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Study Guides > Precalculus II

Solutions for Solving Trigonometric Equations with Identities

Solutions to Try Its

1. cscθcosθtanθ=(1sinθ)cosθ(sinθcosθ) =cosθsinθ(sinθcosθ) =sinθcosθsinθcosθ =1\begin{array}{l}\csc \theta \cos \theta \tan \theta =\left(\frac{1}{\sin \theta }\right)\cos \theta \left(\frac{\sin \theta }{\cos \theta }\right)\hfill \\ \text{ }=\frac{\cos \theta }{\sin \theta }\left(\frac{\sin \theta }{\cos \theta }\right)\hfill \\ \text{ }=\frac{\sin \theta \cos \theta }{\sin \theta \cos \theta }\hfill \\ \text{ }=1\hfill \end{array} 2. cotθcscθ=cosθsinθ1sinθ =cosθsinθsinθ1 =cosθ\begin{array}{l}\frac{\cot \theta }{\csc \theta }=\frac{\frac{\cos \theta }{\sin \theta }}{\frac{1}{\sin \theta }}\hfill \\ \text{ }=\frac{\cos \theta }{\sin \theta }\cdot \frac{\sin \theta }{1}\hfill \\ \text{ }=\cos \theta \hfill \end{array} 3. sin2θ1tanθsinθtanθ=(sinθ+1)(sinθ1)tanθ(sinθ1)=sinθ+1tanθ\begin{array}{c}\frac{{\sin }^{2}\theta -1}{\tan \theta \sin \theta -\tan \theta }=\frac{\left(\sin \theta +1\right)\left(\sin \theta -1\right)}{\tan \theta \left(\sin \theta -1\right)}\\ =\frac{\sin \theta +1}{\tan \theta }\end{array} 4. This is a difference of squares formula: 259sin2θ=(53sinθ)(5+3sinθ)25 - 9{\sin }^{2}\theta =\left(5 - 3\sin \theta \right)\left(5+3\sin \theta \right). 5. cosθ1+sinθ(1sinθ1sinθ)=cosθ(1sinθ)1sin2θ =cosθ(1sinθ)cos2θ =1sinθcosθ\begin{array}{l}\frac{\cos \theta }{1+\sin \theta }\left(\frac{1-\sin \theta }{1-\sin \theta }\right)=\frac{\cos \theta \left(1-\sin \theta \right)}{1-{\sin }^{2}\theta }\hfill \\ \text{ }=\frac{\cos \theta \left(1-\sin \theta \right)}{{\cos }^{2}\theta }\hfill \\ \text{ }=\frac{1-\sin \theta }{\cos \theta }\hfill \end{array}

Solutions to Odd-Numbered Exercises

1. All three functions, F,GF,G, and HH, 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)F\left(-x\right)=\sin \left(-x\right)\sin \left(-x\right)=\left(-\sin x\right)\left(-\sin x\right)={\sin }^{2}x=F\left(x\right),G\left(-x\right)=\cos \left(-x\right)\cos \left(-x\right)=\cos x\cos x={\cos }^{2}x=G\left(x\right) and H(x)=tan(x)tan(x)=(tanx)(tanx)=tan2x=H(x)H\left(-x\right)=\tan \left(-x\right)\tan \left(-x\right)=\left(-\tan x\right)\left(-\tan x\right)={\tan }^{2}x=H\left(x\right). 3. When cost=0\cos t=0, then sect=10\sec t=\frac{1}{0}, which is undefined. 5. sinx\sin x 7. secx\sec x 9. csct\csc t 11. 1-1 13. sec2x{\sec }^{2}x 15. sin2x+1{\sin }^{2}x+1 17. 1sinx\frac{1}{\sin x} 19. 1cotx\frac{1}{\cot x} 21. tanx\tan x 23. 4secxtanx-4\sec x\tan x 25. ±1cot2x+1\pm \sqrt{\frac{1}{{\cot }^{2}x}+1} 27. ±1sin2xsinx\frac{\pm \sqrt{1-{\sin }^{2}x}}{\sin x} 29. Answers will vary. Sample proof: cosxcos3x=cosx(1cos2x)\cos x-{\cos }^{3}x=\cos x\left(1-{\cos }^{2}x\right) =cosxsin2x=\cos x{\sin }^{2}x 31. Answers will vary. Sample proof: 1+sin2xcos2x=1cos2x+sin2xcos2x=sec2x+tan2x=tan2x+1+tan2x=1+2tan2x\frac{1+{\sin }^{2}x}{{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}+\frac{{\sin }^{2}x}{{\cos }^{2}x}={\sec }^{2}x+{\tan }^{2}x={\tan }^{2}x+1+{\tan }^{2}x=1+2{\tan }^{2}x 33. Answers will vary. Sample proof: cos2xtan2x=1sin2x(sec2x1)=1sin2xsec2x+1=2sin2xsec2x{\cos }^{2}x-{\tan }^{2}x=1-{\sin }^{2}x-\left({\sec }^{2}x - 1\right)=1-{\sin }^{2}x-{\sec }^{2}x+1=2-{\sin }^{2}x-{\sec }^{2}x 35. False 37. False 39. Proved with negative and Pythagorean identities 41. True 3sin2θ+4cos2θ=3sin2θ+3cos2θ+cos2θ=3(sin2θ+cos2θ)+cos2θ=3+cos2θ3{\sin }^{2}\theta +4{\cos }^{2}\theta =3{\sin }^{2}\theta +3{\cos }^{2}\theta +{\cos }^{2}\theta =3\left({\sin }^{2}\theta +{\cos }^{2}\theta \right)+{\cos }^{2}\theta =3+{\cos }^{2}\theta

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