Question

Sin 2 pi/8 +sin 2 3pi/8 + sin 2 5pi/8 +sin 2 7pi/8 = 2

Answer 1

sin²π/8+sin²3π/8+sin²5π/8+sin²7π/8
=sin²π/8+sin²3π/8+sin²(π/2+π/8)+sin²(π/2+3π/8)
=sin²π/8+sin²3π/8+cos²π/8+cos²3π/8
=(sin²π/8+cos²π/8)+(sin²3π/8+cos²3π/8) [∵, sin²Ф+cos²Ф=1]
=1+1
=2 Ans.

Answer 2

To prove this take LHS and solve the form to get the RHS

LHS  

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

Since we know the formula of allied angles that is $\sin \frac{\pi}{2}-\theta=\cos \theta$

Using this in last two terms  

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

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

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

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

Using the formula $\sin ^{2} \theta+\cos ^{2} \theta=1$

= 1+1  

=2 = RHS  

Therefore LHS = RHS  

Hence proved.