Solutions to Odd-Numbered Exercises

1. Substitute $\alpha$ into cosine and $\beta$ into sine and evaluate.

3. Answers will vary. There are some equations that involve a sum of two trig expressions where when converted to a product are easier to solve. For example: $\frac{\sin \left(3x\right)+\sin x}{\cos x}=1$. When converting the numerator to a product the equation becomes: $\frac{2\sin \left(2x\right)\cos x}{\cos x}=1\\$

5. $8\left(\cos \left(5x\right)-\cos \left(27x\right)\right)$

7. $\sin \left(2x\right)+\sin \left(8x\right)$

9. $\frac{1}{2}\left(\cos \left(6x\right)-\cos \left(4x\right)\right)$

11. $2\cos \left(5t\right)\cos t$

13. $2\cos \left(7x\right)$

15. $2\cos \left(6x\right)\cos \left(3x\right)$

17. $\frac{1}{4}\left(1+\sqrt{3}\right)$

19. $\frac{1}{4}\left(\sqrt{3}-2\right)$

21. $\frac{1}{4}\left(\sqrt{3}-1\right)$

23. $\cos \left(80^\circ \right)-\cos \left(120^\circ \right)$

25. $\frac{1}{2}\left(\sin \left(221^\circ \right)+\sin \left(205^\circ \right)\right)$

27. $\sqrt{2}\cos \left(31^\circ \right)$

29. $2\cos \left(66.5^\circ \right)\sin \left(34.5^\circ \right)$

31. $2\sin \left(-1.5^\circ \right)\cos \left(0.5^\circ \right)$

33. ${2}\sin \left({7x}\right){-2}\sin{ x}={ 2}\sin \left({4x}+{ 3x }\right)-{ 2 }\sin\left({4x } - { 3x }\right)=\\ {2}\left(\sin\left({ 4x }\right)\cos\left({ 3x }\right)+\sin\left({ 3x }\right)\cos\left({ 4x }\right)\right)-{ 2 }\left(\sin\left({ 4x }\right)\cos\left({ 3x }\right)-\sin \left({ 3x }\right)\cos\left({ 4x }\right)\right)=\\{2}\sin\left({ 4x }\right)\cos\left({ 3x }\right)+{2}\sin\left({ 3x }\right)\cos\left({ 4x }\right)-{ 2 }\sin\left({ 4x }\right)\cos\left({ 3x }\right)+{ 2 }\sin\left({ 3x }\right)\cos\left({ 4x }\right)=\\{ 4 }\sin\left({ 3x }\right)\cos\left({ 4x }\right)\\$

35. $\sin x+\sin \left(3x\right)=2\sin \left(\frac{4x}{2}\right)\cos \left(\frac{-2x}{2}\right)=$
$2\sin \left(2x\right)\cos x=2\left(2\sin x\cos x\right)\cos x=$
$4\sin x{\cos }^{2}x$

37. $2\tan x\cos \left(3x\right)=\frac{2\sin x\cos \left(3x\right)}{\cos x}=\frac{2\left(.5\left(\sin \left(4x\right)-\sin \left(2x\right)\right)\right)}{\cos x}$
$=\frac{1}{\cos x}\left(\sin \left(4x\right)-\sin \left(2x\right)\right)=\sec x\left(\sin \left(4x\right)-\sin \left(2x\right)\right)$

39. $2\cos \left({35}^{\circ }\right)\cos \left({23}^{\circ }\right),\text{ 1}\text{.5081}$

41. $-2\sin \left({33}^{\circ }\right)\sin \left({11}^{\circ }\right),\text{ }-0.2078$

43. $\frac{1}{2}\left(\cos \left({99}^{\circ }\right)-\cos \left({71}^{\circ }\right)\right),\text{ }-0.2410$

45. It is an identity.

47. It is not an identity, but $2{\cos }^{3}x$ is.

49. $\tan \left(3t\right)$

51. $2\cos \left(2x\right)$

53. $-\sin \left(14x\right)$

55. Start with $\cos x+\cos y$. Make a substitution and let $x=\alpha +\beta$ and let $y=\alpha -\beta$, so $\cos x+\cos y$ becomes

$\cos \left(\alpha +\beta \right)+\cos \left(\alpha -\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta +\cos \alpha \cos \beta +\sin \alpha \sin \beta =2\cos \alpha \cos \beta$

Since $x=\alpha +\beta$ and $y=\alpha -\beta$, we can solve for $\alpha$ and $\beta$ in terms of x and y and substitute in for $2\cos \alpha \cos \beta$ and get $2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)$.

57. $\frac{\cos \left(3x\right)+\cos x}{\cos \left(3x\right)-\cos x}=\frac{2\cos \left(2x\right)\cos x}{-2\sin \left(2x\right)\sin x}=-\cot \left(2x\right)\cot x$

59. $\begin{array}{l}\frac{\cos \left(2y\right)-\cos \left(4y\right)}{\sin \left(2y\right)+\sin \left(4y\right)}=\frac{-2\sin \left(3y\right)\sin \left(-y\right)}{2\sin \left(3y\right)\cos y}=\\ \frac{2\sin \left(3y\right)\sin \left(y\right)}{2\sin \left(3y\right)\cos y}=\tan y\end{array}$

61. $\begin{array}{l}\cos x-\cos \left(3x\right)=-2\sin \left(2x\right)\sin \left(-x\right)=\\ 2\left(2\sin x\cos x\right)\sin x=4{\sin }^{2}x\cos x\end{array}$

63. $\tan \left(\frac{\pi }{4}-t\right)=\frac{\tan \left(\frac{\pi }{4}\right)-\tan t}{1+\tan \left(\frac{\pi }{4}\right)\tan \left(t\right)}=\frac{1-\tan t}{1+\tan t}$