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.$\frac{1}{3}$

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,
$(\sin {12^ \circ }\sin {48^ \circ })\sin {54^ \circ } \\ \Rightarrow \frac{1}{2}(\cos ({48^ \circ } - {12^ \circ }) - \cos ({48^ \circ } + {12^ \circ }))\sin ({90^ \circ } - {36^ \circ })\;{\text{ [Using }}\sin a\sin b{\text{ and sin(}}{90^ \circ } - \theta {\text{) formula]}} \\ \Rightarrow \frac{1}{2}(\cos {36^ \circ } - \cos {60^ \circ })\cos {36^ \circ }{\text{ [}}\cos ( - x) = \cos x{\text{ and }}\sin ({90^ \circ } - x) = \cos x{\text{]}} \\ \Rightarrow \frac{1}{2}(\frac{{\sqrt 5 + 1}}{4} - \frac{1}{2})(\frac{{\sqrt 5 + 1}}{4}){\text{ [}}\cos {36^ \circ } = \frac{{\sqrt 5 + 1}}{4}{\text{]}} \\ \Rightarrow \frac{1}{{2 \times 4 \times 4}}(\sqrt 5 + 1 - 2)(\sqrt 5 + 1) \\ \Rightarrow \frac{1}{{32}}(\sqrt 5 - 1)(\sqrt 5 + 1) \\ \Rightarrow \frac{1}{{32}}({(\sqrt 5 )^2} - {1^2}){\text{ [}}{a^2} - {b^2} = (a + b)(a - b){\text{]}} \\ \Rightarrow \frac{1}{{32}}(5 - 1) \\ \Rightarrow \frac{1}{{32}} \times 4 \\ \Rightarrow \frac{1}{8} \\$
And hence,$\sin {12^ \circ }\sin {48^ \circ }\sin {54^ \circ } = \frac{1}{8}$