Question

$\sin {20}^{\circ }\sin {24}^{\circ }\sin {48}^{\circ }\sin {84}^{\circ }$ is equal to

• A. $\cos {20}^{\circ }\cos {40}^{\circ }$ $\cos {60}^{\circ }\cos {80}^{\circ }$
• B. $\sin {20}^{\circ }\sin {40}^{\circ }\sin {60}^{\circ }\sin {80}^{\circ }$
• C. $\frac{3}{15}$
• D. $\frac{1}{16}$

Answer 1

The correct option is D $\frac{1}{16}$

Given: $\sin 12°\sin 24°\sin 48°\sin 84°$

$\Rightarrow \frac{1}{4}\left(2\sin 84°\sin 24°\right)\left(2\sin 48°\sin 12°\right)$

Formula: $sinA.sinB=\frac{1}{2}\left[cos\right(A-B)-cos(A+B\left)\right]$

$\Rightarrow \frac{1}{4}(\cos 60°-\cos 108°)(\cos 36°-\cos 60°)$

$\Rightarrow \frac{1}{4}(\frac{1}{2}-\cos (90°+18°))(\cos 36°-\frac{1}{2})$

$\Rightarrow \frac{1}{4}(\frac{1}{2}+\sin 18°)(\cos 36°-\frac{1}{2})$ $[\because \cos (90°+x)=-\sin x]$

$\Rightarrow \frac{1}{4}(\frac{1}{2}+\frac{\sqrt{5}-1}{4})(\frac{\sqrt{5}+1}{4}-\frac{1}{2})$

$\Rightarrow \frac{1}{64}(2+\sqrt{5}-1)(\sqrt{5}+1-2)$

$=\frac{1}{64}(\sqrt{5}+1)(\sqrt{5}-1)=\frac{1}{64}\left(\right(\sqrt{5}{)}^2-1)$

$=\frac{1}{64}\times 4=\frac{1}{16}$