Q60 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Mathematics
$\begin{aligned} & \sin ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}-\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8} \\ &+\cos ^4 \frac{7 \pi}{8}-\sin ^4 \frac{7 \pi}{8}=\end{aligned}$
- A$\frac{1}{4}$
- B$\frac{1}{2}$
- C0
- Correct answerD$\frac{3}{4}$
Solution
We have,
$\begin{aligned}
& \sin ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}-\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8} \\
& +\cos ^4 \frac{7 \pi}{8}-\sin ^4 \frac{7 \pi}{8} \\
& =\sin ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}-\sin ^4 \frac{3 \pi}{8}+\sin ^4\left(\pi-\frac{3 \pi}{8}\right) \\
& +\cos ^4 \frac{7 \pi}{8}-\sin ^4\left(\pi-\frac{\pi}{8}\right) \\
& =\sin ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}-\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{3 \pi}{8} \\
& +\cos ^4 \frac{7 \pi}{8}-\sin ^4 \frac{\pi}{8} \\
& =\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{7 \pi}{8}=\cos ^4 \frac{3 \pi}{8}+\cos ^4\left(\pi-\frac{\pi}{8}\right) \\
& =\cos ^4 \frac{3 \pi}{8}+\left(-\cos \frac{\pi}{8}\right)^4=\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{\pi}{8} \\
&
\end{aligned}$
$\begin{aligned} & =\left(\frac{1+\cos \frac{3 \pi}{4}}{2}\right)^2+\left(\frac{1+\cos \frac{\pi}{4}}{2}\right)^2 \\ & =\frac{1}{4}\left[\left(1-\frac{1}{\sqrt{2}}\right)^2+\left(1+\frac{1}{\sqrt{2}}\right)^2\right] \\ & =\frac{1}{4}\left[1+\frac{1}{2}-\frac{2}{\sqrt{2}}+1+\frac{1}{2}+\frac{2}{\sqrt{2}}\right]=\frac{1}{4}[3]=\frac{3}{4}\end{aligned}$
Hence, the correct answer is option: 'D'.