Question

Solve: $\sin {48}^0.\sin {12}^0=$

Answer 1

$\sin 48°\sin 12°$

$=\sin (30+18)\sin (30-18)$

$=(\sin 30\cos 18+\cos 30\sin 18)(\sin 30\cos 18-\cos 30\sin 18)$

$={\left(\sin 30\cos 18\right)}^2-{\left(\cos 30\sin 18\right)}^2$

$={\sin }^{2}30{\cos }^{2}18-{\cos }^{2}30{\sin }^{2}18$

$=\frac{1}{4}{\cos }^{2}18-\frac{3}{4}{\sin }^{2}18$

$=\frac{(1-{\sin }^{2}18)-3{\sin }^{2}18}{4}$

$=\frac{1-4{\sin }^{2}18}{4}$

$=\frac{1}{4}-{\sin }^{2}18$

Hence $\sin 48°\sin 12°=\frac{1}{4}-{\sin }^{2}18$.