Solutions to Odd-Numbered Answers

1. The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures $x$, the second angle measures $\frac{\pi }{2}-x$. Then $\sin x=\cos \left(\frac{\pi }{2}-x\right)$. The same holds for the other cofunction identities. The key is that the angles are complementary.

3. $\sin \left(-x\right)=-\sin x$, so $\sin x$ is odd. $\cos \left(-x\right)=\cos \left(0-x\right)=\cos x$, so $\cos x$ is even.

5. $\frac{\sqrt{2}+\sqrt{6}}{4}$

7. $\frac{\sqrt{6}-\sqrt{2}}{4}$

9. $-2-\sqrt{3}$

11. $-\frac{\sqrt{2}}{2}\sin x-\frac{\sqrt{2}}{2}\cos x$

13. $-\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x$

15. $\csc \theta$

17. $\cot x$

19. $\tan \left(\frac{x}{10}\right)$

21. $\sin \left(a-b\right)=\left(\frac{4}{5}\right)\left(\frac{1}{3}\right)-\left(\frac{3}{5}\right)\left(\frac{2\sqrt{2}}{3}\right)=\frac{4 - 6\sqrt{2}}{15}$
$\cos \left(a+b\right)=\left(\frac{3}{5}\right)\left(\frac{1}{3}\right)-\left(\frac{4}{5}\right)\left(\frac{2\sqrt{2}}{3}\right)=\frac{3 - 8\sqrt{2}}{15}$

23. $\frac{\sqrt{2}-\sqrt{6}}{4}$

25. $\sin x$

27. $\cot \left(\frac{\pi }{6}-x\right)$

29. $\cot \left(\frac{\pi }{4}+x\right)$

31. $\frac{\sin x}{\sqrt{2}}+\frac{\cos x}{\sqrt{2}}$

33. They are the same.

35. They are the different, try $g\left(x\right)=\sin \left(9x\right)-\cos \left(3x\right)\sin \left(6x\right)$.

37. They are the same.

39. They are the different, try $g\left(\theta \right)=\frac{2\tan \theta }{1-{\tan }^{2}\theta }$.

41. They are different, try $g\left(x\right)=\frac{\tan x-\tan \left(2x\right)}{1+\tan x\tan \left(2x\right)}$.

43. $-\frac{\sqrt{3}-1}{2\sqrt{2}},\text{ or }-0.2588$

45. $\frac{1+\sqrt{3}}{2\sqrt{2}}$, or 0.9659

47. $\begin{array}{c}\tan \left(x+\frac{\pi }{4}\right)=\\ \frac{\tan x+\tan \left(\frac{\pi }{4}\right)}{1-\tan x\tan \left(\frac{\pi }{4}\right)}=\\ \frac{\tan x+1}{1-\tan x\left(1\right)}=\frac{\tan x+1}{1-\tan x}\end{array}$

49. $\begin{array}{c}\frac{\cos \left(a+b\right)}{\cos a\cos b}=\\ \frac{\cos a\cos b}{\cos a\cos b}-\frac{\sin a\sin b}{\cos a\cos b}=1-\tan a\tan b\end{array}$

51. $\begin{array}{c}\frac{\cos \left(x+h\right)-\cos x}{h}=\\ \frac{\cos x\mathrm{cosh}-\sin x\mathrm{sinh}-\cos x}{h}=\\ \frac{\cos x\left(\mathrm{cosh}-1\right)-\sin x\mathrm{sinh}}{h}=\cos x\frac{\cos h - 1}{h}-\sin x\frac{\sin h}{h}\end{array}$

53. True

55. True. Note that $\sin \left(\alpha +\beta \right)=\sin \left(\pi -\gamma \right)$ and expand the right hand side.