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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. $\tan^4 2x$

## Texas Instruments Resources

In the CK-12 Texas Instruments Trigonometry FlexBook® resource, 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.

Sep 26, 2013

## Last Modified:

Dec 16, 2014
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