Question

Solve the question ​

Answer 1

SOLUTION:-

Given:

sin alpha= 1/2

sin alpha= sin30°

alpha= 30°

Therefore,

Take L.H.S

3cos alpha -4cos³alpha

$= > 3cos30 \degree - 4cos {}^{3} 30 \degree \\ \\ = > 3( \frac{ \sqrt{3} }{2} ) - 4( \frac{ \sqrt{3} }{2} ) {}^{3} \\ \\ = > \frac{3 \sqrt{3} }{2} - 4( \frac{3 \sqrt{3} }{8} ) \\ \\ = > \frac{3 \sqrt{3} }{2} - \frac{3 \sqrt{3} }{2} \\ \\ = > \frac{3 \sqrt{3} - 3 \sqrt{3} }{2} \\ \\ = > \frac{0}{2} \\ \\ = > 0 \: \: \: \: \: \: \: \: [R.H.S]$

Proved.

Hope it helps ☺️