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You are reading an older version of this FlexBook® textbook: CK-12 Trigonometry - Second Edition Go to the latest version.

Chapter 3: Trigonometric Identities and Equations

Difficulty Level: At Grade Created by: CK-12

Chapter Outline

Chapter Summary

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), where C=A2+B2,cosD=AC and sinD=BC

Review Questions

  1. Find the sine, cosine, and tangent of an angle with terminal side on (8,15).
  2. If sina=53 and tana<0, find seca.
  3. Simplify: cos4xsin4xcos2xsin2x.
  4. Verify the identity: 1+sinxcosxsinx=secx(cscx+1)

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

  1. sec(x+π2)+2=0
  2. 8sin(x2)8=0
  3. 2sin2x+sin2x=0
  4. 3tan22x=1
  5. Solve the trigonometric equation 1sinx=3sinx over the interval [0,π].
  6. Solve the trigonometric equation 2cos3x1=0 over the interval [0,2π].
  7. Solve the trigonometric equation 2sec2xtan4x=3 for all real values of x.

Find the exact value of:

  1. cos157.5
  2. sin13π12
  3. Write as a product: 4(cos5x+cos9x)
  4. Simplify: cos(xy)cosysin(xy)siny
  5. Simplify: sin(4π3x)+cos(x+5π6)
  6. Derive a formula for sin6x.
  7. If you solve cos2x=2cos2x1 for cos2x, you would get cos2x=12(cos2x+1). This new formula is used to reduce powers of cosine by substituting in the right part of the equation for cos2x. Try writing cos4x in terms of the first power of cosine.
  8. If you solve cos2x=12sin2x for sin2x, you would get sin2x=12(1cos2x). Similar to the new formula above, this one is used to reduce powers of sine. Try writing sin4x in terms of the first power of cosine.
  9. Rewrite in terms of the first power of cosine:
    1. sin2xcos22x
    2. tan42x

Review Answers

  1. If the terminal side is on (8,15), then the hypotenuse of this triangle would be 17 (by the Pythagorean Theorem, c=(8)2+152). Therefore, sinx=1517,cosx=817, and tanx=158.
  2. If sina=53 and \begin{align*}\tan a < 0\end{align*}tana<0, then \begin{align*}a\end{align*}a is in Quadrant II. Therefore \begin{align*}\sec a\end{align*}seca is negative. To find the third side, we need to do the Pythagorean Theorem. \begin{align*}\left (\sqrt{5} \right )^2 + b^2 & = 3^2 \\ 5 + b^2 & = 9\\ b^2 & = 4 \\ b & = 2\end{align*}
    (5)2+b25+b2b2b=32=9=4=2
    So \begin{align*}\sec a = -\frac{3}{2}\end{align*}seca=32.
  3. Factor top, cancel like terms, and use the Pythagorean Theorem Identity. Note that this simplification doesn't hold true for values of \begin{align*}x\end{align*}x that are \begin{align*}\frac{\pi}{4} + \frac{n\pi}{2}\end{align*}π4+nπ2, where \begin{align*}n\end{align*}n is a positive integer,, since the original expression is undefined for these values of \begin{align*}x\end{align*}x. \begin{align*}& \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\end{align*}
    cos4xsin4xcos2xsin2x(cos2x+sin2x)(cos2xsin2x)cos2xsin2xcos2x+sin2x1111
  4. Change secant and cosecant into terms of sine and cosine, then find a common denominator. \begin{align*}\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}\end{align*}
    1+sinxcosxsinx=secx(cscx+1)=1cosx(1sinx+1)=1cosx(1+sinxsinx)=1+sinxcosxsinx
  5. \begin{align*}\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}\end{align*}
    sec(x+π2)+2sec(x+π2)cos(x+π2)x+π2xx=0=2=12=2π3,4π3=2π3π2,4π3π2=π6,5π6
  6. \begin{align*}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\end{align*}
    8sin(x2)88sinx2sinx2x2x=0=8=1=x2=π
  7. \begin{align*}& \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}\end{align*}
      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. \begin{align*}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}\end{align*}
    3tan22xtan22xtan2x2xx=1=13=±33=π6,5π6,7π6,11π6,13π6,17π6,19π6,23π6=π12,5π12,7π12,11π12,13π12,17π12,19π12,23π12
  9. \begin{align*}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 \end{align*}
    1sinx1111+3=3sinx=sinx+3sinx=sinx(1+3)=sinx
    \begin{align*}\sin^{-1} \left (\frac{1}{1 + \sqrt{3}} \right ) = x\end{align*}sin1(11+3)=x or \begin{align*}x = .3747\end{align*}x=.3747 radians and \begin{align*}x = 2.7669\end{align*}x=2.7669 radians
  10. Because this is \begin{align*}\cos 3x\end{align*}cos3x, 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 \begin{align*}2\pi\end{align*}2π when finding \begin{align*}3x\end{align*}3x. \begin{align*}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} \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, \begin{align*}1 + \tan^2 \theta = \sec^2 \theta\end{align*}1+tan2θ=sec2θ. \begin{align*}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\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 \begin{align*}x\end{align*}x. \begin{align*}\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\end{align*}
    tan2x1tan2xtanxx=0=1=±1=π4+πk and 3π4+πk
    Where \begin{align*}k\end{align*}k is any integer.
  12. Use the half angle formula with \begin{align*}315^\circ\end{align*}315. \begin{align*}\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} \end{align*}
    cos157.5=cos3152=1+cos3152=1+222=2+24=2+22
  13. Use the sine sum formula. \begin{align*}\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}\end{align*}
    sin13π12=sin(10π12+3π12)=sin(5π6+π4)=sin5π6cosπ4+cos5π6sinπ4=1222+(32)22=264
  14. \begin{align*}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\end{align*}
    4(cos5x+cos9x)=4[2cos(5x+9x2)cos(5x9x2)]=8cos7xcos(2x)=8cos7xcos2x
  15. \begin{align*}& \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\end{align*}
    cos(xy)cosysin(xy)sinycosy(cosxcosy+sinxsiny)siny (sinxcosycosxsiny)cosxcos2y+sinxsinycosysinxsinycosy+cosxsin2ycosxcos2y+cosxsin2ycosx(cos2y+sin2y)cosx
  16. Use the sine and cosine sum formulas. \begin{align*}& \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\end{align*}
    sin(4π3x)+cos(x+5π6)sin4π3cosxcos4π3sinx+cosxcos5π6sinxsin5π632cosx+12sinx32cosx12sinx3cosx
  17. Use the sine sum formula as well as the double angle formula. \begin{align*} \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\end{align*}
    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, \begin{align*}\cos^4 x = \left [\frac{1}{2}(\cos 2x + 1) \right ]^2 \end{align*}cos4x=[12(cos2x+1)]2 Now, our final answer needs to be in the first power of cosine, so we need to find a formula for \begin{align*}\cos^2 2x\end{align*}cos22x. For this, we substitute \begin{align*}2x\end{align*}2x everywhere there is an \begin{align*}x\end{align*}x and the formula translates to \begin{align*}\cos^2 2x = \frac{1}{2}(\cos 4x + 1)\end{align*}cos22x=12(cos4x+1). Now we can write \begin{align*}\cos^4 x\end{align*}cos4x in terms of the first power of cosine as follows. \begin{align*}\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} \end{align*}
    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, \begin{align*}\sin^4 x = \left [\frac{1}{2}(1 - \cos 2x) \right ]^2 \end{align*}sin4x=[12(1cos2x)]2 Now, our final answer needs to be in the first power of cosine, so we need to find a formula for \begin{align*}\cos^2 2x\end{align*}cos22x. For this, we substitute \begin{align*}2x\end{align*}2x everywhere there is an \begin{align*}x\end{align*}x and the formula translates to \begin{align*}\cos^2 2x = \frac{1}{2}(\cos 4x + 1)\end{align*}cos22x=12(cos4x+1). Now we can write \begin{align*}\sin^4 x\end{align*}sin4x in terms of the first power of cosine as follows. \begin{align*}\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} \end{align*}
    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: \begin{align*}\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) \end{align*}
    (b) For tangent, we use the identity \begin{align*}\tan x = \frac{\sin x}{\cos x}\end{align*} and then substitute in our new formulas. \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. \begin{align*}\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} \end{align*}

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.

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