Question

Find the value of $\cos \pi /8$.

• A. $\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}$
• B. $\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}$
• C. $\sqrt{\frac{\sqrt{2}+1}{2\sqrt{3}}}$
• D. $\sqrt{\frac{\sqrt{2}+1}{\sqrt{2}}}$

Answer 1

The correct option is B $\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}$

Given:

Let $\frac{\pi }{8}=A$ and 2A =$\frac{\pi }{4}$.

We know that, $\cos 2A=2{\cos }^{2}A-1$

Also, $\cos \left(\frac{\pi }{4}\right)=2{\cos }^{2}A-1$.

${\cos }^{2}A=1+\frac{1}{\sqrt{2}}=\frac{\sqrt{2}+1}{\sqrt{2}}$

${\cos }^{2}A=\sqrt{2}\left(\frac{\sqrt{2}+1}{2}\right)$

Multiply numerator and denominator by $\sqrt{2}$.

${\cos }^{2}A=\left(\frac{2+\sqrt{2}}{4}\right)$

$\cos A=\sqrt{\left(\frac{2+\sqrt{2}}{4}\right)}$

$\cos \left(\frac{\pi }{8}\right)=\frac{\sqrt{2+\sqrt{2}}}{2}$

Thus, $\cos \left(\frac{\pi }{8}\right)=\frac{\sqrt{2+\sqrt{2}}}{2}$.