Question

The value of $\sin {12}^{\circ }\sin {48}^{\circ }\sin {54}^{\circ }$ is equal to.

• A. $\frac{2}{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{8}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C $\frac{1}{8}$

Consider, $\sin {12}^o\sin {48}^o\sin {54}^o$

$=\left(\sin {12}^o\sin {48}^o\right)\sin {54}^o$

$=\frac{1}{2}\left(\cos \right(12-48)-\cos (12+48\left)\right)\sin ({90}^o-{36}^o)$ ..... [Using $\sin a\sin b$ formula]

$=\frac{1}{2}(\cos {36}^o-\cos {60}^o)\cos {36}^o$ ........ $[\because \cos (-x)=\cos x\text{and}\sin ({90}^o-x)=\cos x]$

$=\frac{1}{2}(\frac{\sqrt{5}+1}{4}-\frac{1}{2})\left(\frac{\sqrt{5}+1}{4}\right)$ $[\cos {36}^o=\frac{\sqrt{5}+1}{4},\sin {36}^o=\frac{\sqrt{5}-1}{4}]$

$=\frac{1}{2\times 4\times 4}(\sqrt{5}+1-2)\left(\sqrt{5}+1\right)$

$=\frac{1}{32}(\sqrt{5}-1)\left(\sqrt{5}+1\right)$

$=\frac{1}{32}(5-1)$

$=\frac{4}{32}=\frac{1}{8}$

Hence, $\sin {12}^o\sin {48}^o\sin {54}^o=\frac{1}{8}$