Prove that:

Question:

Prove that:

$\sin ^{2} \frac{\pi}{8}+\sin ^{2} \frac{3 \pi}{8}+\sin ^{2} \frac{5 \pi}{8}+\sin ^{2} \frac{7 \pi}{8}=2$

Solution:

$\mathrm{LHS}=\sin ^{2} \frac{\pi}{8}+\sin ^{2} \frac{3 \pi}{8}+\sin ^{2} \frac{5 \pi}{8}+\sin ^{2} \frac{7 \pi}{8}$

$=\sin ^{2}\left(\frac{\pi}{2}-\frac{3 \pi}{8}\right)+\sin ^{2}\left(\frac{\pi}{2}-\frac{\pi}{8}\right)+\sin ^{2} \frac{5 \pi}{8}+\sin ^{2} \frac{7 \pi}{8}$

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

$=\cos ^{2} \frac{3 \pi}{8}+\sin ^{2} \frac{\pi}{8}+\sin ^{2} \frac{3 \pi}{8}+\cos ^{2} \frac{\pi}{8}$

$=\left(\cos ^{2} \frac{\pi}{8}+\sin ^{2} \frac{\pi}{8}\right)+\left(\cos ^{2} \frac{3 \pi}{8}+\sin ^{2} \frac{3 \pi}{8}\right)$

$=1+1=2=$ RHS

Hence proved.

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now