Question

The value of ${\sin }^4\frac{\pi }{16}+{\sin }^4\frac{3\pi }{16}+{\sin }^4\frac{5\pi }{16}+{\sin }^4\frac{7\pi }{16}$ is

• A. $\frac{1}{2}$
• B. $1$
• C. $\frac{3}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $\frac{3}{2}$

Let
$y={\sin }^4\frac{\pi }{16}+{\sin }^4\frac{3\pi }{16}+{\sin }^4\frac{5\pi }{16}+{\sin }^4\frac{7\pi }{16}$

$\frac{7\pi }{16}=\frac{\pi }{2}-\frac{\pi }{16}$
$\frac{5\pi }{16}=\frac{\pi }{2}-\frac{3\pi }{16}$
$y={\sin }^4\frac{\pi }{16}+{\cos }^4\frac{\pi }{16}+{\sin }^4\frac{3\pi }{16}+{\cos }^4\frac{3\pi }{16}$
Now ${\sin }^4\frac{\pi }{16}+{\cos }^4\frac{\pi }{16}=1-2{\sin }^2\frac{\pi }{16}{\cos }^2\frac{\pi }{16}=1-\frac{1}{2}{\sin }^2\frac{\pi }{8}$
Similarly, ${\sin }^4\frac{3\pi }{16}+{\cos }^4\frac{3\pi }{16}=1-\frac{1}{2}{\sin }^2\frac{3\pi }{8}$
But $\sin \frac{3\pi }{8}=\sin (\frac{\pi }{2}-\frac{\pi }{8})=\cos \frac{\pi }{8}$

$y=2-\frac{1}{2}({\sin }^2\frac{\pi }{8}+{\cos }^2\frac{\pi }{8})$$=\frac{3}{2}$
Hence, option 'C' is correct.