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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

tanθcscθ=sinθcosθcotθ=cosθsinθ=1sinθ secθ=1cosθ cotθ=1tanθ

Pythagorean Identities

sin2θ+cos2θ=11+cot2θ=csc2θtan2θ+1=sec2θ

Even & Odd Identities

sin(x)csc(x)=sinx=cscxcos(x)=cosxsec(x)=secxtan(x)=tanxcot(x)=cotx

Co-Function Identities

sin(π2θ)=cosθcsc(π2θ)=secθcos(π2θ)=sinθsec(π2θ)=cscθtan(π2θ)=cotθcot(π2θ)=tanθ

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

Double Angle Identities

cos(2α)sin(2α)tan(2α)=cos2αsin2α=2cos2α1=12sin2α=2sinαcosβ=2tanα1tan2α

Half Angle Identities

cosα2=±1+cosα2sinα2=±1cosα2tanα2=1cosαsinα=sinα1+cosα

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)]

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: cos(xy)cosysin(xy)siny\begin{align*}\cos(x - y) \cos y - \sin(x - y) \sin y\end{align*}
5. Simplify: sin(4π3x)+cos(x+5π6)\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 sin6x\begin{align*}\sin 6x\end{align*}.
7. If you solve cos2x=2cos2x1\begin{align*}\cos 2x = 2 \cos^2x - 1\end{align*} for cos2x\begin{align*}\cos^2 x\end{align*}, you would get cos2x=12(cos2x+1)\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 cos2x\begin{align*}\cos^2 x\end{align*}. Try writing cos4x\begin{align*}\cos^4 x\end{align*} in terms of the first power of cosine.
8. If you solve cos2x=12sin2x\begin{align*}\cos 2x = 1 - 2 \sin^2 x\end{align*} for sin2x\begin{align*}\sin^2x\end{align*}, you would get sin2x=12(1cos2x)\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 sin4x\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. sin2xcos22x\begin{align*}\sin^2x \cos^2 2x\end{align*}
2. tan42x\begin{align*}\tan^4 2x\end{align*}

1. If the terminal side is on (8,15)\begin{align*}(-8,15)\end{align*}, then the hypotenuse of this triangle would be 17 (by the Pythagorean Theorem, c=(8)2+152\begin{align*}c = \sqrt{(-8)^2 + 15^2}\end{align*}). Therefore, sinx=1517,cosx=817\begin{align*}\sin x = \frac{15}{17}, \cos x = - \frac{8}{17}\end{align*}, and tanx=158\begin{align*}\tan x = - \frac{15}{8}\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*}, then a\begin{align*}a\end{align*} is in Quadrant II. Therefore seca\begin{align*}\sec a\end{align*} is negative. To find the third side, we need to do the Pythagorean Theorem.
(5)2+b25+b2b2b=32=9=4=2
So seca=32\begin{align*}\sec a = -\frac{3}{2}\end{align*}.
3. Factor top, cancel like terms, and use the Pythagorean Theorem Identity. Note that this simplification doesn't hold true for values of x\begin{align*}x\end{align*} that are π4+nπ2\begin{align*}\frac{\pi}{4} + \frac{n\pi}{2}\end{align*}, where n\begin{align*}n\end{align*} is a positive integer,, since the original expression is undefined for these values of x\begin{align*}x\end{align*}.
cos4xsin4xcos2xsin2x(cos2x+sin2x)(cos2xsin2x)cos2xsin2xcos2x+sin2x1111
4. Change secant and cosecant into terms of sine and cosine, then find a common denominator.
1+sinxcosxsinx=secx(cscx+1)=1cosx(1sinx+1)=1cosx(1+sinxsinx)=1+sinxcosxsinx
5. sec(x+π2)+2sec(x+π2)cos(x+π2)x+π2xx=0=2=12=2π3,4π3=2π3π2,4π3π2=π6,5π6
6. 8sin(x2)88sinx2sinx2x2x=0=8=1=x2=π
7.   2sin2x+sin2x=02sin2x+2sinxcosx=0 2sinx(sinx+cosx)=0 So, 2sinx=0orsinx+cosx=02sinx=0sinx+cosx=0  sinx=0 sinx=cosx x=0,πx=3π4,7π4
8. 3tan22xtan22xtan2x2xx=1=13=±33=π6,5π6,7π6,11π6,13π6,17π6,19π6,23π6=π12,5π12,7π12,11π12,13π12,17π12,19π12,23π12
9. 1sinx1111+3=3sinx=sinx+3sinx=sinx(1+3)=sinx
sin1(11+3)=x\begin{align*}\sin^{-1} \left (\frac{1}{1 + \sqrt{3}} \right ) = x\end{align*} or x=.3747\begin{align*}x = .3747\end{align*} radians and x=2.7669\begin{align*}x = 2.7669\end{align*} radians
10. Because this is cos3x\begin{align*}\cos 3x\end{align*}, 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 2π\begin{align*}2\pi\end{align*} when finding 3x\begin{align*}3x\end{align*}.
2cos3x12cos3xcos3x3xx=0=1=12=cos1(12)=π3,5π3,7π3,11π3,13π3,17π3=π9,5π9,7π9,11π9,13π9,17π9
11. Rewrite the equation in terms of tan by using the Pythagorean identity, 1+tan2θ=sec2θ\begin{align*}1 + \tan^2 \theta = \sec^2 \theta\end{align*}.
2sec2xtan4x2(1+tan2x)tan4x2+2tan2xtan4xtan4x2tan2x+1(tan2x1)(tan2x1)=3=3=3=0=0
Because these factors are the same, we only need to solve one for x\begin{align*}x\end{align*}.
tan2x1tan2xtanxx=0=1=±1=π4+πk and 3π4+πk
Where k\begin{align*}k\end{align*} is any integer.
12. Use the half angle formula with 315\begin{align*}315^\circ\end{align*}.
cos157.5=cos3152=1+cos3152=1+222=2+24=2+22
13. Use the sine sum formula.
sin13π12=sin(10π12+3π12)=sin(5π6+π4)=sin5π6cosπ4+cos5π6sinπ4=1222+(32)22=264
14. 4(cos5x+cos9x)=4[2cos(5x+9x2)cos(5x9x2)]=8cos7xcos(2x)=8cos7xcos2x
15. cos(xy)cosysin(xy)sinycosy(cosxcosy+sinxsiny)siny (sinxcosycosxsiny)cosxcos2y+sinxsinycosysinxsinycosy+cosxsin2ycosxcos2y+cosxsin2ycosx(cos2y+sin2y)cosx
16. Use the sine and cosine sum formulas.
sin(4π3x)+cos(x+5π6)sin4π3cosxcos4π3sinx+cosxcos5π6sinxsin5π632cosx+12sinx32cosx12sinx3cosx
17. Use the sine sum formula as well as the double angle formula.
sin6x=sin(4x+2x)=sin4xcos2x+cos4xsin2x=sin(2x+2x)cos2x+cos(2x+2x)sin2x=cos2x (sin2xcos2x+cos2xsin2x)+sin2x(cos2xcos2xsin2x sin2x)=2sin2xcos22x+sin2xcos22xsin32x=3sin2x cos22xsin32x=sin2x(3cos22xsin22x)=2sinxcosx[3(cos2xsin2x)2(2sinxcosx)2=2sinxcosx[3(cos4x2sin2xcos2x+sin4x)4 sin2xcos2x]=2sinxcosx[3cos4x6sin2xcos2x+3sin4x4 sin2xcos2x]=2sinxcosx[3cos4x+3sin4x10sin2xcos2x]=6sinxcos5x+6sin5xcosx20sin3xcos3x
18. Using our new formula, cos4x=[12(cos2x+1)]2\begin{align*}\cos^4 x = \left [\frac{1}{2}(\cos 2x + 1) \right ]^2 \end{align*} Now, our final answer needs to be in the first power of cosine, so we need to find a formula for cos22x\begin{align*}\cos^2 2x\end{align*}. For this, we substitute 2x\begin{align*}2x\end{align*} everywhere there is an x\begin{align*}x\end{align*} and the formula translates to cos22x=12(cos4x+1)\begin{align*}\cos^2 2x = \frac{1}{2}(\cos 4x + 1)\end{align*}. Now we can write cos4x\begin{align*}\cos^4 x\end{align*} in terms of the first power of cosine as follows.
cos4x=[12(cos2x+1)]2=14(cos22x+2cos2x+1)=14(12(cos4x+1)+2cos2x+1)=18cos4x+18+12cos2x+14=18cos4x+12cos2x+38
19. Using our new formula, sin4x=[12(1cos2x)]2\begin{align*}\sin^4 x = \left [\frac{1}{2}(1 - \cos 2x) \right ]^2 \end{align*} Now, our final answer needs to be in the first power of cosine, so we need to find a formula for cos22x\begin{align*}\cos^2 2x\end{align*}. For this, we substitute 2x\begin{align*}2x\end{align*} everywhere there is an x\begin{align*}x\end{align*} and the formula translates to cos22x=12(cos4x+1)\begin{align*}\cos^2 2x = \frac{1}{2}(\cos 4x + 1)\end{align*}. Now we can write sin4x\begin{align*}\sin^4 x\end{align*} in terms of the first power of cosine as follows.
sin4x=[12(1cos2x)]2=14(12cos2x+cos22x)=14(12cos2x+12(cos4x+1))=1412cos2x+18cos4x+18=18cos4x12cos2x+38
20. (a) First, we use both of our new formulas, then simplify:
sin2xcos22x=12(1cos2x)12(cos4x+1)=(1212cos2x)(12cos4x+12)=14cos4x+1414cos2xcos4x14cos2x=14(1cos2x+cos4xcos2xcos4x)
(b) For tangent, we use the identity tanx=sinxcosx\begin{align*}\tan x = \frac{\sin x}{\cos x}\end{align*} and then substitute in our new formulas. tan42x=sin42xcos42x\begin{align*}\tan^4 2x = \frac{\sin^4 2x}{\cos^4 2x} \rightarrow\end{align*} Now, use the formulas we derived in #18 and #19.
21. tan42x=sin42xcos42x=18cos8x12cos4x+3818cos8x+12cos4x+38=cos8x4cos4x+3cos8x+4cos4x+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.

Sep 26, 2013