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# Chapter 3: Trigonometric Identities and Equations

Difficulty Level: At Grade Created by: CK-12

## Chapter Summary

Here are the identities studied in this chapter:

Quotient & Reciprocal Identities

$\tan \theta &= \frac{\sin \theta}{\cos \theta} \qquad \cot \theta = \frac{\cos \theta}{\sin \theta}\\\csc \theta &= \frac{1}{\sin \theta} \ \sec \theta = \frac{1}{\cos \theta} \ \cot \theta = \frac{1}{\tan \theta}$

Pythagorean Identities

$\sin^2 \theta + \cos^2 \theta = 1 && 1 + \cot^2 \theta = \csc^2 \theta && \tan^2 \theta + 1 = \sec^2 \theta$

Even & Odd Identities

$\sin (-x) &= -\sin x && \cos (-x) = \cos x && \tan (-x) = -\tan x\\\csc (-x) & = -\csc x && \sec (-x) = \sec x && \cot (-x) = -\cot x$

Co-Function Identities

$& \sin \left ( \frac{\pi}{2} - \theta \right ) = \cos \theta && \cos \left ( \frac{\pi}{2} - \theta \right ) = \sin \theta && \tan \left ( \frac{\pi}{2} - \theta \right ) = \cot \theta\\& \csc \left ( \frac{\pi}{2} - \theta \right )= \sec \theta && \sec \left ( \frac{\pi}{2} - \theta \right ) = \csc \theta && \cot \left ( \frac{\pi}{2} - \theta \right ) = \tan \theta$

Sum and Difference Identities

$\cos(\alpha + \beta) & = \cos \alpha \cos \beta - \sin \alpha \sin \beta && \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \\\sin(\alpha + \beta) & = \sin \alpha \cos \beta + \cos \alpha \sin \beta && \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\\tan (\alpha + \beta) & = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} && \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$

Double Angle Identities

$\cos(2 \alpha) & = \cos^2 \alpha - \sin^2 \alpha = 2 \cos^2 \alpha - 1 = 1 - 2 \sin^2 \alpha \\\sin(2 \alpha) & = 2 \sin \alpha \cos \beta \\\tan (2 \alpha) & = \frac{2 \tan \alpha}{1 - \tan^2 \alpha}$

Half Angle Identities

$\cos \frac{\alpha}{2} = \pm \sqrt{\frac{1 + \cos \alpha}{2}} && \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \cos \alpha}{2}} && \tan \frac{\alpha }{2} = \frac{1 - \cos \alpha}{\sin \alpha} = \frac{\sin \alpha}{1 + \cos \alpha}$

Product to Sum & Sum to Product Identities

$\sin a + \sin b & = 2 \sin \frac{a+b}{2} \cos \frac{a-b}{2} && \sin a \sin b = \frac{1}{2} [\cos (a-b) - \cos (a+b)] \\\sin a - \sin b& = 2 \sin \frac{a-b}{2} \cos \frac{a+b}{2} && \cos a \cos b = \frac{1}{2} [\cos (a-b) + \cos (a+b)] \\\cos a + \cos b & = 2 \cos \frac{a+b}{2} \cos \frac{a-b}{2} && \sin a \cos b = \frac{1}{2} [\sin (a+b) + \sin (a-b)] \\\cos a - \cos b & = 2 - 2 \sin \frac{a+b}{2} \sin \frac{a-b}{2} && \cos a \sin b = \frac{1}{2} [\sin (a+b) - \sin (a-b)]$

Linear Combination Formula

$A \cos x + B \sin x = C \cos (x - D)$, where $C = \sqrt{A^2 + B^2}, \cos D = \frac{A}{C}$ and $\sin D = \frac{B}{C}$

## Review Questions

1. Find the sine, cosine, and tangent of an angle with terminal side on $(-8, 15)$.
2. If $\sin a = \frac{\sqrt{5}}{3}$ and $\tan a < 0$, find $\sec a$.
3. Simplify: $\frac{\cos^4 x - \sin^4 x}{\cos^2 x - \sin^2 x}$.
4. Verify the identity: $\frac{1 + \sin x}{\cos x \sin x} = \sec x (\csc x + 1)$

For problems 5-8, find all the solutions in the interval $[0, 2\pi)$.

1. $\sec \left (x + \frac{\pi}{2} \right ) + 2 = 0$
2. $8 \sin \left (\frac{x}{2} \right ) - 8 = 0$
3. $2 \sin^2 x + \sin 2x =0$
4. $3 \tan^2 2x = 1$
5. Solve the trigonometric equation $1 - \sin x = \sqrt{3} \sin x$ over the interval $[0, \pi]$.
6. Solve the trigonometric equation $2 \cos 3x - 1 = 0$ over the interval $[0, 2\pi]$.
7. Solve the trigonometric equation $2 \sec^2 x - \tan^4 x = 3$ for all real values of $x$.

Find the exact value of:

1. $\cos 157.5^\circ$
2. $\sin \frac{13 \pi}{12}$
3. Write as a product: $4(\cos 5x + \cos 9x)$
4. Simplify: $\cos(x - y) \cos y - \sin(x - y) \sin y$
5. Simplify: $\sin \left (\frac{4 \pi}{3} - x \right ) + \cos \left (x + \frac{5 \pi}{6} \right )$
6. Derive a formula for $\sin 6x$.
7. If you solve $\cos 2x = 2 \cos^2x - 1$ for $\cos^2 x$, you would get $\cos^2 x = \frac{1}{2} (\cos 2x + 1)$. This new formula is used to reduce powers of cosine by substituting in the right part of the equation for $\cos^2 x$. Try writing $\cos^4 x$ in terms of the first power of cosine.
8. If you solve $\cos 2x = 1 - 2 \sin^2 x$ for $\sin^2x$, you would get $\sin^2 x = \frac{1}{2} (1 - \cos 2x)$. Similar to the new formula above, this one is used to reduce powers of sine. Try writing $\sin^4x$ in terms of the first power of cosine.
9. Rewrite in terms of the first power of cosine:
1. $\sin^2x \cos^2 2x$
2. $\tan^4 2x$

1. If the terminal side is on $(-8,15)$, then the hypotenuse of this triangle would be 17 (by the Pythagorean Theorem, $c = \sqrt{(-8)^2 + 15^2}$). Therefore, $\sin x = \frac{15}{17}, \cos x = - \frac{8}{17}$, and $\tan x = - \frac{15}{8}$.
2. If $\sin a = \frac{\sqrt{5}}{3}$ and $\tan a < 0$, then $a$ is in Quadrant II. Therefore $\sec a$ is negative. To find the third side, we need to do the Pythagorean Theorem. $\left (\sqrt{5} \right )^2 + b^2 & = 3^2 \\5 + b^2 & = 9\\b^2 & = 4 \\b & = 2$ So $\sec a = -\frac{3}{2}$.
3. Factor top, cancel like terms, and use the Pythagorean Theorem Identity. Note that this simplification doesn't hold true for values of $x$ that are $\frac{\pi}{4} + \frac{n\pi}{2}$, where $n$ is a positive integer,, since the original expression is undefined for these values of $x$. $& \qquad \qquad \frac{\cos^4 x - \sin^4 x}{\cos^2 x - \sin^2 x} \\& \frac{(\cos^2 x + \sin^2 x)(\cos^2 x - \sin^2 x)}{\cos^2 x - \sin^2 x} \\& \qquad \qquad \frac{\cos^2 x + \sin^2 x}{1} \\& \qquad \qquad \frac{1}{1} \\& \qquad \qquad 1$
4. Change secant and cosecant into terms of sine and cosine, then find a common denominator. $\frac{1 + \sin x}{\cos x \sin x} & = \sec x (\csc x + 1) \\& = \frac{1}{\cos x} \left (\frac{1}{\sin x} + 1 \right ) \\& = \frac{1}{\cos x} \left (\frac{1 + \sin x}{\sin x} \right ) \\& = \frac{1 + \sin x}{\cos x \sin x}$
5. $\sec \left (x + \frac{\pi}{2} \right ) + 2 & = 0 \\\sec \left (x + \frac{\pi}{2} \right ) & = -2 \\\cos \left (x + \frac{\pi}{2} \right ) & = - \frac{1}{2} \\x + \frac{\pi}{2} & = \frac{2 \pi}{3}, \frac{4 \pi}{3} \\x & = \frac{2 \pi}{3} - \frac{\pi}{2}, \frac{4 \pi}{3} - \frac{\pi}{2} \\x & = \frac{\pi}{6}, \frac{5 \pi}{6}$
6. $8 \sin \left (\frac{x}{2} \right ) - 8 & = 0 \\8 \sin \frac{x}{2} & = 8 \\\sin \frac{x}{2} & = 1 \\\frac{x}{2} & = \frac{x}{2} \\x & = \pi$
7. $& \qquad \ \ 2 \sin^2 x + \sin 2x = 0 \\& 2 \sin^2 x + 2 \sin x \cos x = 0 \\& \ 2 \sin x (\sin x + \cos x) = 0 \\& \ \text{So},\ 2 \sin x = 0 \qquad \quad \text{or}\qquad \quad \sin x + \cos x = 0 \\& \qquad 2 \sin x = 0 \qquad \qquad \qquad \quad \sin x + \cos x = 0 \\& \qquad \ \ \sin x = 0 \qquad \qquad \qquad \qquad \qquad \ \sin x = - \cos x \\& \qquad \qquad \ x = 0, \pi \qquad \qquad \qquad \qquad \qquad \quad x = \frac{3\pi}{4}, \frac{7 \pi}{4}$
8. $3\tan^2 2x & = 1 \\\tan^2 2x & = \frac{1}{3} \\\tan 2x & = \pm\frac{\sqrt{3}}{3} \\2x & = \frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{11 \pi}{6}, \frac{13 \pi}{6}, \frac{17 \pi}{6}, \frac{19 \pi}{6}, \frac{23 \pi}{6} \\x & = \frac{\pi}{12}, \frac{5 \pi}{12}, \frac{7 \pi}{12}, \frac{11 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}, \frac{19 \pi}{12}, \frac{23 \pi}{12}$
9. $1 - \sin x & = \sqrt{3} \sin x \\1 & = \sin x + \sqrt{3} \sin x \\1 & = \sin x \left (1 + \sqrt{3} \right ) \\\frac{1}{1 + \sqrt{3}} & = \sin x$ $\sin^{-1} \left (\frac{1}{1 + \sqrt{3}} \right ) = x$ or $x = .3747$ radians and $x = 2.7669$ radians
10. Because this is $\cos 3x$, you will need to divide by 3 at the very end to get the final answer. This is why we went beyond the limit of $2\pi$ when finding $3x$. $2 \cos 3x - 1 & = 0 \\2 \cos 3x & = 1 \\\cos 3x & = \frac{1}{2} \\3x & = \cos^{-1} \left (\frac{1}{2} \right ) = \frac{\pi}{3},\frac{5 \pi}{3}, \frac{7 \pi}{3}, \frac{11 \pi}{3}, \frac{13 \pi}{3}, \frac{17 \pi}{3} \\x & = \frac{\pi}{9},\frac{5 \pi}{9}, \frac{7 \pi}{9}, \frac{11 \pi}{9}, \frac{13 \pi}{9}, \frac{17 \pi}{9}$
11. Rewrite the equation in terms of tan by using the Pythagorean identity, $1 + \tan^2 \theta = \sec^2 \theta$. $2 \sec^2 x - \tan^4 x & = 3 \\2 (1 + \tan^2 x) - \tan^4 x & = 3 \\2 + 2 \tan^2 x - \tan^4 x & = 3 \\\tan^4 x - 2 \tan^2 x + 1 & = 0 \\(\tan^2 x - 1)(\tan^2 x - 1) & = 0$ Because these factors are the same, we only need to solve one for $x$. $\tan^2 x - 1 & = 0 \\\tan^2 x & = 1 \\\tan x & = \pm 1 \\x & = \frac{\pi}{4} + \pi k \ \text{and}\ \frac{3 \pi}{4} + \pi k$ Where $k$ is any integer.
12. Use the half angle formula with $315^\circ$. $\cos 157.5^\circ & = \cos \frac{315^\circ}{2} \\& = - \sqrt{\frac{1 + \cos 315^\circ}{2}} \\& = - \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} \\& = - \sqrt{\frac{2 + \sqrt{2}}{4}} \\& = - \frac{\sqrt{2 + \sqrt{2}}}{2}$
13. Use the sine sum formula. $\sin \frac{13 \pi}{12} & = \sin \left (\frac{10 \pi}{12} + \frac{3 \pi}{12} \right ) \\& = \sin \left (\frac{5 \pi}{6} + \frac{\pi}{4} \right ) \\& = \sin \frac{5 \pi}{6} \cos \frac{\pi}{4} + \cos \frac{5 \pi}{6} \sin \frac{\pi}{4} \\& = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + (-\frac{\sqrt{3}}{2}) \cdot \frac{\sqrt{2}}{2} \\& = \frac{\sqrt{2} - \sqrt{6}}{4}$
14. $4 (\cos 5x + \cos 9x) & = 4 \left [2 \cos \left (\frac{5x + 9x}{2} \right ) \cos \left (\frac{5x - 9x}{2} \right ) \right ] \\& = 8 \cos 7x \cos (-2x) \\& = 8 \cos 7x \cos 2x$
15. $& \qquad \qquad \qquad \cos(x - y) \cos y - \sin(x - y) \sin y \\ & \cos y (\cos x \cos y + \sin x \sin y) - \sin y\ (\sin x \cos y - \cos x \sin y) \\& \cos x \cos^2 y + \sin x \sin y \cos y - \sin x \sin y \cos y + \cos x \sin^2 y \\ & \qquad \qquad \qquad \quad \cos x \cos^2y + \cos x \sin^2 y \\ & \qquad \qquad \qquad \quad \cos x (\cos^2y + \sin^2 y) \\ & \qquad \qquad \qquad \qquad \qquad \cos x$
16. Use the sine and cosine sum formulas. $& \qquad \qquad \qquad \sin \left (\frac{4 \pi}{3} - x \right ) + \cos \left (x + \frac{5 \pi}{6} \right ) \\& \sin \frac{4 \pi}{3} \cos x - \cos \frac{4 \pi}{3} \sin x + \cos x \cos \frac{5 \pi}{6} - \sin x \sin \frac{5 \pi}{6} \\& \qquad - \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x \\& \qquad \qquad \qquad \qquad \qquad - \sqrt{3} \cos x$
17. Use the sine sum formula as well as the double angle formula. $\sin 6x & = \sin (4x + 2x) \\& = \sin 4x \cos 2x + \cos 4x \sin 2x \\& = \sin(2x + 2x) \cos 2x + \cos (2x + 2x) \sin 2x \\& = \cos 2x\ (\sin 2x \cos 2x + \cos 2x \sin 2x) + \sin 2x(\cos 2x \cos 2x - \sin 2x\ \sin 2x) \\& = 2 \sin 2x \cos^2 2x + \sin 2x \cos^2 2x - \sin^3 2x \\& = 3 \sin 2x\ \cos^2 2x - \sin^3 2x \\& = \sin 2x(3 \cos^2 2x - \sin^2 2x) \\& = 2 \sin x \cos x [3(\cos^2 x - \sin^2 x)^2 - (2 \sin x \cos x)^2 \\& = 2 \sin x \cos x [3(\cos^4 x - 2 \sin^2 x \cos^2 x + \sin^4 x) - 4\ \sin^2 x \cos^2 x] \\& = 2 \sin x \cos x [3 \cos^4 x - 6 \sin^2 x \cos^2 x + 3 \sin^4 x - 4\ \sin^2 x \cos^2 x] \\& = 2 \sin x \cos x [3 \cos^4 x + 3 \sin^4 x - 10 \sin^2 x \cos^2 x] \\& = 6 \sin x \cos^5 x + 6 \sin^5 x \cos x - 20 \sin^3x \cos^3x$
18. Using our new formula, $\cos^4 x = \left [\frac{1}{2}(\cos 2x + 1) \right ]^2$ Now, our final answer needs to be in the first power of cosine, so we need to find a formula for $\cos^2 2x$. For this, we substitute $2x$ everywhere there is an $x$ and the formula translates to $\cos^2 2x = \frac{1}{2}(\cos 4x + 1)$. Now we can write $\cos^4 x$ in terms of the first power of cosine as follows. $\cos^4 x&=[\frac{1}{2}(\cos 2x+1)]^2 \\&=\frac{1}{4}(\cos^2 2x+2\cos 2x+1) \\&=\frac{1}{4}(\frac{1}{2}(\cos 4x+1)+2\cos 2x+1) \\&=\frac{1}{8}\cos 4x+\frac{1}{8}+\frac{1}{2}\cos 2x+\frac{1}{4} \\&=\frac{1}{8}\cos 4x+\frac{1}{2}\cos 2x+\frac{3}{8}$
19. Using our new formula, $\sin^4 x = \left [\frac{1}{2}(1 - \cos 2x) \right ]^2$ Now, our final answer needs to be in the first power of cosine, so we need to find a formula for $\cos^2 2x$. For this, we substitute $2x$ everywhere there is an $x$ and the formula translates to $\cos^2 2x = \frac{1}{2}(\cos 4x + 1)$. Now we can write $\sin^4 x$ in terms of the first power of cosine as follows. $\sin^4 x&=[\frac{1}{2}(1-\cos 2x)]^2 \\&=\frac{1}{4}(1-2\cos 2x + \cos^2 2x) \\&=\frac{1}{4}(1-2\cos 2x + \frac{1}{2}(\cos 4x + 1)) \\&=\frac{1}{4} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x + \frac{1}{8} \\&=\frac{1}{8}\cos 4x - \frac{1}{2}\cos 2x + \frac{3}{8}$
20. (a) First, we use both of our new formulas, then simplify: $\sin^2 x \cos^2 2x & = \frac{1}{2}(1 - \cos 2x) \frac{1}{2}(\cos 4x + 1) \\& = \left (\frac{1}{2} - \frac{1}{2} \cos 2x \right ) \left (\frac{1}{2} \cos 4x + \frac{1}{2} \right ) \\& = \frac{1}{4} \cos 4x + \frac{1}{4} - \frac{1}{4} \cos 2x \cos 4x - \frac{1}{4} \cos 2x \\& = \frac{1}{4} (1 - \cos 2x + \cos 4x - \cos 2x \cos 4x)$ (b) For tangent, we use the identity $\tan x = \frac{\sin x}{\cos x}$ and then substitute in our new formulas. $\tan^4 2x = \frac{\sin^4 2x}{\cos^4 2x} \rightarrow$ Now, use the formulas we derived in #18 and #19.
21. $\tan^4 2x &= \frac{\sin^4 2x}{\cos^4 2x} \\&=\frac{\frac{1}{8}\cos 8x - \frac{1}{2}\cos 4x + \frac{3}{8}}{\frac{1}{8}\cos 8x+\frac{1}{2}\cos 4x+\frac{3}{8}} \\&=\frac{\cos 8x - 4\cos 4x + 3}{\cos 8x+4\cos 4x+3}$

## Texas Instruments Resources

In the CK-12 Texas Instruments Trigonometry FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9701.

Feb 23, 2012