Question

4sin a cos a cos 2a=sin 4a

Answer 1

sin2α=2sinαcosα
sinα*cosβ=1/2(sin(α+β)+sin(α-β))

$4sin \alpha *cos \alpha *cos2 \alpha =2*2sin \alpha *cos \alpha*cos2 \alpha =2 *sin2 \alpha * cos2 \alpha = \\ =2* \frac{1}{2} (sin( 2 \alpha +2 \alpha )+sin( 2 \alpha -2 \alpha ))= 1*(sin4 \alpha +sin0)= \\ =1*(sin4 \alpha +0)=1*sin4 \alpha =sin4 \alpha$