Question

The value of ${\sin }^4\frac{\pi }{8}+{\sin }^4\frac{3\pi }{8}+{\sin }^4\frac{5\pi }{8}+{\sin }^4\frac{7\pi }{8}$ is

• A. $\frac{\sqrt{3}}{4}$
• B. $\frac{3}{4}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{3}{2}$

Answer 1

Answer: (D)

We have, $\frac{7\pi }{8}=\pi -\frac{\pi }{8}$
and $\frac{5\pi }{8}=\pi -\frac{3\pi }{8}$
$\Rightarrow \sin \frac{7\pi }{8}=\sin \left(\pi -\frac{\pi }{8}\right)$
and $\sin \frac{5\pi }{8}=\sin \left(\pi -\frac{3\pi }{8}\right)$
$\Rightarrow \sin \frac{7\pi }{8}=\sin \frac{\pi }{8}$ and $\sin \frac{5\pi }{8}=\sin \frac{3\pi }{8}$
$\Rightarrow {\sin }^4\left(\frac{7\pi }{8}\right)={\sin }^4\left(\frac{\pi }{8}\right)$
and ${\sin }^4\left(\frac{5\pi }{8}\right)={\sin }^4\left(\frac{3\pi }{8}\right)$
Now, ${\sin }^4\left(\frac{\pi }{8}\right)+{\sin }^4\left(\frac{3\pi }{8}\right)+{\sin }^4\left(\frac{5\pi }{8}\right)+{\sin }^4\left(\frac{7\pi }{8}\right)$
$={\sin }^4\left(\frac{\pi }{8}\right)+{\sin }^4\left(\frac{3\pi }{8}\right)+{\sin }^4\left(\frac{3\pi }{8}\right)+{\sin }^4\left(\frac{\pi }{8}\right)$
$=2{\sin }^4\left(\frac{\pi }{8}\right)+2{\sin }^4\left(\frac{3\pi }{8}\right)$
$=2\left[{\left({\sin }^2\frac{\pi }{8}\right)}^2+{\left({\sin }^2\left(\frac{3\pi }{8}\right)\right)}^2\right]$
$=2\left[{\left(\frac{1-\cos 2\left(\frac{\pi }{8}\right)}{2}\right)}^2+{\left(\frac{1-\cos 2\left(\frac{3\pi }{8}\right)}{2}\right)}^2\right]$
$=2\left[\frac{{\left(1-\cos \frac{\pi }{4}\right)}^2}{4}+\frac{{\left(1-\cos \frac{3\pi }{4}\right)}^2}{4}\right]$
$=\frac{1}{2}\left[{\left(1-\frac{1}{\sqrt{2}}\right)}^2+{\left(1+\frac{1}{\sqrt{2}}\right)}^2\right]$
$\left[\because \cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}\text{and}\cos \frac{3\pi }{4}=\frac{-1}{\sqrt{2}}\right]$
$=\frac{1}{2}\left[1+\frac{1}{2}-\frac{2}{\sqrt{2}}+1+\frac{1}{2}+\frac{2}{\sqrt{2}}\right]$
$=\frac{1}{2}\left[2+\frac{2}{2}\right]=\frac{1}{2}\left[2+1\right]=\frac{3}{2}$