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أدلة الدراسة > Precalculus II

Solutions for Double-Angle, Half-Angle, and Reduction Formulas

Solutions to Try Its

1. cos⁡(2α)=732\cos \left(2\alpha \right)=\frac{7}{32} 2. cos⁡4θ−sin⁡4θ=(cos⁡2θ+sin⁡2θ)(cos⁡2θ−sin⁡2θ)=cos⁡(2θ){\cos }^{4}\theta -{\sin }^{4}\theta =\left({\cos }^{2}\theta +{\sin }^{2}\theta \right)\left({\cos }^{2}\theta -{\sin }^{2}\theta \right)=\cos \left(2\theta \right) 3. cos⁡(2θ)cos⁡θ=(cos⁡2θ−sin⁡2θ)cos⁡θ=cos⁡3θ−cos⁡θsin⁡2θ\cos \left(2\theta \right)\cos \theta =\left({\cos }^{2}\theta -{\sin }^{2}\theta \right)\cos \theta ={\cos }^{3}\theta -\cos \theta {\sin }^{2}\theta 4. 10cos⁡4x=10cos⁡4x=10(cos⁡2x)2 =10[1+cos⁡(2x)2]2Substitute reduction formula for cos2x. =104[1+2cos⁡(2x)+cos⁡2(2x)] =104+102cos⁡(2x)+104(1+cos⁡2(2x)2)Substitute reduction formula for cos2x. =104+102cos⁡(2x)+108+108cos⁡(4x) =308+5cos⁡(2x)+108cos⁡(4x) =154+5cos⁡(2x)+54cos⁡(4x)\begin{array}{ll}10{\cos }^{4}x=10{\cos }^{4}x=10{\left({\cos }^{2}x\right)}^{2}\hfill & \hfill \\ \text{ }=10{\left[\frac{1+\cos \left(2x\right)}{2}\right]}^{2}\hfill & {\text{Substitute reduction formula for cos}}^{2}x.\hfill \\ \text{ }=\frac{10}{4}\left[1+2\cos \left(2x\right)+{\cos }^{2}\left(2x\right)\right]\hfill & \hfill \\ \text{ }=\frac{10}{4}+\frac{10}{2}\cos \left(2x\right)+\frac{10}{4}\left(\frac{1+\cos 2\left(2x\right)}{2}\right)\hfill & {\text{Substitute reduction formula for cos}}^{2}x.\hfill \\ \text{ }=\frac{10}{4}+\frac{10}{2}\cos \left(2x\right)+\frac{10}{8}+\frac{10}{8}\cos \left(4x\right)\hfill & \hfill \\ \text{ }=\frac{30}{8}+5\cos \left(2x\right)+\frac{10}{8}\cos \left(4x\right)\hfill & \hfill \\ \text{ }=\frac{15}{4}+5\cos \left(2x\right)+\frac{5}{4}\cos \left(4x\right)\hfill & \hfill \end{array} 5. −25-\frac{2}{\sqrt{5}}

Solution to Odd-Numbered Exercises

1. Use the Pythagorean identities and isolate the squared term. 3. 1−cos⁡xsin⁡x,sin⁡x1+cos⁡x\frac{1-\cos x}{\sin x},\frac{\sin x}{1+\cos x}, multiplying the top and bottom by 1−cos⁡x\sqrt{1-\cos x} and 1+cos⁡x\sqrt{1+\cos x}, respectively. 5. a) 3732\frac{3\sqrt{7}}{32} b) 3132\frac{31}{32} c) 3731\frac{3\sqrt{7}}{31} 7. a) 32\frac{\sqrt{3}}{2} b) −12-\frac{1}{2} c) −3-\sqrt{3} 9. cos⁡θ=−255,sin⁡θ=55,tan⁡θ=−12,csc⁡θ=5,sec⁡θ=−52,cot⁡θ=−2\cos \theta =-\frac{2\sqrt{5}}{5},\sin \theta =\frac{\sqrt{5}}{5},\tan \theta =-\frac{1}{2},\csc \theta =\sqrt{5},\sec \theta =-\frac{\sqrt{5}}{2},\cot \theta =-2 11. 2sin⁡(π2)2\sin \left(\frac{\pi }{2}\right) 13. 2−22\frac{\sqrt{2-\sqrt{2}}}{2} 15. 2−32\frac{\sqrt{2-\sqrt{3}}}{2} 17. 2+32+\sqrt{3} 19. −1−2-1-\sqrt{2} 21. a) 31313\frac{3\sqrt{13}}{13} b) −21313-\frac{2\sqrt{13}}{13} c) −32-\frac{3}{2} 23. a) 104\frac{\sqrt{10}}{4} b) 64\frac{\sqrt{6}}{4} c) 153\frac{\sqrt{15}}{3} 25. 120169,−119169,−120119\frac{120}{169},-\frac{119}{169},-\frac{120}{119} 27. 21313,31313,23\frac{2\sqrt{13}}{13},\frac{3\sqrt{13}}{13},\frac{2}{3} 29. cos⁡(74∘)\cos \left({74}^{\circ }\right) 31. cos⁡(18x)\cos \left(18x\right) 33. 3sin⁡(10x)3\sin \left(10x\right) 35. −2sin⁡(−x)cos⁡(−x)=−2(−sin⁡(x)cos⁡(x))=sin⁡(2x)-2\sin \left(-x\right)\cos \left(-x\right)=-2\left(-\sin \left(x\right)\cos \left(x\right)\right)=\sin \left(2x\right) 37. sin⁡(2θ)1+cos⁡(2θ)tan⁡2θ=2sin⁡(θ)cos⁡(θ)1+cos⁡2θ−sin⁡2θtan⁡2θ=2sin⁡(θ)cos⁡(θ)2cos⁡2θtan⁡2θ=sin⁡(θ)cos⁡θtan⁡2θ=cot⁡(θ)tan⁡2θ=tan⁡θ\begin{array}{l}\frac{\sin \left(2\theta \right)}{1+\cos \left(2\theta \right)}{\tan }^{2}\theta =\frac{2\sin \left(\theta \right)\cos \left(\theta \right)}{1+{\cos }^{2}\theta -{\sin }^{2}\theta }{\tan }^{2}\theta =\\ \frac{2\sin \left(\theta \right)\cos \left(\theta \right)}{2{\cos }^{2}\theta }{\tan }^{2}\theta =\frac{\sin \left(\theta \right)}{\cos \theta }{\tan }^{2}\theta =\\ \cot \left(\theta \right){\tan }^{2}\theta =\tan \theta \end{array} 39. 1+cos⁡(12x)2\frac{1+\cos \left(12x\right)}{2} 41. 3+cos⁡(12x)−4cos⁡(6x)8\frac{3+\cos \left(12x\right)-4\cos \left(6x\right)}{8} 43. 2+cos⁡(2x)−2cos⁡(4x)−cos⁡(6x)32\frac{2+\cos \left(2x\right)-2\cos \left(4x\right)-\cos \left(6x\right)}{32} 45. 3+cos⁡(4x)−4cos⁡(2x)3+cos⁡(4x)+4cos⁡(2x)\frac{3+\cos \left(4x\right)-4\cos \left(2x\right)}{3+\cos \left(4x\right)+4\cos \left(2x\right)} 47. 1−cos⁡(4x)8\frac{1-\cos \left(4x\right)}{8} 49. 3+cos⁡(4x)−4cos⁡(2x)4(cos⁡(2x)+1)\frac{3+\cos \left(4x\right)-4\cos \left(2x\right)}{4\left(\cos \left(2x\right)+1\right)} 51. (1+cos⁡(4x))sin⁡x2\frac{\left(1+\cos \left(4x\right)\right)\sin x}{2} 53. 4sin⁡xcos⁡x(cos⁡2x−sin⁡2x)4\sin x\cos x\left({\cos }^{2}x-{\sin }^{2}x\right) 55. 2tan⁡x1+tan⁡2x=2sin⁡xcos⁡x1+sin⁡2xcos⁡2x=2sin⁡xcos⁡xcos⁡2x+sin⁡2xcos⁡2x=\frac{2\tan x}{1+{\tan }^{2}x}=\frac{\frac{2\sin x}{\cos x}}{1+\frac{{\sin }^{2}x}{{\cos }^{2}x}}=\frac{\frac{2\sin x}{\cos x}}{\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}}= 2sin⁡xcos⁡x.cos⁡2x1=2sin⁡xcos⁡x=sin⁡(2x)\frac{2\sin x}{\cos x}.\frac{{\cos }^{2}x}{1}=2\sin x\cos x=\sin \left(2x\right) 57. 2sin⁡xcos⁡x2cos⁡2x−1=sin⁡(2x)cos⁡(2x)=tan⁡(2x)\frac{2\sin x\cos x}{2{\cos }^{2}x - 1}=\frac{\sin \left(2x\right)}{\cos \left(2x\right)}=\tan \left(2x\right) 59. sin⁡(x+2x)=sin⁡xcos⁡(2x)+sin⁡(2x)cos⁡x=sin⁡x(cos⁡2x−sin⁡2x)+2sin⁡xcos⁡xcos⁡x=sin⁡xcos⁡2x−sin⁡3x+2sin⁡xcos⁡2x=3sin⁡xcos⁡2x−sin⁡3x\begin{array}{l}\sin \left(x+2x\right)=\sin x\cos \left(2x\right)+\sin \left(2x\right)\cos x\hfill \\ =\sin x\left({\cos }^{2}x-{\sin }^{2}x\right)+2\sin x\cos x\cos x\hfill \\ =\sin x{\cos }^{2}x-{\sin }^{3}x+2\sin x{\cos }^{2}x\hfill \\ =3\sin x{\cos }^{2}x-{\sin }^{3}x\hfill \end{array} 61. 1+cos⁡(2t)sin⁡(2t)−cos⁡t=1+2cos⁡2t−12sin⁡tcos⁡t−cos⁡t=2cos⁡2tcos⁡t(2sin⁡t−1)=2cos⁡t2sin⁡t−1\begin{array}{l}\frac{1+\cos \left(2t\right)}{\sin \left(2t\right)-\cos t}=\frac{1+2{\cos }^{2}t - 1}{2\sin t\cos t-\cos t}\hfill \\ =\frac{2{\cos }^{2}t}{\cos t\left(2\sin t - 1\right)}\hfill \\ =\frac{2\cos t}{2\sin t - 1}\hfill \end{array} 63. (cos⁡2(4x)−sin⁡2(4x)−sin⁡(8x))(cos⁡2(4x)−sin⁡2(4x)+sin⁡(8x))= =(cos⁡(8x)−sin⁡(8x))(cos⁡(8x)+sin⁡(8x)) =cos⁡2(8x)−sin⁡2(8x) =cos⁡(16x)\begin{array}{l}\left({\cos }^{2}\left(4x\right)-{\sin }^{2}\left(4x\right)-\sin \left(8x\right)\right)\left({\cos }^{2}\left(4x\right)-{\sin }^{2}\left(4x\right)+\sin \left(8x\right)\right)=\hfill \\ \text{ }=\left(\cos \left(8x\right)-\sin \left(8x\right)\right)\left(\cos \left(8x\right)+\sin \left(8x\right)\right)\hfill \\ \text{ }={\cos }^{2}\left(8x\right)-{\sin }^{2}\left(8x\right)\hfill \\ \text{ }=\cos \left(16x\right)\hfill \\ \hfill \end{array}

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