MCQ
${\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $
- A$\frac{1}{2}$
- B$\frac{1}{4}$
- ✓$\frac{3}{2}$
- D$\frac{3}{4}$
$=\frac{1}{4}\left[\left(2 \sin ^{2} \frac{\pi}{8}\right)^{2}+\left(2 \sin ^{2} \frac{3 \pi}{8}\right)^{2}\right]$
$\quad+\frac{1}{4}\left[\left(2 \sin ^{2} \frac{\pi}{8}\right)^{2}+\left(2 \sin ^{2} \frac{3 \pi}{8}\right)^{2}\right]$
$=\frac{1}{4}\left[\left(1-\cos \frac{\pi}{4}\right)^{2}+\left(1-\cos \frac{3 \pi}{4}\right)^{2}\right]$
$\quad+\frac{1}{4}\left[\left(1-\cos \frac{\pi}{4}\right)^{2}+\left(1-\cos \frac{3 \pi}{4}\right)^{2}\right]$
$=\frac{1}{4}\left[\left(1-\frac{1}{\sqrt{2}}\right)^{2}+\left(1+\frac{1}{\sqrt{2}}\right)^{2}\right]$
$+\frac{1}{4}\left[\left(1-\frac{1}{\sqrt{2}}\right)^{2}+\left(1+\frac{1}{\sqrt{2}}\right)^{2}\right]$
$=\frac{1}{4}(3)+\frac{1}{4}(3)=\frac{3}{2}$
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