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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


Even & Odd Identities


Co-Function Identities


Sum and Difference Identities


Double Angle Identities


Half Angle Identities


Product to Sum & Sum to Product Identities


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 tana<0, then a is in Quadrant II. Therefore seca is negative. To find the third side, we need to do the Pythagorean Theorem.
    So 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 x that are π4+nπ2, where n is a positive integer,, since the original expression is undefined for these values of x.
  4. Change secant and cosecant into terms of sine and cosine, then find a common denominator.
  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 or x=.3747 radians and x=2.7669 radians
  10. Because this is 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 2π when finding 3x.
  11. Rewrite the equation in terms of tan by using the Pythagorean identity, 1+tan2θ=sec2θ.
    Because these factors are the same, we only need to solve one for x.
    tan2x1tan2xtanxx=0=1=±1=π4+πk and 3π4+πk
    Where k is any integer.
  12. Use the half angle formula with 315.
  13. Use the sine sum formula.
  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.
  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 Now, our final answer needs to be in the first power of cosine, so we need to find a formula for cos22x. For this, we substitute 2x everywhere there is an x and the formula translates to cos22x=12(cos4x+1). Now we can write cos4x in terms of the first power of cosine as follows.
  19. Using our new formula, 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 cos22x. For this, we substitute 2x everywhere there is an x and the formula translates to cos22x=12(cos4x+1). Now we can write sin4x in terms of the first power of cosine as follows.
  20. (a) First, we use both of our new formulas, then simplify:
    (b) For tangent, we use the identity tanx=sinxcosx and then substitute in our new formulas. tan42x=sin42xcos42x 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.

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