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Question: The value of \(\sin {12^ \circ }\sin {48^ \circ }\sin {54^ \circ }\)is equal to: A.\(\frac{2}{3}\)...

The value of sin⁡12∘sin⁡48∘sin⁡54∘\sin {12^ \circ }\sin {48^ \circ }\sin {54^ \circ }is equal to:
A.23\frac{2}{3}
B.12\frac{1}{2}
C.18\frac{1}{8}
D.13\frac{1}{3}

Explanation

Solution

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

Consider the given expression: sin⁡12∘sin⁡48∘sin⁡54∘\sin {12^ \circ }\sin {48^ \circ }\sin {54^ \circ }.We know the formula:
sin⁡asin⁡b=12[cos⁡(a−b)−cos⁡(a+b)]\sin a\sin b = \frac{1}{2}[\cos (a - b) - \cos (a + b)], where consider,a=48∘,b=12∘a = {48^ \circ },b = {12^ \circ }. Putting the values in the given expression will give us,
(sin⁡12∘sin⁡48∘)sin⁡54∘ ⇒12(cos⁡(48∘−12∘)−cos⁡(48∘+12∘))sin⁡(90∘−36∘)   [Using sin⁡asin⁡b and sin(90∘−θ) formula] ⇒12(cos⁡36∘−cos⁡60∘)cos⁡36∘ [cos⁡(−x)=cos⁡x and sin⁡(90∘−x)=cos⁡x] ⇒12(5+14−12)(5+14) [cos⁡36∘=5+14] ⇒12×4×4(5+1−2)(5+1) ⇒132(5−1)(5+1) ⇒132((5)2−12) [a2−b2=(a+b)(a−b)] ⇒132(5−1) ⇒132×4 ⇒18  (\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∘sin⁡48∘sin⁡54∘=18\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.