Question

Sin27° * sin 63° = ?

Answer 1

$sin27^\circ \cdot sin63^\circ =sin27^\circ \cdot sin(\underbrace {90^\circ-27^\circ}_{63})=\underbrace {sin27^\circ \cdot cos27^\circ }_{\frac{1}{2}sin(2\cdot 27)}=\\\\=\frac{1}{2}sin54^\circ \\\\\\P.S.\; \; \; \; \; 2\cdot sinx\cdot cosx=sin2x\; \; \to \; \; sinx\cdot cosx=\frac{1}{2}\cdot sin2x$

$sin(90^\circ-x)=cosx$