Chapter 5 - Review Exercises - Page 248: 16

(a) $sin ~\frac{\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}$ $cos ~\frac{\pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4}$ $tan ~\frac{\pi}{12} = 2-\sqrt{3}$ (b) $sin ~\frac{\pi}{12} = \frac{\sqrt{2-\sqrt{3}}}{2}$ $cos ~\frac{\pi}{12} = \frac{\sqrt{2+\sqrt{3}}}{2}$ $tan ~\frac{\pi}{12} = 2-\sqrt{3}$

(a) $sin ~\frac{\pi}{12} = sin~(\frac{\pi}{4}-\frac{\pi}{6})$ $sin ~\frac{\pi}{12} = sin~\frac{\pi}{4}~cos(-\frac{\pi}{6})+cos~\frac{\pi}{4}~sin~(-\frac{\pi}{6})$ $sin ~\frac{\pi}{12} = (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2})+(\frac{\sqrt{2}}{2})(-\frac{1}{2})$ $sin ~\frac{\pi}{12} = \frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}$ $sin ~\frac{\pi}{12} = \frac{\sqrt{6}-\sqrt{2}}{4}$ $cos ~\frac{\pi}{12} = cos~(\frac{\pi}{4}-\frac{\pi}{6})$ $cos ~\frac{\pi}{12} = cos~\frac{\pi}{4}~cos(-\frac{\pi}{6})-sin~\frac{\pi}{4}~sin~(-\frac{\pi}{6})$ $cos ~\frac{\pi}{12} = (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2})-(\frac{\sqrt{2}}{2})(-\frac{1}{2})$ $cos ~\frac{\pi}{12} = \frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}$ $cos ~\frac{\pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4}$ $tan ~\frac{\pi}{12} = tan~(\frac{\pi}{4}-\frac{\pi}{6})$ $tan ~\frac{\pi}{12} = \frac{tan~\frac{\pi}{4}-tan~\frac{\pi}{6}}{1+tan~\frac{\pi}{4}~tan~\frac{\pi}{6}}$ $tan ~\frac{\pi}{12} = \frac{1-\frac{1}{\sqrt{3}}}{1+(1)~(\frac{1}{\sqrt{3}})}$ $tan ~\frac{\pi}{12} = \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}$ $tan ~\frac{\pi}{12} = \frac{\sqrt{3}-1}{\sqrt{3}+1}\cdot \frac{\sqrt{3}-1}{\sqrt{3}-1}$ $tan ~\frac{\pi}{12} = \frac{4-2\sqrt{3}}{2}$ $tan ~\frac{\pi}{12} = 2-\sqrt{3}$ (b) $sin ~\frac{\pi}{12} = sin~(\frac{\pi/6}{2})$ $sin ~\frac{\pi}{12} = \sqrt{\frac{1-cos~\frac{\pi}{6}}{2}}$ $sin ~\frac{\pi}{12} = \sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}$ $sin ~\frac{\pi}{12} = \sqrt{\frac{2-\sqrt{3}}{4}}$ $sin ~\frac{\pi}{12} = \frac{\sqrt{2-\sqrt{3}}}{2}$ $cos ~\frac{\pi}{12} = cos~(\frac{\pi/6}{2})$ $cos ~\frac{\pi}{12} = \sqrt{\frac{1+cos~\frac{\pi}{6}}{2}}$ $cos ~\frac{\pi}{12} = \sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}$ $cos ~\frac{\pi}{12} = \sqrt{\frac{2+\sqrt{3}}{4}}$ $cos ~\frac{\pi}{12} = \frac{\sqrt{2+\sqrt{3}}}{2}$ $tan ~\frac{\pi}{12} = tan~(\frac{\pi/6}{2})$ $tan ~\frac{\pi}{12} = \frac{sin~\frac{\pi}{6}}{1+cos~\frac{\pi}{6}}$ $tan ~\frac{\pi}{12} = \frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}$ $tan ~\frac{\pi}{12} = \frac{\frac{1}{2}}{\frac{2+\sqrt{3}}{2}}$ $tan ~\frac{\pi}{12} = \frac{1}{2+\sqrt{3}}\cdot \frac{2-\sqrt{3}}{2-\sqrt{3}}$ $tan ~\frac{\pi}{12} = \frac{2-\sqrt{3}}{1}$ $tan ~\frac{\pi}{12} = 2-\sqrt{3}$