Question

докажите, помогите, пожалуйста

$1-4sin^2x=4sin (frac{pi}{6} - x ) sin( frac{pi}{6}+x) 3-4cos^2y= -4sin( frac{pi}{6} +y)sin( frac{pi}{6} -y)$

Answer 1

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

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