Question

Using Trigonometric functions of $\frac{\pi }{4}$. Find the value of $\sin \frac{\pi }{8}=?$

Answer 1

$\cos 2A=1-2{\sin }^{2}A$

$⟹\cos A=1-2{\sin }^2\frac{A}{2}$

Putting, $A=\frac{\pi }{4}$

$⟹\cos \frac{\pi }{4}=1-2{\sin }^2\frac{\pi }{8}$

$⟹\frac{1}{\sqrt{2}}=1-2{\sin }^2\frac{\pi }{8}(\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}})$

$⟹2{\sin }^2\frac{\pi }{8}=1-\frac{1}{\sqrt{2}}$

$⟹{\sin }^2\frac{\pi }{8}=\frac{\sqrt{2}-1}{2\sqrt{2}}$

$\therefore \sin \frac{\pi }{8}=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}$