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

Use $\sin a\sin b$ formula in the first pair and $\sin ({90}^{\circ }-\theta )$formula In the third try and try to solve.

 Consider the given expression: $\sin {12}^{\circ }\sin {48}^{\circ }\sin {54}^{\circ }$.We know the formula:
$\sin a\sin b=\frac{1}{2}[\cos (a-b)-\cos (a+b)]$, where consider,$a={48}^{\circ },b={12}^{\circ }$. Putting the values in the given expression will give us,
$$\begin{gathered} (\sin {12}^{\circ }\sin {48}^{\circ })\sin {54}^{\circ } \\ \Rightarrow \frac{1}{2}(\cos ({48}^{\circ }-{12}^{\circ })-\cos ({48}^{\circ }+{12}^{\circ }))\sin ({90}^{\circ }-{36}^{\circ })\text{ [Using }\sin a\sin b\text{ and sin(}{90}^{\circ }-\theta \text{) formula]} \\ \Rightarrow \frac{1}{2}(\cos {36}^{\circ }-\cos {60}^{\circ })\cos {36}^{\circ }\text{ [}\cos (-x)=\cos x\text{ and }\sin ({90}^{\circ }-x)=\cos x\text{]} \\ \Rightarrow \frac{1}{2}(\frac{\sqrt{5}+1}{4}-\frac{1}{2})(\frac{\sqrt{5}+1}{4})\text{ [}\cos {36}^{\circ }=\frac{\sqrt{5}+1}{4}\text{]} \\ \Rightarrow \frac{1}{2\times 4\times 4}(\sqrt{5}+1-2)(\sqrt{5}+1) \\ \Rightarrow \frac{1}{32}(\sqrt{5}-1)(\sqrt{5}+1) \\ \Rightarrow \frac{1}{32}((\sqrt{5}{)}^2-1^2)\text{ [}{a}^2-b^2=(a+b)(a-b)\text{]} \\ \Rightarrow \frac{1}{32}(5-1) \\ \Rightarrow \frac{1}{32}\times 4 \\ \Rightarrow \frac{1}{8} \end{gathered}$$
And hence,$\sin {12}^{\circ }\sin {48}^{\circ }\sin {54}^{\circ }=\frac{1}{8}$

Note: Always try to use pairing of angles and find, which formula is suitable to start with. Once you start with the correct formula solution becomes easy.