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# Chapter 3: Trigonometric Identities and Equations

Difficulty Level: At Grade Created by: CK-12

Chapter Outline

## Chapter Summary

Here are the identities studied in this chapter:

Quotient & Reciprocal Identities

tanθcscθ=sinθcosθcotθ=cosθsinθ=1sinθ secθ=1cosθ cotθ=1tanθ\begin{align*}\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} \end{align*}

Pythagorean Identities

sin2θ+cos2θ=11+cot2θ=csc2θtan2θ+1=sec2θ\begin{align*}\sin^2 \theta + \cos^2 \theta = 1 && 1 + \cot^2 \theta = \csc^2 \theta && \tan^2 \theta + 1 = \sec^2 \theta \end{align*}

Even & Odd Identities

sin(x)csc(x)=sinx=cscxcos(x)=cosxsec(x)=secxtan(x)=tanxcot(x)=cotx\begin{align*}\sin (-x) &= -\sin x && \cos (-x) = \cos x && \tan (-x) = -\tan x\\ \csc (-x) & = -\csc x && \sec (-x) = \sec x && \cot (-x) = -\cot x\end{align*}

Co-Function Identities

sin(π2θ)=cosθcsc(π2θ)=secθcos(π2θ)=sinθsec(π2θ)=cscθtan(π2θ)=cotθcot(π2θ)=tanθ\begin{align*}& \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\end{align*}

Sum and Difference Identities

cos(α+β)sin(α+β)tan(α+β)=cosαcosβsinαsinβ=sinαcosβ+cosαsinβ=tanα+tanβ1tanαtanβcos(αβ)=cosαcosβ+sinαsinβsin(αβ)=sinαcosβcosαsinβtan(αβ)=tanαtanβ1+tanαtanβ\begin{align*}\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}\end{align*}

Double Angle Identities

cos(2α)sin(2α)tan(2α)=cos2αsin2α=2cos2α1=12sin2α=2sinαcosβ=2tanα1tan2α\begin{align*}\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}\end{align*}

Half Angle Identities

cosα2=±1+cosα2sinα2=±1cosα2tanα2=1cosαsinα=sinα1+cosα\begin{align*}\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}\end{align*}

Product to Sum & Sum to Product Identities

sina+sinbsinasinbcosa+cosbcosacosb=2sina+b2cosab2=2sinab2cosa+b2=2cosa+b2cosab2=22sina+b2sinab2sinasinb=12[cos(ab)cos(a+b)]cosacosb=12[cos(ab)+cos(a+b)]sinacosb=12[sin(a+b)+sin(ab)]cosasinb=12[sin(a+b)sin(ab)]\begin{align*}\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)] \end{align*}

Linear Combination Formula

Acosx+Bsinx=Ccos(xD)\begin{align*}A \cos x + B \sin x = C \cos (x - D)\end{align*}, where C=A2+B2,cosD=AC\begin{align*}C = \sqrt{A^2 + B^2}, \cos D = \frac{A}{C}\end{align*} and sinD=BC\begin{align*}\sin D = \frac{B}{C}\end{align*}

## Review Questions

1. Find the sine, cosine, and tangent of an angle with terminal side on (8,15)\begin{align*}(-8, 15)\end{align*}.
2. If sina=53\begin{align*}\sin a = \frac{\sqrt{5}}{3}\end{align*} and tana<0\begin{align*}\tan a < 0\end{align*}, find seca\begin{align*}\sec a\end{align*}.
3. Simplify: cos4xsin4xcos2xsin2x\begin{align*}\frac{\cos^4 x - \sin^4 x}{\cos^2 x - \sin^2 x}\end{align*}.
4. Verify the identity: 1+sinxcosxsinx=secx(cscx+1)\begin{align*}\frac{1 + \sin x}{\cos x \sin x} = \sec x (\csc x + 1)\end{align*}

For problems 5-8, find all the solutions in the interval [0,2π)\begin{align*}[0, 2\pi)\end{align*}.

1. sec(x+π2)+2=0\begin{align*}\sec \left (x + \frac{\pi}{2} \right ) + 2 = 0\end{align*}
2. 8sin(x2)8=0\begin{align*}8 \sin \left (\frac{x}{2} \right ) - 8 = 0\end{align*}
3. 2sin2x+sin2x=0\begin{align*}2 \sin^2 x + \sin 2x =0\end{align*}
4. 3tan22x=1\begin{align*}3 \tan^2 2x = 1\end{align*}
5. Solve the trigonometric equation 1sinx=3sinx\begin{align*}1 - \sin x = \sqrt{3} \sin x\end{align*} over the interval [0,π]\begin{align*}[0, \pi]\end{align*}.
6. Solve the trigonometric equation 2cos3x1=0\begin{align*}2 \cos 3x - 1 = 0\end{align*} over the interval [0,2π]\begin{align*}[0, 2\pi]\end{align*}.
7. Solve the trigonometric equation 2sec2xtan4x=3\begin{align*}2 \sec^2 x - \tan^4 x = 3\end{align*} for all real values of x\begin{align*}x\end{align*}.

Find the exact value of:

1. cos157.5\begin{align*}\cos 157.5^\circ\end{align*}
2. sin13π12\begin{align*}\sin \frac{13 \pi}{12}\end{align*}
3. Write as a product: 4(cos5x+cos9x)\begin{align*}4(\cos 5x + \cos 9x)\end{align*}
4. Simplify: \begin{align*}\cos(x - y) \cos y - \sin(x - y) \sin y\end{align*}
5. Simplify: \begin{align*}\sin \left (\frac{4 \pi}{3} - x \right ) + \cos \left (x + \frac{5 \pi}{6} \right )\end{align*}
6. Derive a formula for \begin{align*}\sin 6x\end{align*}.
7. If you solve \begin{align*}\cos 2x = 2 \cos^2x - 1\end{align*} for \begin{align*}\cos^2 x\end{align*}, you would get \begin{align*}\cos^2 x = \frac{1}{2} (\cos 2x + 1)\end{align*}. This new formula is used to reduce powers of cosine by substituting in the right part of the equation for \begin{align*}\cos^2 x\end{align*}. Try writing \begin{align*}\cos^4 x\end{align*} in terms of the first power of cosine.
8. If you solve \begin{align*}\cos 2x = 1 - 2 \sin^2 x\end{align*} for \begin{align*}\sin^2x\end{align*}, you would get \begin{align*}\sin^2 x = \frac{1}{2} (1 - \cos 2x)\end{align*}. Similar to the new formula above, this one is used to reduce powers of sine. Try writing \begin{align*}\sin^4x\end{align*} in terms of the first power of cosine.
9. Rewrite in terms of the first power of cosine:
1. \begin{align*}\sin^2x \cos^2 2x\end{align*}
2. \begin{align*}\tan^4 2x\end{align*}

## 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.

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