Question

The value of $tan\frac{\pi }{16}$

• A. $\sqrt{4+2\sqrt{2}}+(\sqrt{2}+1)$
• B. $\sqrt{4+2\sqrt{2}}-(\sqrt{2}+1)$
• C. $\sqrt{4-2\sqrt{2}}-(\sqrt{2}+1)$
• D. $\sqrt{4+2\sqrt{2}}-(\sqrt{2}-1)$

Answer 1

The correct option is B $\sqrt{4+2\sqrt{2}}-(\sqrt{2}+1)$

$\tan \frac{\pi }{16}=\frac{\sin \frac{\pi }{16}}{\cos \frac{\pi }{16}}$

$=\frac{2{\sin }^2\frac{\pi }{16}}{2\sin \frac{\pi }{16}\cos \frac{\pi }{16}}$

$=\frac{1-\cos \frac{\pi }{8}}{\sin \frac{\pi }{8}}$ $(\because \sin 2x=2\sin x\cos x,\cos 2x=1-2{\sin }^{2}x)$

$=\frac{1-\sqrt{\frac{1+\cos \frac{\pi }{4}}{2}}}{\sqrt{\frac{1-\cos \frac{\pi }{4}}{2}}}$

$=\frac{1-\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}}{\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}}$

$=\frac{\sqrt{2\sqrt{2}}-\sqrt{\sqrt{2}+1}}{\sqrt{\sqrt{2}-1}}$

$=\frac{\sqrt{2\sqrt{2}(\sqrt{2}+1)}-(\sqrt{2}+1)}{\sqrt{(\sqrt{2}-1)(\sqrt{2}+1)}}$

$=\sqrt{4+2\sqrt{2}}-(\sqrt{2}+1)$