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Question: The value of \(\sin {20^ \circ }\sin {40^ \circ }\sin {60^ \circ }\sin {100^ \circ }\) is equal to ...

The value of sin⁡20∘sin⁡40∘sin⁡60∘sin⁡100∘\sin {20^ \circ }\sin {40^ \circ }\sin {60^ \circ }\sin {100^ \circ } is equal to
A. 34{\text{A}}{\text{. }}\dfrac{3}{4}
B. 18{\text{B}}{\text{. }}\dfrac{1}{8}
C. 32{\text{C}}{\text{. }}\dfrac{3}{2}
D. 316{\text{D}}{\text{. }}\dfrac{3}{{16}}

Explanation

Solution

Hint: Assume the given equation as, S = sin⁡20∘sin⁡40∘sin⁡60∘sin⁡100∘\sin {20^ \circ }\sin {40^ \circ }\sin {60^ \circ }\sin {100^ \circ } and use the product identity, i.e. 2sin⁡xsin⁡y=cos⁡(x−y)−cos⁡(x+y)2\sin x\sin y = \cos (x - y) - \cos (x + y), and then solve the question.

Complete step-by-step answer:

Let us assume, S=sin⁡20∘sin⁡40∘sin⁡60∘sin⁡100∘S = \sin {20^ \circ }\sin {40^ \circ }\sin {60^ \circ }\sin {100^ \circ }.
Dividing and multiplying by 2, we get-
S=12(sin⁡60∘sin⁡20∘)(2sin⁡100∘sin⁡40∘)S = \dfrac{1}{2}(\sin {60^ \circ }\sin {20^ \circ })(2\sin {100^ \circ }\sin {40^ \circ })
Using the product identity, 2sin⁡xsin⁡y=cos⁡(x−y)−cos⁡(x+y)2\sin x\sin y = \cos (x - y) - \cos (x + y) for x=100∘,y=40∘x = {100^ \circ },y = {40^ \circ }.
∴S=12(sin⁡60∘sin⁡20∘)(cos⁡(100∘−40∘)−cos⁡(100∘+40∘)) =12(32sin⁡20∘)cos⁡60∘−cos⁡140∘ =34sin⁡20∘(12−cos⁡140∘) =34(12sin⁡20∘−sin⁡20∘cos⁡140∘)−(1)  \therefore S = \dfrac{1}{2}(\sin {60^ \circ }\sin {20^ \circ })(\cos ({100^ \circ } - {40^ \circ }) - \cos ({100^ \circ } + {40^ \circ })) \\\ = \dfrac{1}{2}\left( {\dfrac{{\sqrt 3 }}{2}\sin {{20}^ \circ }} \right)\\{ \cos {60^ \circ } - \cos {140^ \circ }\\} \\\ = \dfrac{{\sqrt 3 }}{4}\sin {20^ \circ }\left( {\dfrac{1}{2} - \cos {{140}^ \circ }} \right) \\\ = \dfrac{{\sqrt 3 }}{4}\left( {\dfrac{1}{2}\sin {{20}^ \circ } - \sin {{20}^ \circ }\cos {{140}^ \circ }} \right) - (1) \\\
Dividing and multiplying by 2 in equation (1), we get-
S=38(sin⁡20∘−2sin⁡20∘cos⁡140∘)S = \dfrac{{\sqrt 3 }}{8}\left( {\sin {{20}^ \circ } - 2\sin {{20}^ \circ }\cos {{140}^ \circ }} \right)
Using the product identity, 2sin⁡xcos⁡y=sin⁡(x+y)+sin⁡(x−y)2\sin x\cos y = \sin (x + y) + \sin (x - y) for x=20∘,y=140∘x = {20^ \circ },y = {140^ \circ }.
∴S=38(sin⁡20∘−sin⁡(20∘+140∘)−sin⁡(20∘−140∘)) =38(sin⁡20∘−sin⁡(160∘)+sin⁡(120∘)) =38(sin⁡20∘−sin⁡(180∘−20∘)+sin⁡(120∘)) =38(sin⁡20∘−sin⁡(20∘)+32) =316  \therefore S = \dfrac{{\sqrt 3 }}{8}\left( {\sin {{20}^ \circ } - \sin ({{20}^ \circ } + {{140}^ \circ }) - \sin ({{20}^ \circ } - {{140}^ \circ })} \right) \\\ = \dfrac{{\sqrt 3 }}{8}\left( {\sin {{20}^ \circ } - \sin ({{160}^ \circ }) + \sin ({{120}^ \circ })} \right) \\\ = \dfrac{{\sqrt 3 }}{8}\left( {\sin {{20}^ \circ } - \sin ({{180}^ \circ } - {{20}^ \circ }) + \sin ({{120}^ \circ })} \right) \\\ = \dfrac{{\sqrt 3 }}{8}\left( {\sin {{20}^ \circ } - \sin ({{20}^ \circ }) + \dfrac{{\sqrt 3 }}{2}} \right) \\\ = \dfrac{3}{{16}} \\\
Therefore, the correct answer is option (D).

Note: While solving such types of questions, always use the correct trigonometric identity, as used in the question. Also, the steps are more, so be careful while writing each step, to avoid mistakes, and then solve the question.