Question

6 или. 9, какой сможете.

Answer 1

$frac{68sin(3pi -56^0)cos(pi +56^0)}{5sin68^0}=frac{68sin56^0cdot (-cos56^0)}{5sin68^0}=frac{-34sin112^0}{5sin68^0}=\\=frac{-34sin(90+22)}{5sin(90-22)}=frac{-34cos22}{5sin22}=-frac{34}{5}=-6,8$

$2.; ; sin(frac{3pi }{2}+ alpha )=-cos alpha =0,8; ; to ; ; cos alpha =-0,8\\Formyla:; ; cos2 alpha =cos^2 alpha -sin^2 alpha =\\=cos^2 alpha -(1-cos^2 alpha )=2cos^2alpha -1$

$frac{3}{4}cos^2(pi -2 alpha )=frac{3}{4}cos^22 alpha =frac{3}{4}(2cos^2 alpha -1)^2=$

$=frac{3}{4}(2cdot 0,64-1)^2=0,0588$

Answer 2

Волшбник(ца),спасибо!!!

Answer 3

ца!