Question

please answer this​

Answer 1

We have been given to prove that:-

$sin^{2}\bigg( \dfrac{2 \pi}{7} \bigg) + sin^{2} \bigg( \dfrac{3 \pi}{14} \bigg) + sin^{2} \bigg( \dfrac{11 \pi}{14}\bigg) + sin^{2} \bigg( \dfrac{5 \pi}{7} \bigg) = 2$

As, we know that $sin (90 \degree + \theta) = cos \theta$

Left hand side of the expression can be re-written as:-

$= sin^{2}\bigg( \dfrac{2 \pi}{7} \bigg) + sin^{2} \bigg( \dfrac{3 \pi}{14} \bigg) + sin^{2} \bigg(\dfrac{\pi}{2} + \dfrac{2\pi}{7}\bigg) + sin^{2} \bigg(\dfrac{\pi}{2}+\dfrac{3 \pi}{14} \bigg)$

$= sin^{2} \bigg( \dfrac{2 \pi}{7} \bigg) +sin^{2} \bigg( \dfrac{3\pi}{14} \bigg) + cos^{2} \bigg( \dfrac{2 \pi}{7}\bigg) + cos^{2} \bigg( \dfrac{3 \pi}{14} \bigg)$

Now, group the terms as follows:-

$= \Bigg[cos^{2} \bigg( \dfrac{5 \pi}{7} \bigg) + sin^{2} \bigg( \dfrac{5 \pi}{7} \bigg) \Bigg] + \Bigg[ cos^{2} \bigg( \dfrac{11 \pi}{14} \bigg) + sin^{2} \bigg( \dfrac{11 \pi}{14}\bigg) \Bigg]$

As we know, $sin^{2} \theta + cos^{2} \theta = 1$

This expression gets reduced to:-

= 1 + 1

= 2

This is equal to the right hand side.

L.H.S = R.H.S

Hence Proved !