90 = 90 - x \\[3ex] \cos(x) = \cos(-x) ...Even\:\: Identity \\[3ex] \cos(x - 90) = \cos(90 - x) ...Even\:\: Identity \\[3ex] = \sin x ...YES \\[5ex] K.\:\: \sqrt{1 - (\cos x^\circ)^2} \\[3ex] (\cos x)^2 = \cos^2 x \\[3ex] 1 - \cos^2 x = \sin^2 x ...Pythagorean\:\: Identity \\[3ex] = \sqrt{\sin^2 x} = \sin x ...YES$