Помогите, пожалуйста, с алгеброй.

Question 1

Question 2

Используются формулы синуса двойного угла
$sin(2x)=2sin(x)cos(x)$
формула приведения
$sin(\pi+x)=-sin x$
$sin(90^0-x)=cos x$
$sin(180^0-x)=sin x$

$cos\frac{\pi}{7}cos\frac{2\pi}{7}cos\frac{4\pi}{7}=$
$\frac{2sin\frac{\pi}{7}cos\frac{\pi}{7}cos\frac{2\pi}{7}cos\frac{4\pi}{7}}{2sin\frac{\pi}{7}}$
$\frac{sin(2*\frac{\pi}{7})cos\frac{2\pi}{7}cos\frac{4\pi}{7}}{2sin\frac{\pi}{7}}$
$\frac{sin(\frac{2\pi}{7})cos\frac{2\pi}{7}cos\frac{4\pi}{7}}{2sin\frac{\pi}{7}}$
$\frac{2sin(\frac{2\pi}{7})cos\frac{2\pi}{7}cos\frac{4\pi}{7}}{4sin\frac{\pi}{7}}$
$\frac{sin(2*\frac{2\pi}{7})cos\frac{4\pi}{7}}{4sin\frac{\pi}{7}}$
$\frac{sin(\frac{4\pi}{7})cos\frac{4\pi}{7}}{4sin\frac{\pi}{7}}$
$\frac{2sin(\frac{4\pi}{7})cos\frac{4\pi}{7}}{8sin\frac{\pi}{7}}$
$\frac{sin(2*\frac{4\pi}{7})}{8sin\frac{\pi}{7}}$
$\frac{sin(\frac{8\pi}{7})}{8sin\frac{\pi}{7}}$
$\frac{sin(\pi+\frac{\pi}{7})}{8sin\frac{\pi}{7}}$
$\frac{-sin\frac{\pi}{7}}{8sin\frac{\pi}{7}}$
$-\frac{1}{8}$
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$8sin10^0sin50^0sin 70^0=$
$8sin(90^0-80^0)sin(90^0-40^0)sin(90^0-20^0)=$
$8cos80^0cos40^0cos20^0=$
$8cos20^0cos40^0cos80^0=$
$\frac{4*2cos20^0*sin20^0cos40^0cos80^0}{sin20^0}=$
$\frac{4*sin(2*20^0)cos40^0cos80^0}{sin20^0}=$
$\frac{4*sin(40^0)cos40^0cos80^0}{sin20^0}=$
$\frac{2*2sin(40^0)cos40^0cos80^0}{sin20^0}=$
$\frac{2*sin(2*40^0)cos80^0}{sin20^0}=$
$\frac{2*sin(80^0)cos80^0}{sin20^0}=$
$\frac{sin(2*80^0)}{sin20^0}=$
$\frac{sin(160^0)}{sin20^0}=$
$\frac{sin(180^0-20^0)}{sin20^0}=$
$\frac{sin20^0}{sin 20^0}=1$