Question

Alpha equals to 1 by 2 then find the value of a 3 sin alpha minus 4 sin cube Alpha

Answer 1

sin(α)=21​ 
Suppose a right angled triangle with perpendicular =1 and hypotenuse =2

Then base = root 3

\cos( \alpha ) = \frac{ \sqrt{3} }{2}cos(α)=23​​ 
Therefore,
3 \cos( \alpha ) = \frac{3 \sqrt{3} }{2}3cos(α)=233​​ 
and
4 \cos {}^{3} ( { \alpha }) = \frac{3 \sqrt{3} }{2}4cos3(α)=233​​