Solutions for Solving Trigonometric Equations with Identities
Solutions to Try Its
1. cscθcosθtanθ=(sinθ1)cosθ(cosθsinθ)=sinθcosθ(cosθsinθ)=sinθcosθsinθcosθ=1
2. cscθcotθ=sinθ1sinθcosθ=sinθcosθ⋅1sinθ=cosθ
3. tanθsinθ−tanθsin2θ−1=tanθ(sinθ−1)(sinθ+1)(sinθ−1)=tanθsinθ+1
4. This is a difference of squares formula: 25−9sin2θ=(5−3sinθ)(5+3sinθ).
5. 1+sinθcosθ(1−sinθ1−sinθ)=1−sin2θcosθ(1−sinθ)=cos2θcosθ(1−sinθ)=cosθ1−sinθ
Solutions to Odd-Numbered Exercises
1. All three functions, F,G, and H, are even.
This is because F(−x)=sin(−x)sin(−x)=(−sinx)(−sinx)=sin2x=F(x),G(−x)=cos(−x)cos(−x)=cosxcosx=cos2x=G(x) and H(−x)=tan(−x)tan(−x)=(−tanx)(−tanx)=tan2x=H(x).
3. When cost=0, then sect=01, which is undefined.
5. sinx
7. secx
9. csct
11. −1
13. sec2x
15. sin2x+1
17. sinx1
19. cotx1
21. tanx
23. −4secxtanx
25. ±cot2x1+1
27. sinx±1−sin2x
29. Answers will vary. Sample proof:
cosx−cos3x=cosx(1−cos2x)=cosxsin2x
31. Answers will vary. Sample proof:
cos2x1+sin2x=cos2x1+cos2xsin2x=sec2x+tan2x=tan2x+1+tan2x=1+2tan2x
33. Answers will vary. Sample proof:
cos2x−tan2x=1−sin2x−(sec2x−1)=1−sin2x−sec2x+1=2−sin2x−sec2x
35. False
37. False
39. Proved with negative and Pythagorean identities
41. True
3sin2θ+4cos2θ=3sin2θ+3cos2θ+cos2θ=3(sin2θ+cos2θ)+cos2θ=3+cos2θ