Question

Помогите, прошу! Вычислите: 1)Sin23*sin53+sin67*sin37 2(sin75*cos35+cos125*cos75)/sin40

Answer 1

$1)\; \; sin23\cdot sin53+sin67\cdot sin37=\\\\=sin(90-67)\cdot sin53+sin67\cdot sin(90-53)=\\\\=cos67\cdot sin53+sin67\cdot cos53=sin(67+53)=\\\\=sin120=sin(180-60)=sin60=\frac{\sqrt3}{2}$

$2)\; \; \frac{sin75\cdot cos35+cos125\cdot cos75}{cos40} =[\; cos125=cos(90+35)=-sin35\; ]=\\\\= \frac{sin75\cdot cos35-sin35\cdot cos75}{cos40} = \frac{sin(75-35)}{cos40}=\frac{sin40}{cos40}=tg40$