Question

Помогите, пожалуйста, с алгеброй.

Answer 1

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

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