Докажите, помогите, пожалуйста

Решите задачу:

$1-4sin^2x=4sin(\frac{\pi}{6}-x)sin(\frac{\pi}{6}+x)\\\\ sin(\frac{\pi}{6}-x)sin(\frac{\pi}{6}+x) = \frac{cos2x-cos\frac{\pi}{3}}{2}=\frac{cos2x-0.5}{2}\\ 4*\frac{cos2x-0.5}{2}=2(cos2x-0.5)=2cos2x-1\\ 2(1-2sin^2a)-1=2-4sin^2a-1=1-4sin^2a$

$3-4cos^2y=-4sin(\frac{\pi}{6}+y)sin(\frac{\pi}{6}-y)\\ -4*\frac{cos2y-cos\frac{\pi}{3}}{2}=-2(cos2y-0.5)=-2(2cos^2y-1.5)=-4cos^2y+3$