6 или. 9, какой сможете.

Решите задачу:

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

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

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

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