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Solved Examples on Trigonometric Expressions

Samuel Dominic Chukwuemeka (SamDom For Peace) Verify your answers with these Calculators as applicable.

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Simplify the trigonometric expressions completely
State the reasons for every step
Show all work
There should be no negative angles

(1.) ACT For all $x$ such that $0 \le x \le 90$, which of the following expressions is NOT equal to $\sin x^\circ$?

$ F.\:\: -\sin(-x^\circ) \\[3ex] G.\:\: \sin(-x^\circ) \\[3ex] H.\:\: \cos(90 - x)^\circ \\[3ex] J.\:\: \cos(x - 90)^\circ \\[3ex] K.\:\: \sqrt{1 - (\cos x^\circ)^2} $


$ 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 $
(2.) ACT The expression $(180 - x)$ is the degree measure of a nonzero acute angle if and only if:

$ A.\;\; 0 \lt x \lt 45 \\[3ex] B.\;\; 0 \lt x \lt 90 \\[3ex] C.\;\; 45 \lt x \lt 90 \\[3ex] D.\;\; 90 \lt x \lt 135 \\[3ex] E.\;\; 90 \lt x \lt 180 \\[3ex] $

Second Quadrant Identities or Identities of Supplements

$90 \lt \theta \lt 180$ ... Angle in Degrees
Reference Angle = $180 - \theta$ ... Angle in Degrees

The expression $(180 - x)$ is the degree measure of a nonzero acute angle if and only if: $90 \lt x \lt 180$
(3.) $(\sin \theta + \csc \theta)(\sin \theta - \csc \theta)$


$ (\sin \theta + \csc \theta)(\sin \theta - \csc \theta) \\[3ex] = \sin^2 \theta - \csc^2 \theta ... Difference\:\: of\:\: Two\:\: Squares $
(4.) $\cos \theta \sin \theta(\tan \theta + \csc \theta)$


$ \cos \theta \sin \theta(\tan \theta + \csc \theta) \\[3ex] = \cos \theta \sin \theta \tan \theta + \cos \theta \sin \theta \csc \theta \\[3ex] But;\:\: \tan \theta = \dfrac{\sin \theta}{\cos \theta} ... \:Quotient \:Identity \\[5ex] \csc \theta = \dfrac{1}{\sin \theta} ... \:Reciprocal \:Identity \\[5ex] Substitute \:these\: identities \\[3ex] = \cos \theta \sin \theta * \dfrac{\sin \theta}{\cos \theta} + \cos \theta \sin \theta * \dfrac{1}{\sin \theta} \\[5ex] = \sin \theta \sin \theta + \cos \theta \\[3ex] = \sin^2 \theta + \cos \theta $
(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.) $\dfrac{\sin^2 \theta - 25}{2\cos \theta + 1} * \dfrac{4\cos \theta + 2}{5\sin \theta + 25}$


$ \dfrac{\sin^2 \theta - 25}{2\cos \theta + 1} * \dfrac{4\cos \theta + 2}{5\sin \theta + 25} \\[5ex] \sin^2 \theta - 25 = \sin^2 \theta - 5^2 \\[3ex] \sin^2 \theta - 5^2 = (\sin \theta + 5)(\sin \theta - 5) ... \:Difference \:of \:Two \:Squares \\[3ex] 4\cos \theta + 2 = 2(2\cos \theta + 1) \\[3ex] 5\sin \theta + 25 = 5(\sin \theta + 5) \\[3ex] = \dfrac{(\sin \theta + 5)(\sin \theta - 5)}{2\cos \theta + 1} * \dfrac{2(2\cos \theta + 1)}{5(\sin \theta + 5)} \\[5ex] = \dfrac{1 * (\sin \theta - 5)}{1} * \dfrac{2 * 1}{5 * 1} \\[5ex] = \dfrac{2(\sin \theta - 5)}{5} $
(9.) ACT Which of the following is equivalent to $\dfrac{1 - \cos^2 \theta}{\cos^2 \theta}$

$ F.\:\: \sec^2 \theta \\[3ex] G.\:\: (\csc^2 \theta) - 1 \\[3ex] H.\:\: \tan^2 \theta \\[3ex] J.\:\: \sin^2 \theta \\[3ex] K.\:\: -\dfrac{1}{\sin^2 \theta} $


We can solve this question in two ways.
Use whatever way you prefer

$ \underline{First\:\: Method} \\[3ex] \dfrac{1 - \cos^2 \theta}{\cos^2 \theta} \\[5ex] \sin^2 \theta + \cos^2 \theta = 1 ...Pythagorean\:\: Identity \\[3ex] \implies \sin^2 \theta = 1 - \cos^2 \theta \\[3ex] = \dfrac{\sin^2 \theta}{\cos^2 \theta} \\[5ex] \tan \theta = \dfrac{\sin \theta}{\cos \theta} \\[5ex] = \left(\dfrac{\sin \theta}{\cos \theta}\right)^2 \\[5ex] = \tan^2 \theta \\[3ex] \underline{Second\:\: Method} \\[3ex] \dfrac{1 - \cos^2 \theta}{\cos^2 \theta} \\[5ex] = \dfrac{1}{\cos^2\theta} - \dfrac{\cos^2\theta}{\cos^2\theta} \\[5ex] \dfrac{1}{\cos\theta} = \sec\theta \\[5ex] \rightarrow \dfrac{1}{\cos^2\theta} = \sec^2\theta \\[5ex] = \sec^2\theta - 1 \\[3ex] 1 + \tan^2\theta = \sec^2\theta ... Pythagorean\:\: Identity \\[3ex] \tan^2\theta = \sec^2\theta - 1 \\[3ex] = \tan^2\theta $
(10.) $\dfrac{\sin \theta}{1 - \cos \theta} * \dfrac{1 + \cos \theta}{1 + \cos \theta}$


$ \dfrac{\sin \theta}{1 - \cos \theta} * \dfrac{1 + \cos \theta}{1 + \cos \theta} \\[5ex] = \dfrac{\sin \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] 1 - \cos^2 \theta = \sin^2 \theta ... \:Pythagorean\: Identity \\[3ex] = \dfrac{\sin \theta(1 + \cos \theta)}{\sin^2 \theta} \\[5ex] = \dfrac{1 + \cos \theta}{\sin \theta} \\[5ex] = \dfrac{1}{\sin \theta} + \dfrac{\cos \theta}{\sin \theta} \\[5ex] \csc \theta = \dfrac{1}{\sin \theta} ... \:Reciprocal\: Identity \\[5ex] \cot \theta = \dfrac{\cos \theta}{\sin \theta} ... \:Quotient\: Identity \\[5ex] = \csc \theta + \cot \theta $
(11.) $2\sin^3\omega\cos\omega + 2\sin\omega\cos^3\omega$


$ 2\sin^3\omega\cos\omega + 2\sin\omega\cos^3\omega \\[3ex] = 2\sin\omega\cos\omega(\sin^2\omega + \cos^2\omega) \\[3ex] \sin^2\omega + \cos^2\omega ... Pythagorean\:\: Identity \\[3ex] 2\sin\omega\cos\omega = \cos2\omega ... Double-Angle\:\: Formula \\[3ex] = \cos2\omega * 1 \\[3ex] = \cos2\omega $
(12.) $\dfrac{(\cos \theta - \sin \theta)(\cos \theta - \sin \theta) - 1}{\sin \theta \cos \theta}$


$ \dfrac{(\cos \theta - \sin \theta)(\cos \theta - \sin \theta) - 1}{\sin \theta \cos \theta} \\[5ex] = \dfrac{(\cos^2 \theta - \sin \theta \cos \theta - \sin \theta \cos \theta + \sin^2 \theta) - 1}{\sin \theta \cos \theta} \\[5ex] = \dfrac{(\cos^2 \theta - 2\sin \theta \cos \theta + \sin^2 \theta) - 1}{\sin \theta \cos \theta} \\[5ex] = \dfrac{(\cos^2 \theta + \sin^2 \theta - 2\sin \theta \cos \theta) - 1}{\sin \theta \cos \theta} \\[5ex] \cos^2 \theta + \sin^2 \theta = 1 ... \:Pythagorean\: Identity \\[3ex] = \dfrac{(1 - 2\sin \theta \cos \theta) - 1}{\sin \theta \cos \theta} \\[5ex] = \dfrac{1 - 2\sin \theta \cos \theta - 1}{\sin \theta \cos \theta} \\[5ex] = \dfrac{- 2\sin \theta \cos \theta}{\sin \theta \cos \theta} \\[5ex] = -2 $
(13.) $\dfrac{8\cot \theta \csc \theta - 2\csc \theta}{4\cot \theta \csc \theta + 2\csc \theta}$


$ \dfrac{8\cot \theta \csc \theta - 2\csc \theta}{4\cot \theta \csc \theta + 2\csc \theta} \\[5ex] = \dfrac{2\csc \theta(4\cot \theta - 1)}{2\csc \theta(2\cot \theta + 1)} \\[5ex] = \dfrac{4\cot \theta - 1}{2\cot \theta + 1} $
(14.) ACT The expression $4 \sin x \cos x$ is equivalent to which of the following?
(Note: $\sin(x + y) = \sin x \cos y + \cos x \sin y$)

$ F.\:\: 2\sin 2x \\[3ex] G.\:\: 2\cos 2x \\[3ex] H.\:\: 2\sin 4x \\[3ex] J.\:\: 8\sin 2x \\[3ex] K.\:\: 8\cos 2x $


$ 4\sin x \cos x = 2\sin x \cos x + 2\sin x \cos x \\[3ex] = 2(\sin x \cos x + \sin x \cos x) \\[3ex] \sin(x + x) = \sin x \cos x + \cos x \sin x = \sin x \cos x + \sin x \cos x \\[3ex] = 2\sin(x + x) \\[3ex] = 2\sin 2x $
(15.) ACT Whenever $\dfrac{\tan \theta}{\sin \theta}$ is defined, it is equivalent to:

$ F.\:\: \cos \theta \\[3ex] G.\:\: \dfrac{1}{\cos \theta} \\[5ex] H.\:\: \dfrac{1}{\sin \theta} \\[5ex] J.\:\: \dfrac{1}{\sin^2 \theta} \\[5ex] K.\:\: \dfrac{\cos \theta}{\sin^2 \theta} $


$ \dfrac{\tan \theta}{\sin \theta} \\[5ex] = \tan \theta \div \sin \theta \\[3ex] = \tan \theta * \dfrac{1}{\sin \theta} \\[5ex] \tan \theta = \dfrac{\sin \theta}{\cos \theta} ...Reciprocal\:\: Identity \\[5ex] = \dfrac{\sin \theta}{\cos \theta} * \dfrac{1}{\sin \theta} \\[5ex] = \dfrac{1}{\cos \theta} \\[5ex] = \sec \theta $
(16.) $\sin \theta(\sin \theta - 2\cos \theta) + \cos^2 \theta$


$ \sin \theta(\sin \theta - 2\cos \theta) + \cos^2 \theta \\[3ex] = \sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta \\[3ex] = \sin^2 \theta + \cos^2 \theta - 2\sin \theta \cos \theta \\[3ex] \sin^2 \theta + \cos^2 \theta = 1 ... \:Pythagorean\: Identity \\[3ex] 2\sin \theta \cos \theta = \sin 2\theta ... \:Double-Angle\: Formulas \\[3ex] = 1 - \sin 2\theta $
(17.) $(\cos \theta + \sec \theta)(3\cos^2 \theta + 3\sec^2 \theta - 3)$


$ (\cos \theta + \sec \theta)(3\cos^2 \theta + 3\sec^2 \theta - 3) \\[3ex] = 3\cos^3 \theta + 3\cos \theta \sec^2 \theta - 3\cos \theta + 3\cos^2 \theta \sec \theta + 3\sec^3\theta -3\sec\theta \\[3ex] \sec\theta = \dfrac{1}{\cos\theta} ... Reciprocal\:\: Identity \\[3ex] 3\cos\theta\ sec^2\theta = 3 * \cos\theta * \sec\theta * \sec\theta = 3 * \cos\theta * \dfrac{1}{\cos\theta} * \sec\theta = 3\sec\theta \\[5ex] 3\cos^2 \theta \sec \theta = 3 * \cos\theta * \cos\theta * \dfrac{1}{\cos\theta} = 3\cos\theta \\[5ex] = 3\cos^3 \theta + 3\sec\theta - 3\cos \theta + 3\cos\theta + 3\sec^3\theta -3\sec\theta \\[3ex] = 3\cos^3 \theta + 3\sec^3\theta + 3\sec\theta - 3\sec\theta - 3\cos \theta + 3\cos\theta \\[3ex] = 3\cos^3\theta + 3\sec^3\theta $
(18.) $\tan^3 \theta + 27$


$ \tan^3 \theta + 27 \\[3ex] = \tan^3 \theta + 3^3 \\[3ex] x^3 +y^3 = (x + y)(x^2 - xy + y^2) ...\:Sum\: of\: Two\: Cubes \\[3ex] Compare\: and\: Substitute \\[3ex] x = \tan \theta \\[3ex] y = 3 \\[3ex] = (\tan \theta + 3)(\tan^2 \theta - 3\tan \theta + 3^2) \\[3ex] = (\tan \theta + 3)(\tan^2 \theta - 3\tan \theta + 9) $
(19.) Rationalize the numerator

$\sqrt{\dfrac{7 + 7\cos \theta}{7 - 7\cos \theta}}$


We want to rationalize the numerator
This means that we want to make the numerator a rational number
What do we have to multiply by $1$ (the exponent at the denominator is $1$) in terms of exponent so that it is divisible by $2$
Why $2$? ... it is because of the square root.
Remember: Square root means an exponent of $\dfrac{1}{2}$... division by $2$
$\sqrt{x} = x^\frac{1}{2}$
So, we have to multiply $7 + 7\cos \theta$ by $7 + 7\cos \theta$ so we can get the squared of it $(7 + 7\cos \theta)^2$ which is divisible by $2$
Whatever you do to the numerator, you have to do the same thing to the denominator
This is necessary in order to get an equivalent fraction to the original fraction
$ \sqrt{\dfrac{7 + 7\cos \theta}{7 - 7\cos \theta}} \\[5ex] = \dfrac{\sqrt{7 + 7\cos \theta}}{\sqrt{7 - 7\cos \theta}} \\[5ex] = \dfrac{\sqrt{7 + 7\cos \theta}}{\sqrt{7 - 7\cos \theta}} * \dfrac{\sqrt{7 + 7\cos \theta}}{\sqrt{7 + 7\cos \theta}} \\[5ex] = \dfrac{\sqrt{(7 + 7\cos \theta) * (7 + 7\cos \theta)}}{\sqrt{(7 - 7\cos \theta)* (7 + 7\cos \theta)}} \\[5ex] (7 - 7\cos \theta) * (7 + 7\cos \theta) = 7^2 - (7\cos \theta)^2 ...\:Difference\: of\: Two\: Squares \\[3ex] (7 - 7\cos \theta) * (7 + 7\cos \theta) = 49 - 49\cos^2 \theta \\[3ex] = \dfrac{\sqrt{(7 + 7\cos \theta)^2}}{\sqrt{49 - 49\cos^2 \theta}} \\[5ex] = \dfrac{7 + 7\cos \theta}{\sqrt{49(1 - \cos^2 \theta)}} \\[5ex] 1 - \cos^2 \theta = \sin^2 \theta ...\:Pythagorean\: Identity \\[3ex] = \dfrac{7(1 + \cos \theta)}{\sqrt{49\sin^2 \theta}} \\[5ex] = \dfrac{7(1 + \cos \theta)}{\sqrt{49} * \sqrt{\sin^2 \theta}} \\[5ex] = \dfrac{7(1 + \cos \theta)}{7\sin \theta} \\[5ex] = \dfrac{1 + \cos \theta}{\sin \theta} \\[5ex] $ The numerator is now a rational number.
(20.) $12\cos^2\alpha + \cos\alpha - 13$


This is a quadratic expression in $\cos\alpha$
Let us make this as simple as possible.
Let us use substitution
Then, try to use the Factoring method
Remember to substitute back

$ 12\cos^2\alpha + \cos\alpha - 13 \\[3ex] Let\:\: \cos\alpha = p \\[3ex] 12p^2 + p - 13 \\[3ex] 12p^2 + 13p - 12p - 13 \\[3ex] p(12p + 13) - 1(12p + 13) \\[3ex] (12p + 13)(p - 1) \\[3ex] Substitute\:\: back \\[3ex] (12\cos\alpha + 13)(\cos\alpha - 1) $




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(21.) $\dfrac{1}{\sin\omega - 1} - \dfrac{1}{\sin\omega + 1}$


$ \dfrac{1}{\sin\omega - 1} - \dfrac{1}{\sin\omega + 1} \\[5ex] = \dfrac{1(\sin\omega + 1) - 1(\sin\omega - 1)}{(\sin\omega - 1)(\sin\omega + 1)} \\[5ex] (\sin\omega - 1)(\sin\omega + 1) = \sin^2\omega - 1^2 ... Difference\:\: of\:\: Two\:\: Squares \\[3ex] = \dfrac{\sin\omega + 1 - \sin\omega + 1}{\sin^2\omega - 1^2} \\[5ex] = \dfrac{2}{\sin^2\omega - 1} \\[5ex] \sin^2\omega + \cos^2\omega = 1 ... Pythagorean\:\: Identity \\[3ex] \rightarrow \sin^2\omega - 1 = -\cos^2\omega \\[3ex] = \dfrac{2}{-\cos^2\omega} \\[5ex] = \dfrac{2}{-1 * \cos^2\omega} \\[5ex] \dfrac{1}{\cos\omega} = \sec\omega \\[5ex] \rightarrow \dfrac{1^2}{\cos^2\omega} = \dfrac{1}{\cos^2\omega} = \sec^2\omega \\[5ex] = 2 * -1 * \sec^2\omega \\[3ex] = -2\sec^2\omega $
(22.) Rationalize the numerator

$\sqrt{\dfrac{\sin \theta}{\cos^7 \theta}}$


We want to rationalize the numerator
This means that we want to make the numerator a rational number
What do we have to multiply by $1$ (the exponent at the numerator is $1$) in terms of exponent so that it is divisible by $2$
Why $2$? ... it is because of the square root.
Remember: Square root means an exponent of $\dfrac{1}{2}$... division by $2$
$\sqrt{x} = x^\frac{1}{2}$
So, we have to multiply $\sin \theta$ by $\sin \theta$ so we can get $\sin^2 \theta$ because $2$ is divisible by $2$
Whatever you do to the numerator, you have to do the same thing to the denominator
This is necessary in order to get an equivalent fraction to the original fraction
$ \sqrt{\dfrac{\sin \theta}{\cos^7 \theta}} \\[5ex] = \dfrac{\sqrt{\sin \theta}}{\sqrt{\cos^7 \theta}} \\[5ex] = \dfrac{\sqrt{\sin \theta}}{\sqrt{\cos^7 \theta}} * \dfrac{\sqrt{\sin \theta}}{\sqrt{\sin \theta}} \\[5ex] = \dfrac{\sqrt{\sin \theta * \sin \theta}}{\sqrt{\cos^7 \theta * \sin \theta}} \\[5ex] = \dfrac{\sqrt{\sin^2 \theta}}{\sqrt{\cos^6 \theta * \cos \theta * \sin \theta}} \\[5ex] = \dfrac{\sin \theta}{\cos^3 \theta \sqrt{\cos \theta \sin \theta}} \\[5ex] $ The numerator is now a rational number.
(23.) Rationalize the denominator

$\sqrt{\dfrac{7 + 7\cos \theta}{7 - 7\cos \theta}}$


We want to rationalize the denominator
This means that we want to make the denominator a rational number
What do we have to multiply by $1$ (the exponent at the denominator is $1$) in terms of exponent so that it is divisible by $2$
Why $2$? ... it is because of the square root.
Remember: Square root means an exponent of $\dfrac{1}{2}$... division by $2$
$\sqrt{x} = x^\frac{1}{2}$
So, we have to multiply $7 - 7\cos \theta$ by $7 - 7\cos \theta$ so we can get the squared of it $(7 - 7\cos \theta)^2$ which is divisible by $2$
Whatever you do to the denominator, you have to do the same thing to the numerator
This is necessary in order to get an equivalent fraction to the original fraction
$ \sqrt{\dfrac{7 + 7\cos \theta}{7 - 7\cos \theta}} \\[5ex] = \dfrac{\sqrt{7 + 7\cos \theta}}{\sqrt{7 - 7\cos \theta}} \\[5ex] = \dfrac{\sqrt{7 + 7\cos \theta}}{\sqrt{7 - 7\cos \theta}} * \dfrac{\sqrt{7 - 7\cos \theta}}{\sqrt{7 - 7\cos \theta}} \\[5ex] = \dfrac{\sqrt{(7 + 7\cos \theta) * (7 - 7\cos \theta)}}{\sqrt{(7 - 7\cos \theta)* (7 - 7\cos \theta)}} \\[5ex] (7 + 7\cos \theta) * (7 - 7\cos \theta) = 7^2 - (7\cos \theta)^2 ...\:Difference\: of\: Two\: Squares \\[3ex] (7 + 7\cos \theta) * (7 - 7\cos \theta) = 49 - 49\cos^2 \theta \\[3ex] = \dfrac{\sqrt{(49 - 49\cos^2 \theta)}}{\sqrt{(7 - 7\cos \theta)^2}} \\[5ex] = \dfrac{\sqrt{(49(1 - \cos^2 \theta)}}{(7 - 7\cos \theta)} \\[5ex] 1 - \cos^2 \theta = \sin^2 \theta ...\:Pythagorean\: Identity \\[3ex] = \dfrac{\sqrt{49\sin^2 \theta}}{7(1 - \cos \theta)} \\[5ex] = \dfrac{\sqrt{49} * \sqrt{\sin^2 \theta}}{7(1 - \cos \theta)} \\[5ex] = \dfrac{7\sin \theta}{7(1 - \cos \theta)} \\[5ex] = \dfrac{\sin \theta}{1 - \cos \theta} \\[5ex] $ The denominator is now a rational number.
(24.) $\dfrac{3 \sec \theta}{\csc \theta} + \dfrac{4\sin \theta}{\cos \theta}$


$ \dfrac{3 \sec \theta}{\csc \theta} + \dfrac{4\sin \theta}{\cos \theta} \\[5ex] = \dfrac{3\sec \theta \cos \theta + 4\sin \theta \csc \theta}{\csc \theta \cos \theta} \\[5ex] \sec \theta = \dfrac{1}{\cos \theta} ... \: Reciprocal\: Identity \\[5ex] \csc \theta = \dfrac{1}{\sin \theta} ... \: Reciprocal\: Identity \\[5ex] = \dfrac{3 * \dfrac{1}{\cos \theta} * \cos \theta + 4 * \sin \theta * \dfrac{1}{\sin \theta}}{\dfrac{1}{\sin \theta} \cos \theta} \\[7ex] \tan \theta = \dfrac{\sin \theta}{\cos \theta} ... \: Quotient\: Identity \\[5ex] \cot \theta = \dfrac{\cos \theta}{\sin \theta} ... \: Quotient\: Identity \\[5ex] = \dfrac{7}{\dfrac{\cos \theta}{\sin \theta}} \\[7ex] = \dfrac{7}{\cot \theta} \\[5ex] \tan \theta = \dfrac{1}{\cot \theta} ... \:Reciprocal\: Identity \\[5ex] = 7\tan \theta $
(25.) $\cos3\omega + \cos6\omega$


$ \cos3\omega + \cos6\omega \\[3ex] = 2\cos\left(\dfrac{3\omega + 6\omega}{2}\right)\cos\left(\dfrac{3\omega - 6\omega}{2}\right) ...Sum-to-Product\:\: Formula \\[5ex] = 2\cos\left(\dfrac{9\omega}{2}\right)\cos\left(\dfrac{-3\omega}{2}\right) \\[5ex] \cos(-\omega) = \cos(\omega) ... Even\:\: Identity \\[3ex] = 2\cos\left(\dfrac{9\omega}{2}\right)\cos\left(\dfrac{3\omega}{2}\right) $
(26.) $\cos(-\gamma) - \cos\gamma$


$ \cos(-\gamma) - \cos\gamma \\[3ex] \cos(-\gamma) = \cos\gamma ... Even\:\: Identity \\[3ex] = \cos\gamma - \cos\gamma \\[3ex] = 0 $
(27.) $\cos(\alpha + \beta) + \cos(\alpha - \beta)$


$ \cos(\alpha + \beta) + \cos(\alpha - \beta) \\[3ex] \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta ... Sum\:\: Formula \\[3ex] \cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta ... Difference\:\: Formula \\[3ex] = \cos\alpha\cos\beta - \sin\alpha\sin\beta + \cos\alpha\cos\beta + \sin\alpha\sin\beta \\[3ex] = 2\cos\alpha\cos\beta $
(28.) $\dfrac{\sin^5\theta\cos\theta}{\cos^5\theta\sin\theta}$


$ \dfrac{\sin^5\theta\cos\theta}{\cos^5\theta\sin\theta} \\[5ex] = \dfrac{\sin^4\theta}{\cos^4\theta} \\[5ex] = \tan^4\theta $
(29.) $(3\cos\beta - 7\sin\beta)^2 + (7\cos\beta + 3\sin\beta)^2$


$ (3\cos\beta - 7\sin\beta)^2 + (7\cos\beta + 3\sin\beta)^2 \\[3ex] \underline{augend} \\[3ex] (3\cos\beta - 7\sin\beta)^2 = (3\cos\beta - 7\sin\beta)(3\cos\beta - 7\sin\beta) \\[3ex] = 9\cos^2\beta - 21\sin\beta\cos\beta - 21\sin\beta\cos\beta + 49\sin^2\beta \\[3ex] = 9\cos^2\beta - 42\sin\beta\cos\beta + 49\sin^2\beta \\[3ex] \underline{added} \\[3ex] (7\cos\beta + 3\sin\beta)^2 = (7\cos\beta + 3\sin\beta)(7\cos\beta + 3\sin\beta) \\[3ex] = 49\cos^2\beta + 21\sin\beta\cos\beta + 21\sin\beta\cos\beta + 9\sin^2\beta \\[3ex] = 49\cos^2\beta + 42\sin\beta\cos\beta + 9\sin^2\beta \\[3ex] \underline{sum} \\[3ex] = 9\cos^2\beta - 42\sin\beta\cos\beta + 49\sin^2\beta + 49\cos^2\beta + 42\sin\beta\cos\beta + 9\sin^2\beta \\[3ex] = 58\cos^2\beta + 58\sin^2\beta \\[3ex] = 58\sin^2\beta + 58\cos^2\beta \\[3ex] = 58(\sin^2\beta + \cos^2\beta) \\[3ex] \sin^2\beta + \cos^2\beta = 1 ... Pythagorean\:\: Identity \\[3ex] = 58 * 1 \\[3ex] = 58 $
(30.) $\dfrac{3\sec\gamma}{\csc\gamma} + \dfrac{4\sin\gamma}{\cos\gamma}$


$ \dfrac{3\sec\gamma}{\csc\gamma} + \dfrac{4\sin\gamma}{\cos\gamma} \\[5ex] = \dfrac{3\sec\gamma\cos\gamma}{\csc\gamma\cos\gamma} + \dfrac{4\sin\gamma\csc\gamma}{\csc\gamma\cos\gamma} \\[5ex] = \dfrac{3\sec\gamma\cos\gamma + 4\sin\gamma\csc\gamma}{\csc\gamma\cos\gamma} \\[5ex] \underline{Numerator} \\[3ex] \sec\gamma = \dfrac{1}{\cos\gamma} ... Reciprocal\:\: Identity \\[5ex] 3\sec\gamma\cos\gamma = 3 * \dfrac{1}{\cos\gamma} * \cos\gamma = 3 * 1 = 3 \\[5ex] \csc\gamma = \dfrac{1}{\sin\gamma} ... Reciprocal\:\: Identity \\[5ex] 4\sin\gamma\csc\gamma = 4 * \sin\gamma * \dfrac{1}{\sin\gamma} = 4 * 1 = 4 \\[5ex] 3 + 4 = 7 \\[3ex] \underline{Denominator} \\[3ex] \csc\gamma\cos\gamma = \dfrac{1}{\sin\gamma} * \cos\gamma = \dfrac{\cos\gamma}{\sin\gamma} \\[5ex] \dfrac{\cos\gamma}{\sin\gamma} = \cot\gamma ... Quotient\:\: Identity \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{7}{\cot\gamma} \\[5ex] = 7 * \dfrac{1}{\cot\gamma} \\[5ex] \dfrac{1}{\cot\gamma} = \tan\gamma ... Reciprocal\:\: Identity \\[5ex] = 7 * \tan\gamma \\[3ex] = 7\tan\gamma $
(31.) $\dfrac{5\sin\alpha}{\cos^2\alpha} * \dfrac{\cos^2\alpha - \cos\alpha\sin\alpha}{\cos^2\alpha - \sin^2\alpha}$


$ \dfrac{5\sin\alpha}{\cos^2\alpha} * \dfrac{\cos^2\alpha - \cos\alpha\sin\alpha}{\cos^2\alpha - \sin^2\alpha} \\[5ex] \cos^2\alpha - \cos\alpha\sin\alpha = \cos\alpha(\cos\alpha - \sin\alpha) ... Factor\:\: by\:\: GCF \\[3ex] \cos^2\alpha - \sin^2\alpha = (\cos\alpha + \sin\alpha)(\cos\alpha - \sin\alpha) ... Difference \:\:of\:\: Two\:\: Squares \\[3ex] = \dfrac{5\sin\alpha}{\cos^2\alpha} * \dfrac{\cos\alpha(\cos\alpha - \sin\alpha)}{(\cos\alpha + \sin\alpha)(\cos\alpha - \sin\alpha)} \\[5ex] = \dfrac{5\sin\alpha}{\cos\alpha} * \dfrac{1}{\cos\alpha + \sin\alpha} \\[5ex] \dfrac{\sin\alpha}{\cos\alpha} = \tan\alpha ... Quotient\:\: Identity \\[5ex] = 5\tan\alpha * \dfrac{1}{\cos\alpha + \sin\alpha} \\[5ex] = \dfrac{5\tan\alpha * 1}{\cos\alpha + \sin\alpha} \\[5ex] = \dfrac{5\tan\alpha}{\cos\alpha + \sin\alpha} $
(32.) $\sqrt{\sin^4\theta\cos^2\theta} * \sqrt{\cos^2\theta}$


$ \sqrt{\sin^4\theta\cos^2\theta} * \sqrt{\cos^2\theta} \\[5ex] \underline{multiplier} \\[3ex] \sqrt{\sin^4\theta\cos^2\theta} = \left(\sin^4\theta\right)^\dfrac{1}{2} * \left(\cos^2\theta\right)^\dfrac{1}{2} \\[5ex] = \sin^2\theta * \cos\theta \\[3ex] \underline{multiplicand} \\[3ex] \sqrt{\cos^2\theta} = \left(\cos^2\theta\right)^\dfrac{1}{2} \\[5ex] = \cos\theta \\[3ex] \underline{product} \\[3ex] = \sin^2\theta * \cos\theta * \cos\theta \\[3ex] = \sin^2\theta\cos^2\theta $
(33.) For $\sqrt{p^2 - c^2}$, determine $\sin\beta$ if $c = p\cos\beta$


$ c = p\cos\beta \\[3ex] \sqrt{p^2 - c^2} = \sqrt{p^2 - (p\cos\beta)^2} \\[3ex] = \sqrt{p^2 - p^2\cos^2\beta} \\[3ex] = \sqrt{p^2(1 - \cos^2\beta)} \\[3ex] \sin^2\beta + \cos^2\beta = 1 ... Pythagorean\:\: Identity \\[3ex] \sin^2\beta = 1 - \cos^2\beta \\[3ex] = \sqrt{p^2 * \sin^2\beta} \\[3ex] = p\sin\beta \\[3ex] \therefore \sqrt{p^2 - c^2} = p\sin\beta \\[3ex] p\sin\beta = \sqrt{p^2 - c^2} \\[3ex] \sin\beta = \dfrac{\sqrt{p^2 - c^2}}{p} $
(34.) $\sqrt{\dfrac{\sin^2\phi}{2\cos^2\phi}}$


$ \sqrt{\dfrac{\sin^2\phi}{2\cos^2\phi}} \\[5ex] = \dfrac{\sqrt{\sin^2\phi}}{\sqrt{2\cos^2\phi}} \\[5ex] = \dfrac{\sin\phi}{\sqrt{2}\cos\phi} \\[5ex] = \dfrac{\sin\phi}{\cos\phi} * \dfrac{1}{\sqrt{2}} \\[5ex] \dfrac{\sin\phi}{\cos\phi} = \tan\phi ... Quotient\:\: Identity \\[5ex] = \tan\phi * \dfrac{1}{\sqrt{2}} * \dfrac{\sqrt{2}}{\sqrt{2}} \\[5ex] = \dfrac{\sqrt{2}\tan\phi}{2} $
(35.) For $\sqrt{p^2 - c^2}$, determine $\tan\beta$ if $c = p\cos\beta$


$ c = p\cos\beta ...eqn.(1) \\[3ex] \sqrt{p^2 - c^2} = \sqrt{p^2 - (p\cos\beta)^2} \\[3ex] = \sqrt{p^2 - p^2\cos^2\beta} \\[3ex] = \sqrt{p^2(1 - \cos^2\beta)} \\[3ex] \sin^2\beta + \cos^2\beta = 1 ... Pythagorean\:\: Identity \\[3ex] \sin^2\beta = 1 - \cos^2\beta \\[3ex] = \sqrt{p^2 * \sin^2\beta} \\[3ex] = p\sin\beta \\[3ex] \therefore \sqrt{p^2 - c^2} = p\sin\beta \\[3ex] p\sin\beta = \sqrt{p^2 - c^2} \\[3ex] \sin\beta = \dfrac{\sqrt{p^2 - c^2}}{p} ...eqn.(2) \\[5ex] \tan\beta = \dfrac{\sin\beta}{\cos\beta} ... Quotient\:\: Identity \\[5ex] Divide\:\: both\:\: sides\:\: of\:\: eqn.(2) \:\:by\:\: \cos\beta \\[3ex] \dfrac{\sin\beta}{\cos\beta} = \dfrac{\sqrt{p^2 - c^2}}{p} \div \cos\beta \\[5ex] \tan\beta = \dfrac{\sqrt{p^2 - c^2}}{p} \div \cos\beta ...eqn.(3) \\[5ex] From\:\: eqn.(1) \\[3ex] p\cos\beta = c \\[3ex] \cos\beta = \dfrac{c}{p} \\[5ex] Substitute\:\: for\:\: \cos\beta \:\:in\:\: eqn. (3) \\[3ex] \tan\beta = \dfrac{\sqrt{p^2 - c^2}}{p} \div \dfrac{c}{p} \\[5ex] \tan\beta = \dfrac{\sqrt{p^2 - c^2}}{p} * \dfrac{p}{c} \\[5ex] \tan\beta = \dfrac{\sqrt{p^2 - c^2}}{c} $
(36.) $\dfrac{\sin\gamma + \cos\gamma}{\sin\gamma} + \dfrac{\sin\gamma - \cos\gamma}{\cos\gamma}$


$ \dfrac{\sin\gamma + \cos\gamma}{\sin\gamma} + \dfrac{\sin\gamma - \cos\gamma}{\cos\gamma} \\[5ex] = \dfrac{\cos\gamma(\sin\gamma + \cos\gamma) + \sin\gamma(\sin\gamma - \cos\gamma)}{\sin\gamma\cos\gamma} \\[5ex] = \dfrac{\cos\gamma\sin\gamma + \cos^2\gamma + \sin^\gamma - \sin\gamma\cos\gamma)}{\sin\gamma\cos\gamma} \\[5ex] \sin^2\gamma + \cos^2\gamma = 1 ... Pythagorean\:\: Identity \\[3ex] = \dfrac{1}{\sin\gamma\cos\gamma} \\[5ex] = \dfrac{1}{\sin\gamma} * \dfrac{1}{\cos\gamma} \\[5ex] \dfrac{1}{\sin\gamma} = \csc\gamma ... Reciprocal\:\: Identity \\[5ex] \dfrac{1}{\cos\gamma} = \sec\gamma ... Reciprocal\:\: Identity \\[5ex] = \csc\gamma\sec\gamma $
(37.) $\dfrac{1 + \tan\beta}{1 - \tan\beta} + \dfrac{1 + \cot\beta}{1 - \cot\beta}$


$ \dfrac{1 + \tan\beta}{1 - \tan\beta} + \dfrac{1 + \cot\beta}{1 - \cot\beta} \\[5ex] \underline{First\:\: term} \\[3ex] \dfrac{1 + \tan\beta}{1 - \tan\beta} \\[5ex] \tan\beta = \dfrac{\sin\beta}{\cos\beta} ...Quotient\:\: Identity \\[5ex] \underline{Numerator} \\[3ex] 1 + \tan\beta = 1 + \dfrac{\sin\beta}{\cos\beta} \\[5ex] = \dfrac{\cos\beta}{\cos\beta} + \dfrac{\sin\beta}{\cos\beta} \\[5ex] = \dfrac{\cos\beta + \sin\beta}{\cos\beta} \\[5ex] \underline{Denominator} \\[3ex] 1 - \tan\beta = 1 - \dfrac{\sin\beta}{\cos\beta} \\[5ex] = \dfrac{\cos\beta}{\cos\beta} - \dfrac{\sin\beta}{\cos\beta} \\[5ex] = \dfrac{\cos\beta - \sin\beta}{\cos\beta} \\[5ex] \dfrac{Numerator}{Denominator} \\[3ex] = (1 + \tan\beta) \div (1 - \tan\beta) \\[3ex] = \dfrac{\cos\beta + \sin\beta}{\cos\beta} \div \dfrac{\cos\beta - \sin\beta}{\cos\beta} \\[5ex] = \dfrac{\cos\beta + \sin\beta}{\cos\beta} * \dfrac{\cos\beta}{\cos\beta - \sin\beta} \\[5ex] = \dfrac{\cos\beta + \sin\beta}{\cos\beta - \sin\beta} \\[5ex] \underline{Second\:\: term} \\[3ex] \dfrac{1 + \cot\beta}{1 - \cot\beta} \\[5ex] \cot\beta = \dfrac{\cos\beta}{\sin\beta} ...Quotient\:\: Identity \\[5ex] \underline{Numerator} \\[3ex] 1 + \cot\beta = 1 + \dfrac{\cos\beta}{\sin\beta} \\[5ex] = \dfrac{\sin\beta}{\sin\beta} + \dfrac{\cos\beta}{\sin\beta} \\[5ex] = \dfrac{\sin\beta + \cos\beta}{\sin\beta} \\[5ex] \underline{Denominator} \\[3ex] 1 - \cot\beta = 1 - \dfrac{\cos\beta}{\sin\beta} \\[5ex] = \dfrac{\sin\beta}{\sin\beta} - \dfrac{\cos\beta}{\sin\beta} \\[5ex] = \dfrac{\sin\beta - \cos\beta}{\sin\beta} \\[5ex] \dfrac{Numerator}{Denominator} \\[3ex] = (1 + \cot\beta) \div (1 - \cot\beta) \\[3ex] = \dfrac{\sin\beta + \cos\beta}{\sin\beta} \div \dfrac{\sin\beta - \cos\beta}{\sin\beta} \\[5ex] = \dfrac{\sin\beta + \cos\beta}{\sin\beta} * \dfrac{\sin\beta}{\sin\beta - \cos\beta} \\[5ex] = \dfrac{\sin\beta + \cos\beta}{\sin\beta - \cos\beta} \\[5ex] \underline{First\:\: term\:\: + \:\:Second\:\: term} \\[3ex] \dfrac{1 + \tan\beta}{1 - \tan\beta} + \dfrac{1 + \cot\beta}{1 - \cot\beta} \\[5ex] = \dfrac{\cos\beta + \sin\beta}{\cos\beta - \sin\beta} + \dfrac{\sin\beta + \cos\beta}{\sin\beta - \cos\beta} \\[5ex] = \dfrac{(\cos\beta + \sin\beta)(\sin\beta - \cos\beta) + (\sin\beta + \cos\beta)(\cos\beta - \sin\beta)}{(\cos\beta - \sin\beta)(\sin\beta - \cos\beta)} \\[5ex] = \dfrac{\sin\beta\cos\beta - \cos^2\beta + \sin^2\beta - \sin\beta\cos\beta + \sin\beta\cos\beta - \sin^2\beta + \cos^2\beta - \sin\beta\cos\beta}{(\cos\beta - \sin\beta)(\sin\beta - \cos\beta)} \\[5ex] = \dfrac{0}{(\cos\beta - \sin\beta)(\sin\beta - \cos\beta)} \\[5ex] = 0 $
(38.) $(1 + \cot^2\beta)\sin^2\beta$


$ (1 + \cot^2\beta)\sin^2\beta \\[3ex] = \sin^2\beta(1 + \cot^2\beta) \\[3ex] \cot\beta = \dfrac{\cos\beta}{\sin\beta} ... Quotient\:\: Identity \\[5ex] \cot^2\beta = \left(\dfrac{\cos\beta}{\sin\beta}\right)^2 = \dfrac{\cos^2\beta}{\sin^2\beta} \\[5ex] = \sin^2\beta\left(1 + \dfrac{\cos^2\beta}{\sin^2\beta}\right) \\[5ex] = \sin^2\beta(1) + \sin^2\beta\left(\dfrac{\cos^2\beta}{\sin^2\beta}\right) \\[5ex] = \sin^2\beta + \cos^2\beta \\[3ex] = 1 ... Pythagorean\:\: Identity $
(39.) $\cos(\alpha - \beta)\cos\beta - \sin(\alpha - \beta)\sin\beta$


$ \cos(\alpha - \beta)\cos\beta - \sin(\alpha - \beta)\sin\beta \\[3ex] = \cos\beta * \cos(\alpha - \beta) - \sin\beta * \sin(\alpha - \beta) \\[3ex] \cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta ... Difference\:\: Formula \\[3ex] \sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta ... Difference\:\: Formula \\[3ex] = \cos\beta(\cos\alpha\cos\beta + \sin\alpha\sin\beta) - \sin\beta(\sin\alpha\cos\beta - \cos\alpha\sin\beta) \\[3ex] = \cos\alpha\cos^2\beta + \sin\alpha\sin\beta\cos\beta - \sin\alpha\sin\beta\cos\beta + \cos\alpha\sin^2\beta \\[3ex] = \cos\alpha\cos^2\beta + \cos\alpha\sin^2\beta \\[3ex] = \cos\alpha(\cos^\beta + \sin^2\beta) \\[3ex] \cos^2\beta + \sin^2\beta = 1 ... Pythagorean\:\: Identity \\[3ex] = \cos\alpha * 1 \\[3ex] = \cos\alpha $
(40.) $\cot\alpha * \sec\alpha$


$ \cot\alpha * \sec\alpha \\[3ex] \cot\alpha = \dfrac{\cos\alpha}{\sin\alpha} ... Quotient\:\: Identity \\[5ex] \sec\alpha = \dfrac{1}{\cos\alpha} ... Reciprocal\:\: Identity \\[5ex] = \dfrac{\cos\alpha}{\sin\alpha} * \dfrac{1}{\cos\alpha} \\[5ex] = \dfrac{1}{\sin\alpha} \\[5ex] = \csc\alpha $




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(41.) Determine the exact value of
$\sin1^\circ * \sin2^\circ * \sin3^\circ * ... * \sin44^\circ * \sin45^\circ * \sec46^\circ * ... * \sec87^\circ * \sec88^\circ * \sec89^\circ$


Let us figure a way to simplify this expression.
Let us see if we can cancel several terms

Observations
(1.) All angles from $1^\circ$ to $89^\circ$ are included
(2.) The reciprocal of the sine function is the cosecant function
(3.) The cosecant function and the secant function are cofunctions
(4.) $\csc1$ and $\sec89$ are cofunctions; $\csc2$ and $\sec88$ are cofunctions; $\csc3$ and $\sec87$ are cofunctions, $\csc44$ and $\sec46$ are cofunctions, etc.
(5.) It starts from $\sin1$ up to $\sin45$, then it changes to $\sec46$ and continues till $\sec89$
(6.) Only the reciprocal of $\sin45$ does not have an included cofunction
(7.) Every other term besides $\sin45$ will cancel out

$ \sin1^\circ * \sin2^\circ * \sin3^\circ * ... * \sin44^\circ * \sin45^\circ * \sec46^\circ * ... * \sec87^\circ * \sec88^\circ * \sec89^\circ \\[3ex] \dfrac{1}{\sin1} = \csc1 ... Reciprocal\:\: Identity \\[5ex] \sec89 = \csc1 ... Cofunction\:\: Identity \\[3ex] \therefore \sec89 = \dfrac{1}{\sin1} ... Equality \:\:of\:\: \csc1 \\[5ex] Same\:\: applies\:\: to\:\: every\:\: term\:\: besides\:\: \sin45 \\[3ex] = \sin1 * \sin2 * \sin3 * ... * \sin44 * \sin45 * \dfrac{1}{\sin44} * ... * \dfrac{1}{\sin3} * \dfrac{1}{\sin2} * \dfrac{1}{\sin1} \\[5ex] = \sin45 \\[3ex] = \dfrac{\sqrt{2}}{2} ... Unit\:\: Circle\:\: Trigonometry $
(42.) Determine the exact value of
$\cos1^\circ * \cos2^\circ * \cos3^\circ * ... * \cos44^\circ * \cos45^\circ * \csc46^\circ * ... * \csc87^\circ * \csc88^\circ * \csc89^\circ$


Let us figure a way to simplify this expression.
Let us see if we can cancel several terms

Observations
(1.) All angles from $1^\circ$ to $89^\circ$ are included
(2.) The reciprocal of the cosine function is the secant function
(3.) The secant function and the cosecant function are cofunctions
(4.) $\sec1$ and $\csc89$ are cofunctions; $\sec2$ and $\csc88$ are cofunctions; $\sec3$ and $\csc87$ are cofunctions, $\sec44$ and $\csc46$ are cofunctions, etc.
(5.) It starts from $\cos1$ up to $\cos45$, then it changes to $\csc46$ and continues till $\csc89$
(6.) Only the reciprocal of $\cos45$ does not have an included cofunction
(7.) Every other term besides $\cos45$ will cancel out

$ \cos1^\circ * \cos2^\circ * \cos3^\circ * ... * \cos44^\circ * \cos45^\circ * \csc46^\circ * ... * \csc87^\circ * \csc88^\circ * \csc89^\circ \\[3ex] \dfrac{1}{\cos1} = \sec1 ... Reciprocal\:\: Identity \\[5ex] \csc89 = \sec1 ... Cofunction\:\: Identity \\[3ex] \therefore \csc89 = \dfrac{1}{\cos1} ... Equality \:\:of\:\: \sec1 \\[5ex] Same\:\: applies\:\: to\:\: every\:\: term\:\: besides\:\: \cos45 \\[3ex] = \cos1 * \cos2 * \cos3 * ... * \cos44 * \cos45 * \dfrac{1}{\cos44} * ... * \dfrac{1}{\cos3} * \dfrac{1}{\cos2} * \dfrac{1}{\cos1} \\[5ex] = \cos45 \\[3ex] = \dfrac{\sqrt{2}}{2} ... Unit\:\: Circle\:\: Trigonometry $
(43.) Determine the exact value of
$\tan1^\circ * \tan2^\circ * \tan3^\circ * ... * \tan87^\circ * \tan88^\circ * \tan89^\circ$


Let us figure a way to simplify this expression.
Let us see if we can cancel several terms

Observations
(1.) All angles from $1^\circ$ to $89^\circ$ are included
(2.) The reciprocal of the tangent function is the cotangent function
(3.) The tangent function and the cotangent function are cofunctions
(4.) $\tan89$ and $\cot1$ are cofunctions; $\tan88$ and $\cot2$ are cofunctions; $\tan87$ and $\cot3$ are cofunctions etc.
(5.) It starts from $\tan1$ up to $\tan89$
(6.) Only the reciprocal of $\tan45$ does not have an included cofunction
(7.) Every term besides $\tan45$ will cancel out

$ \tan1^\circ * \tan2^\circ * \tan3^\circ * ... * \tan87^\circ * \tan88^\circ * \tan89^\circ \\[3ex] \dfrac{1}{\tan1} = \cot1 ... Reciprocal\:\: Identity \\[5ex] \tan89 = \cot1 ... Cofunction\:\: Identity \\[3ex] \therefore \tan89 = \dfrac{1}{\tan1} ... Equality \:\:of\:\: \cot1 \\[5ex] Similarly\:\: \tan88 = \dfrac{1}{\tan2} \\[5ex] Similarly\:\: \tan87 = \dfrac{1}{\tan3} \\[5ex] Same\:\: applies\:\: to\:\: every\:\: term\:\: besides\:\: \tan45 \\[3ex] = \tan1 * \tan2 * \tan3 * ... * \tan45 * \dfrac{1}{\tan3} * \dfrac{1}{\tan2} * \dfrac{1}{\tan1} \\[5ex] = \tan45 \\[3ex] = 1 ... Unit\:\: Circle\:\: Trigonometry $
(44.) Determine the exact value of
$\cot1^\circ * \cot2^\circ * \cot3^\circ * ... * \cot87^\circ * \cot88^\circ * \cot89^\circ$


Let us figure a way to simplify this expression.
Let us see if we can cancel several terms

Observations
(1.) All angles from $1^\circ$ to $89^\circ$ are included
(2.) The reciprocal of the cotangent function is the tangent function
(3.) The cotangent function and the tangent function are cofunctions
(4.) $\cot89$ and $\tan1$ are cofunctions; $\cot88$ and $\tan2$ are cofunctions; $\cot87$ and $\tan3$ are cofunctions etc.
(5.) It starts from $\cot1$ up to $\cot89$
(6.) Only the reciprocal of $\cot45$ does not have an included cofunction
(7.) Every term besides $\cot45$ will cancel out

$ \cot1^\circ * \cot2^\circ * \cot3^\circ * ... * \cot87^\circ * \cot88^\circ * \cot89^\circ \\[3ex] \dfrac{1}{\cot1} = \tan1 ... From\:\: Reciprocal\:\: Identity \\[5ex] \cot89 = \tan1 ... Cofunction\:\: Identity \\[3ex] \therefore \cot89 = \dfrac{1}{\cot1} ... Equality \:\:of\:\: \tan1 \\[5ex] Similarly\:\: \cot88 = \dfrac{1}{\cot2} \\[5ex] Similarly\:\: \cot87 = \dfrac{1}{\cot3} \\[5ex] Same\:\: applies\:\: to\:\: every\:\: term\:\: besides\:\: \cot45 \\[3ex] = \cot1 * \cot2 * \cot3 * ... * \cot45 * \dfrac{1}{\cot3} * \dfrac{1}{\cot2} * \dfrac{1}{\cot1} \\[5ex] = \cot45 \\[3ex] = \dfrac{1}{\tan45} \\[5ex] \tan45 = 1 ... Unit\:\: Circle\:\: Trigonometry \\[3ex] = \dfrac{1}{1} \\[5ex] = 1 $
(45.) Determine the exact value of
$\csc1^\circ * \csc2^\circ * \csc3^\circ * ... * \csc44^\circ * \csc45^\circ * \cos46^\circ * ... * \cos87^\circ * \cos88^\circ * \cos89^\circ$


Let us figure a way to simplify this expression.
Let us see if we can cancel several terms

Observations
(1.) All angles from $1^\circ$ to $89^\circ$ are included
(2.) The reciprocal of the cosecant function is the sine function
(3.) The sine function and the cosine function are cofunctions
(4.) $\sin1$ and $\cos89$ are cofunctions; $\sin2$ and $\cos88$ are cofunctions; $\sin3$ and $\cos87$ are cofunctions, $\sin44$ and $\cos46$ are cofunctions, etc.
(5.) It starts from $\csc1$ up to $\csc45$, then it changes to $\cos46$ and continues till $\cos89$
(6.) Only the reciprocal of $\csc45$ does not have an included cofunction
(7.) Every other term besides $\csc45$ will cancel out

$ \csc1^\circ * \csc2^\circ * \csc3^\circ * ... * \csc44^\circ * \csc45^\circ * \cos46^\circ * ... * \cos87^\circ * \cos88^\circ * \cos89^\circ \\[3ex] \dfrac{1}{\csc1} = \sin1 ... From\:\: Reciprocal\:\: Identity \\[5ex] \cos89 = \sin1 ... Cofunction\:\: Identity \\[3ex] \therefore \cos89 = \dfrac{1}{\csc1} ... Equality \:\:of\:\: \sin1 \\[5ex] Same\:\: applies\:\: to\:\: every\:\: term\:\: besides\:\: \csc45 \\[3ex] = \csc1 * \csc2 * \csc3 * ... * \csc44 * \csc45 * \dfrac{1}{\csc44} * ... * \dfrac{1}{\csc3} * \dfrac{1}{\csc2} * \dfrac{1}{\csc1} \\[5ex] = \csc45 \\[3ex] = \dfrac{1}{\sin45} ... Reciprocal\:\: Identity \\[5ex] \sin45 = \dfrac{\sqrt{2}}{2} ... Unit\:\: Circle\:\: Trigonometry \\[5ex] = 1 \div \dfrac{\sqrt{2}}{2} \\[5ex] = 1 * \dfrac{2}{\sqrt{2}} \\[5ex] = \dfrac{2}{\sqrt{2}} $
(46.) Determine the exact value of
$\sec1^\circ * \sec2^\circ * \sec3^\circ * ... * \sec44^\circ * \sec45^\circ * \sin46^\circ * ... * \sin87^\circ * \sin88^\circ * \sin89^\circ$


Let us figure a way to simplify this expression.
Let us see if we can cancel several terms

Observations
(1.) All angles from $1^\circ$ to $89^\circ$ are included
(2.) The reciprocal of the secant function is the cosine function
(3.) The cosine function and the sine function are cofunctions
(4.) $\cos1$ and $\sin89$ are cofunctions; $\cos2$ and $\sin88$ are cofunctions; $\cos3$ and $\sin87$ are cofunctions, $\cos44$ and $\sin46$ are cofunctions, etc.
(5.) It starts from $\sec1$ up to $\sec45$, then it changes to $\sin46$ and continues till $\sin89$
(6.) Only the reciprocal of $\sec45$ does not have an included cofunction
(7.) Every other term besides $\sec45$ will cancel out

$ \sec1^\circ * \sec2^\circ * \sec3^\circ * ... * \sec44^\circ * \sec45^\circ * \sin46^\circ * ... * \sin87^\circ * \sin88^\circ * \sin89^\circ \\[3ex] \dfrac{1}{\sec1} = \cos1 ... Reciprocal\:\: Identity \\[5ex] \sin89 = \cos1 ... Cofunction\:\: Identity \\[3ex] \therefore \sin89 = \dfrac{1}{\sec1} ... Equality \:\:of\:\: \cos1 \\[5ex] Same\:\: applies\:\: to\:\: every\:\: term\:\: besides\:\: \sec45 \\[3ex] = \sec1 * \sec2 * \sec3 * ... * \sec44 * \sec45 * \dfrac{1}{\sec44} * ... * \dfrac{1}{\sec3} * \dfrac{1}{\sec2} * \dfrac{1}{\sec1} \\[5ex] = \sec45 \\[3ex] = \dfrac{1}{\cos45} ... Reciprocal\:\: Identity \\[5ex] \cos45 = \dfrac{\sqrt{2}}{2} ... Unit\:\: Circle\:\: Trigonometry \\[5ex] = 1 \div \dfrac{\sqrt{2}}{2} \\[5ex] = 1 * \dfrac{2}{\sqrt{2}} \\[5ex] = \dfrac{2}{\sqrt{2}} $
(47.) $\dfrac{\sin^3\omega + \cos^3\omega}{\sin\omega + \cos\omega}$


$ \dfrac{\sin^3\omega + \cos^3\omega}{\sin\omega + \cos\omega} \\[5ex] \sin^3\omega + \cos^3\omega = (\sin\omega + \cos\omega)(\sin^2\omega - \sin\omega\cos\omega + \cos^2\omega) ... Sum\:\: of\:\: Two\:\: Cubes \\[3ex] = \dfrac{(\sin\omega + \cos\omega)(\sin^2\omega - \sin\omega\cos\omega + \cos^2\omega)}{\sin\omega + \cos\omega} \\[5ex] = \sin^2\omega - \sin\omega\cos\omega + \cos^2\omega \\[3ex] \sin^2\omega + \cos^2\omega = 1 ... Pythagorean\:\: Identity \\[3ex] = 1 - \sin\omega\cos\omega $
(48.) $\dfrac{\sec\omega + \tan\omega}{\tan\omega} - \dfrac{\sec\omega + \tan\omega}{\sec\omega}$

Express your answer as a difference


$ \dfrac{\sec\omega + \tan\omega}{\tan\omega} - \dfrac{\sec\omega + \tan\omega}{\sec\omega} \\[5ex] \sec\omega = \dfrac{1}{\cos\omega} ... Reciprocal\:\: Identity \\[5ex] \tan\omega = \dfrac{\sin\omega}{\cos\omega} ... Quotient\:\: Identity \\[5ex] \sec\omega + \tan\omega = \dfrac{1}{\cos\omega} + \dfrac{\sin\omega}{\cos\omega} = \dfrac{1 + \sin\omega}{\cos\omega} \\[5ex] \underline{Minuend} \\[3ex] \dfrac{\sec\omega + \tan\omega}{\tan\omega} \\[5ex] = (\sec\omega + \tan\omega) \div \tan\omega \\[3ex] = \dfrac{1 + \sin\omega}{\cos\omega} \div \dfrac{\sin\omega}{\cos\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\cos\omega} * \dfrac{\cos\omega}{\sin\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\sin\omega} \\[5ex] \underline{Subtrahend} \\[3ex] \dfrac{\sec\omega + \tan\omega}{\sec\omega} \\[5ex] = (\sec\omega + \tan\omega) \div \sec\omega \\[3ex] = \dfrac{1 + \sin\omega}{\cos\omega} \div \dfrac{1}{\cos\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\cos\omega} * \dfrac{\cos\omega}{1} \\[5ex] = 1 + \sin\omega \\[3ex] \rightarrow \dfrac{\sec\omega + \tan\omega}{\tan\omega} - \dfrac{\sec\omega + \tan\omega}{\sec\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\sin\omega} - (1 + \sin\omega) \\[5ex] = \dfrac{1}{\sin\omega} + \dfrac{\sin\omega}{\sin\omega} - 1 - \sin\omega \\[5ex] = \dfrac{1}{\sin\omega} + 1 - 1 - \sin\omega \\[5ex] = \csc\omega - \sin\omega $
(49.) $\sin7\omega - \sin9\omega$


$ \sin7\omega - \sin9\omega \\[3ex] = 2\sin\left(\dfrac{7\omega - 9\omega}{2}\right)\cos\left(\dfrac{7\omega + 9\omega}{2}\right) ...Sum-to-Product\:\: Formula \\[5ex] = 2\sin\left(\dfrac{-2\omega}{2}\right)\cos\left(\dfrac{16\omega}{2}\right) \\[5ex] = 2\sin(-\omega)\cos(8\omega) \\[3ex] \sin(-\omega) = -\sin(\omega) ... Odd\:\: Identity \\[3ex] = 2 * -\sin(\omega) * \cos(8\omega) \\[3ex] = -2\sin\omega\cos8\omega $
(50.) $\dfrac{\sec\omega + \tan\omega}{\tan\omega} - \dfrac{\sec\omega + \tan\omega}{\sec\omega}$

Express your answer as a product


$ \dfrac{\sec\omega + \tan\omega}{\tan\omega} - \dfrac{\sec\omega + \tan\omega}{\sec\omega} \\[5ex] \sec\omega = \dfrac{1}{\cos\omega} ... Reciprocal\:\: Identity \\[5ex] \tan\omega = \dfrac{\sin\omega}{\cos\omega} ... Quotient\:\: Identity \\[5ex] \sec\omega + \tan\omega = \dfrac{1}{\cos\omega} + \dfrac{\sin\omega}{\cos\omega} = \dfrac{1 + \sin\omega}{\cos\omega} \\[5ex] \underline{Minuend} \\[3ex] \dfrac{\sec\omega + \tan\omega}{\tan\omega} \\[5ex] = (\sec\omega + \tan\omega) \div \tan\omega \\[3ex] = \dfrac{1 + \sin\omega}{\cos\omega} \div \dfrac{\sin\omega}{\cos\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\cos\omega} * \dfrac{\cos\omega}{\sin\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\sin\omega} \\[5ex] \underline{Subtrahend} \\[3ex] \dfrac{\sec\omega + \tan\omega}{\sec\omega} \\[5ex] = (\sec\omega + \tan\omega) \div \sec\omega \\[3ex] = \dfrac{1 + \sin\omega}{\cos\omega} \div \dfrac{1}{\cos\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\cos\omega} * \dfrac{\cos\omega}{1} \\[5ex] = 1 + \sin\omega \\[3ex] \rightarrow \dfrac{\sec\omega + \tan\omega}{\tan\omega} - \dfrac{\sec\omega + \tan\omega}{\sec\omega} \\[5ex] = \dfrac{1 + \sin\omega}{\sin\omega} - (1 + \sin\omega) \\[5ex] = \dfrac{1}{\sin\omega} + \dfrac{\sin\omega}{\sin\omega} - 1 - \sin\omega \\[5ex] = \dfrac{1}{\sin\omega} + 1 - 1 - \sin\omega \\[5ex] = \dfrac{1}{\sin\omega} - \dfrac{\sin\omega}{1} \\[5ex] = \dfrac{1 - \sin\omega(\sin\omega)}{\sin\omega} \\[5ex] = \dfrac{1 - \sin^2\omega}{\sin\omega} \\[5ex] \sin^2\omega + \cos^2\omega = 1 ... Pythagorean\:\: Identity \\[3ex] \rightarrow \cos^2\omega = 1 - \sin^2\omega \\[3ex] = \dfrac{\cos^2\omega}{\sin\omega} \\[5ex] = \dfrac{\cos\omega * \cos\omega}{\sin\omega} \\[5ex] = \cos\omega * \dfrac{\cos\omega}{\sin\omega} \\[5ex] \cot\omega = \dfrac{\cos\omega}{\sin\omega} ... Quotient\:\: Identity \\[5ex] = \cos\omega * \cot\omega \\[3ex] = \cos\omega\cot\omega $
(51.) $7\cos^4\theta - 7\sin^4\theta$


$ 7\cos^4\theta - 7\sin^4\theta \\[3ex] = 7(\cos^4\theta - \sin^4\theta) \\[3ex] \underline{Multiplicand} \\[3ex] \cos^4\theta - \sin^4\theta \\[3ex] = (\cos^2\theta)^2 - (\sin^2\theta)^2 \\[3ex] = (\cos^2\theta + \sin^2\theta)(\cos^2\theta - \sin^2\theta) ... Difference\:\: of\:\: Two\:\: Squares \\[3ex] \cos^2\theta + \sin^2\theta = 1 ... Pythagorean\:\: Identity \\[3ex] = 1 * \cos^2\theta - \sin^2\theta \\[3ex] = \cos^2\theta - \sin^2\theta \\[3ex] = \cos2\theta ... Double-Angle\:\: Identity \\[3ex] \rightarrow 7 * \cos2\theta \\[3ex] = 7\cos2\theta $
(52.) $\tan^2\beta - \sec^2\beta$
Express your answer as a difference


$ \tan^2\beta - \sec^2\beta \\[3ex] 1 + \tan^2\beta = \sec^2\beta ... Pythagorean\:\: Identity \\[3ex] \tan^2\beta = \sec^2\beta - 1 \\[3ex] = \sec^2\beta - 1 - \sec^2\beta \\[3ex] = \sec^2\beta - \sec^2\beta - 1 \\[3ex] = -1 $
(53.) $\sin(\theta + \omega) - \sin(\theta - \omega)$


$ \sin(\theta + \omega) - \sin(\theta - \omega) \\[3ex] \sin(\theta + \omega) = \sin\theta\cos\omega + \cos\theta\sin\omega ... Sum\:\: Formula \\[3ex] \sin(\theta - \omega) = \sin\theta\cos\omega - \cos\theta\sin\omega ... Difference\:\: Formula \\[3ex] = \sin\theta\cos\omega + \cos\theta\sin\omega - (\sin\theta\cos\omega - \cos\theta\sin\omega) \\[3ex] = \sin\theta\cos\omega + \cos\theta\sin\omega - \sin\theta\cos\omega + \cos\theta\sin\omega \\[3ex] = \cos\theta\sin\omega + \cos\theta\sin\omega \\[3ex] = 2\cos\theta\sin\omega $
(54.) $(1 + \cot \beta)^2$


$ (1 + \cot \beta)^2 = (1 + \cot \beta)(1 + \cot \beta) \\[3ex] = 1^2 + \cot \beta + \cot \beta + \cot^2 \beta \\[3ex] = 1 + 2\cot \beta + \cot^2 \beta \\[3ex] = 1 + \cot^2 \beta + 2\cot \beta \\[3ex] 1 + \cot^2 \beta = \csc^2 \beta ... Pythagorean\:\: Identity \\[3ex] = \csc^2 \beta + 2\cot \beta $
(55.) $\dfrac{1 + \sin^2\omega}{1 - \sin^2\omega} + 1$


$ \dfrac{1 + \sin^2\omega}{1 - \sin^2\omega} + 1 \\[5ex] = \dfrac{1 + \sin^2\omega}{1 - \sin^2\omega} + \dfrac{1 - \sin^2\omega}{1 - \sin^2\omega} \\[5ex] = \dfrac{(1 + \sin^2\omega) + (1 - \sin^2\omega)}{1 - \sin^2\omega} \\[5ex] = \dfrac{1 + \sin^2\omega + 1 - \sin^2\omega}{1 - \sin^2\omega} \\[5ex] = \dfrac{2}{1 - \sin^2\omega} \\[5ex] \sin^2\omega + \cos^2\omega = 1 ... Pythagorean\:\: Identity \\[3ex] \rightarrow \cos^2\omega = 1 - \sin^2\omega \\[3ex] = \dfrac{2}{\cos^2\omega} \\[5ex] \sec\omega = \dfrac{1}{\cos\omega} ... Reciprocal\:\: Identity \\[5ex] \rightarrow \sec^2\omega = \dfrac{1^2}{\cos^2\omega} = \dfrac{1}{\cos^2\omega} \\[5ex] = 2 * \dfrac{1}{\cos^2\omega} \\[5ex] = 2\sec^2\omega $
(56.) $\dfrac{\sec\alpha - \csc\alpha}{\sec\alpha\csc\alpha}$


$ \dfrac{\sec\alpha - \csc\alpha}{\sec\alpha\csc\alpha} \\[5ex] \sec\alpha = \dfrac{1}{\cos\alpha} ... Reciprocal\:\: Identity \\[5ex] \csc\alpha = \dfrac{1}{\sin\alpha} ... Reciprocal\:\: Identity \\[5ex] \underline{Numerator} \\[3ex] \sec\alpha - \csc\alpha \\[3ex] = \dfrac{1}{\cos\alpha} - \dfrac{1}{\sin\alpha} \\[5ex] = \dfrac{\sin\alpha - \cos\alpha}{\sin\alpha\cos\alpha} \\[5ex] \underline{Denominator} \\[3ex] \sec\alpha\csc\alpha \\[3ex] = \dfrac{1}{\cos\alpha} * \dfrac{1}{\sin\alpha} \\[5ex] = \dfrac{1}{\sin\alpha\cos\alpha} \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{\sin\alpha - \cos\alpha}{\sin\alpha\cos\alpha} \div \dfrac{1}{\sin\alpha\cos\alpha} \\[5ex] = \dfrac{\sin\alpha - \cos\alpha}{\sin\alpha\cos\alpha} * \dfrac{\sin\alpha\cos\alpha}{1} \\[5ex] = \sin\alpha - \cos\alpha $
(57.) $\dfrac{\cos(\theta - \omega)}{\cos\theta\sin\omega}$


$ \dfrac{\cos(\theta - \omega)}{\cos\theta\sin\omega} \\[5ex] \cos(\theta - \omega) = \cos\theta\cos\omega + \sin\theta\sin\omega ... Difference\:\: Formula \\[3ex] = \dfrac{\cos\theta\cos\omega + \sin\theta\sin\omega}{\cos\theta\sin\omega} \\[5ex] = \dfrac{\cos\theta\cos\omega}{\cos\theta\sin\omega} + \dfrac{\sin\theta\sin\omega}{\cos\theta\sin\omega} \\[5ex] = \dfrac{\cos\omega}{\sin\omega} + \dfrac{\sin\theta}{\cos\theta} \\[5ex] \dfrac{\cos\omega}{\sin\omega} = \cot\omega ... Quotient\:\: Identity \\[5ex] \dfrac{\sin\theta}{\cos\theta} = \tan\theta ... Quotient\:\: Identity \\[5ex] = \cot\omega + \tan\theta $
(58.) $\sec\alpha - \cos\alpha$


$ \sec\alpha - \cos\alpha \\[3ex] \sec\alpha = \dfrac{1}{\cos\alpha} ... Reciprocal\:\: Identity \\[5ex] = \dfrac{1}{\cos\alpha} - \dfrac{\cos\alpha}{1} \\[5ex] = \dfrac{1 - \cos\alpha(\cos\alpha)}{\cos\alpha} \\[5ex] = \dfrac{1 - \cos^2\alpha}{\cos\alpha} \\[5ex] \sin^2\alpha + \cos^2\alpha = 1 ... Pythagorean\:\: Identity \\[3ex] \rightarrow \sin^2\alpha = 1 - \cos^2\alpha \\[3ex] = \dfrac{\sin^2\alpha}{\cos\alpha} \\[5ex] = \sin\alpha * \dfrac{\sin\alpha}{\cos\alpha} \\[5ex] \tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} ... Quotient\:\: Identity \\[5ex] = \sin\alpha * \tan\alpha \\[3ex] = \sin\alpha\tan\alpha $
(59.) $\dfrac{\cos(\theta - \omega)}{\cos\theta\cos\omega}$


$ \dfrac{\cos(\theta - \omega)}{\cos\theta\cos\omega} \\[5ex] \cos(\theta - \omega) = \cos\theta\cos\omega + \sin\theta\sin\omega ... Difference\:\: Formula \\[3ex] = \dfrac{\cos\theta\cos\omega + \sin\theta\sin\omega}{\cos\theta\cos\omega} \\[5ex] = \dfrac{\cos\theta\cos\omega}{\cos\theta\cos\omega} + \dfrac{\sin\theta\sin\omega}{\cos\theta\cos\omega} \\[5ex] = 1 + \dfrac{\sin\theta\sin\omega}{\cos\theta\cos\theta} \\[5ex] \dfrac{\sin\theta}{\cos\theta} = \tan\theta ... Quotient\:\: Identity \\[5ex] \dfrac{\sin\omega}{\cos\omega} = \tan\omega ... Quotient\:\: Identity \\[5ex] = 1 + \tan\theta\tan\omega $
(60.) Rationalize the denominator

$\sqrt{\dfrac{\sin \theta}{\cos^7 \theta}}$


We want to rationalize the denominator
This means that we want to make the denominator a rational number
What do we have to multiply by $7$ (the exponent at the denominator is $7$) in terms of exponent so that it is divisible by $2$
Why $2$? ... it is because of the square root.
Remember: Square root means an exponent of $\dfrac{1}{2}$... division by $2$
$\sqrt{x} = x^\frac{1}{2}$
So, we have to multiply $\cos^7 \theta$ by $\cos \theta$ so we can get $\cos^8 \theta$ because $8$ is divisible by $2$
Whatever you do to the denominator, you have to do the same thing to the numerator
This is necessary in order to get an equivalent fraction to the original fraction
$ \sqrt{\dfrac{\sin \theta}{\cos^7 \theta}} \\[5ex] = \dfrac{\sqrt{\sin \theta}}{\sqrt{\cos^7 \theta}} \\[5ex] = \dfrac{\sqrt{\sin \theta}}{\sqrt{\cos^7 \theta}} * \dfrac{\sqrt{\cos \theta}}{\sqrt{\cos \theta}} \\[5ex] = \dfrac{\sqrt{\sin \theta * \cos \theta}}{\sqrt{\cos^7 \theta * \cos \theta}} \\[5ex] = \dfrac{\sqrt{\sin \theta \cos \theta}}{\sqrt{\cos^8 \theta}} \\[5ex] = \dfrac{\sqrt{\sin \theta \cos \theta}}{\cos^4 \theta} \\[5ex] $ The denominator is now a rational number.




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(61.) $\dfrac{\tan^2\left(\dfrac{\omega}{2}\right) - 1}{\tan^2\left(\dfrac{\omega}{2}\right) + 1}$


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(62.) ACT If $\cos x = -\dfrac{1}{3}$, what is the value of $\cos 2x$

$\left[Note:\:\: (\cos x)^2 = \dfrac{1 + \cos 2x}{2}\right]$


$ (\cos x)^2 = \dfrac{1 + \cos 2x}{2} \\[5ex] Cross\:\:Multiply \\[3ex] 2 * \cos x * \cos x = 1 + \cos2x \\[3ex] 2\cos^2x = 1 + \cos2x \\[3ex] 1 + \cos2x = 2\cos^2x \\[3ex] \cos2x = 2 * \cos x * \cos x - 1 \\[3ex] \cos2x = 2 * -\dfrac{1}{3} * -\dfrac{1}{3} - 1 \\[5ex] \cos2x = \dfrac{2}{9} - \dfrac{9}{9} \\[5ex] \cos2x = \dfrac{2 - 9}{9} \\[5ex] \cos2x = -\dfrac{7}{9} $
(63.)

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(64.) $\sin(-\theta) + \cos(-\theta)$


$ \sin(-\theta) = -\sin \theta ... Odd\;\;Identity \\[3ex] \cos(-\theta) = \cos \theta ... Even\;\;Identity \\[3ex] \sin(-\theta) + \cos(-\theta) \\[3ex] = -\sin \theta + \cos \theta \\[3ex] = \cos \theta - \sin \theta $
(65.)

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(66.)