Q59 / 160 · AP EAMCET 2023 (16 May Shift 1) · Mathematics
$\sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}=$
- A$\frac{1}{4}$
- B$\frac{3}{8}$
- Correct answerC$\frac{3}{2}$
- D$\frac{3}{4}$
Solution
$\sin ^4\left(\frac{\pi}{8}\right)+\sin ^4\left(\frac{3 \pi}{8}\right)+\sin ^4\left(\frac{5 \pi}{8}\right)+\sin ^4\left(\frac{7 \pi}{8}\right)$
$=\sin ^4\left(\frac{\pi}{8}\right)+\sin ^4\left(\frac{3 \pi}{8}\right)+\sin ^4\left(\pi-\frac{3 \pi}{8}\right)+\sin ^4\left(\pi-\frac{\pi}{8}\right)$
$=\sin ^4\left(\frac{\pi}{8}\right)+\sin ^4\left(\frac{3 \pi}{8}\right)+\sin ^4\left(\frac{3 \pi}{8}\right)+\sin ^4\left(\frac{\pi}{8}\right)$
$=2\left[\sin ^4\left(\frac{\pi}{8}\right)+\sin ^4\left(\frac{3 \pi}{8}\right)\right]$
$=2\left[\left(\frac{1-\cos \left(\frac{\pi}{4}\right)}{2}\right)^2+\left(\frac{1-\cos \left(\frac{3 \pi}{4}\right)}{2}\right)^2\right]$
$=\left[\left(\frac{1-\frac{1}{\sqrt{2}}}{2}\right)^2+\left(\frac{1+\frac{1}{\sqrt{2}}}{2}\right)^2\right]$
$=\frac{2}{4}\left[\left(1-\frac{1}{\sqrt{2}}\right)^2+\left(1+\frac{1}{\sqrt{2}}\right)^2\right]=\frac{3}{2}$
$\therefore \quad \sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}=\frac{3}{2}$.
Hence, the correct answer is option: 'C'.