Question

Pls solve this with explaining​

Answer 1

$\sin^4\dfrac {\pi}{8}+\sin^4\dfrac {3\pi}{8}+\sin^4\dfrac {5\pi}{8}+\sin^4\dfrac {7\pi}{8}\\\\=\sin^4\dfrac {\pi}{8}+\sin^4\dfrac {3\pi}{8}+\sin^4\left (\pi-\dfrac {3\pi}{8}\right)+\sin^4\left (\pi-\dfrac {\pi}{8}\right)$

We know that $\sin(\pi-x)=\sin x.$

$\therefore\ \sin^4\dfrac {\pi}{8}+\sin^4\dfrac {3\pi}{8}+\sin^4\left (\pi-\dfrac {3\pi}{8}\right)+\sin^4\left (\pi-\dfrac {\pi}{8}\right)\\\\=\sin^4\dfrac {\pi}{8}+\sin^4\dfrac {3\pi}{8}+\sin^4\dfrac {3\pi}{8}+\sin^4\dfrac {\pi}{8}\\\\=2\left [\sin^4\dfrac {\pi}{8}+\sin^4\dfrac {3\pi}{8}\right]$

But,

$\sin^4\dfrac {\pi}{8}+\sin^4\dfrac {3\pi}{8}\\\\=\sin^4\dfrac {\pi}{8}+\sin^4\dfrac {3\pi}{8}+2\sin^2\dfrac {\pi}{8}\sin^2\dfrac {3\pi}{8}-2\sin^2\dfrac {\pi}{8}\sin^2\dfrac {3\pi}{8}\\\\=\left (\sin^2\dfrac {\pi}{8}+\sin^2\dfrac {3\pi}{8}\right)^2-2\sin^2\dfrac {\pi}{8}\sin^2\dfrac {3\pi}{8}$

$=\left (\sin^2\dfrac {\pi}{8}+\sin^2\left (\dfrac {\pi}{2}-\dfrac {\pi}{8}\right)\right)^2-2\sin^2\dfrac {\pi}{8}\sin^2\left (\dfrac {\pi}{2}-\dfrac {\pi}{8}\right)\\\\=\left (\sin^2\dfrac {\pi}{8}+\cos^2\dfrac {\pi}{8}\right)^2-2\sin^2\dfrac {\pi}{8}\cos^2\dfrac {\pi}{8}\quad\left [\because\ \sin\left (\dfrac {\pi}{2}-x\right)=\cos x\right]$

$=\left (\sin^2\dfrac {\pi}{8}+\cos^2\dfrac {\pi}{8}\right)^2-\dfrac {4}{2}\left(\sin\dfrac {\pi}{8}\cos\dfrac {\pi}{8}\right)^2\\\\=1^2-\dfrac {\left(2\sin\dfrac {\pi}{8}\cos\dfrac {\pi}{8}\right)^2}{2}\quad[\because\ \sin^2x+\cos^2x=1]\\\\=1-\dfrac {\sin^2\dfrac {\pi}{4}}{2}\quad[\because\ 2\sin x\cos x=\sin(2x)]\\\\=1-\dfrac {1}{4}\quad\quad\left[\sin\dfrac {\pi}{4}=\dfrac {1}{\sqrt2}\right]\\\\=\dfrac {3}{4}$

Therefore,

$2\left [\sin^4\dfrac {\pi}{8}+\sin^4\dfrac {3\pi}{8}\right]\\\\=2\cdot\dfrac {3}{4}\\\\=\large\text { \mathbf {\dfrac {3}{2}} }$

Hence (3) is the answer.