# Prove that given equation: \[\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = \dfrac{3}{{16}}\].

Answer

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Hint: Group the sine terms so that you can use the product rule of sines and simplify completely to obtain the right-hand side of the proof.

Complete step-by-step answer:

Let us assign the left hand side of the equation to LHS and solve to get the right hand side, hence proving it.

\[LHS = \sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ \]

We know that the value of sin(60) is \[\dfrac{{\sqrt 3 }}{2}\]. Substituting it in the above equation, we get:

\[LHS = \dfrac{{\sqrt 3 }}{2}\sin 20^\circ \sin 40^\circ \sin 80^\circ \]

We now group the terms sin(40) and sin(80) to get:

\[LHS = \dfrac{{\sqrt 3 }}{2}\sin 20^\circ (\sin 40^\circ \sin 80^\circ ).............(1)\]

We know the formula for the product of sine terms, that is, as follows:

2sin(A) sin(B) = cos(A-B) - cos(A+B) ……………. (2)

Using formula (2) in equation (1), we get:

\[LHS = \dfrac{{\sqrt 3 }}{4}\sin 20^\circ (\cos (40^\circ - 80^\circ ) - \cos (40^\circ + 80^\circ ))\]

Simplifying the expression, we get:

\[LHS = \dfrac{{\sqrt 3 }}{4}\sin 20^\circ (\cos ( - 40^\circ ) - \cos (120^\circ ))\]

We know that the value of cos(-x) is cos(x), hence we have:

\[LHS = \dfrac{{\sqrt 3 }}{4}\sin 20^\circ (\cos 40^\circ - \cos 120^\circ )\]

We know that the value of cos(120) is \[ - \dfrac{1}{2}\], using this in the above equation, we get:

\[LHS = \dfrac{{\sqrt 3 }}{4}\sin 20^\circ \left( {\cos 40^\circ + \dfrac{1}{2}} \right)\]

Simplifying it further, we get:

\[LHS = \dfrac{{\sqrt 3 }}{8}(2\sin 20^\circ \cos 40^\circ + \sin 20^\circ )........(3)\]

We know the formula for the product of sine and cosine terms.

2sinAcosB = sin(A+B) + sin(A-B)

Using this formula in equation (3), we get:

\[LHS = \dfrac{{\sqrt 3 }}{8}((\sin (20^\circ + 40^\circ )\sin (20^\circ - 40^\circ )) + \sin 20^\circ )\]

Simplifying further, we get:

\[LHS = \dfrac{{\sqrt 3 }}{8}(\sin 60^\circ + \sin ( - 20^\circ ) + \sin 20^\circ )\]

We know that the value of sin(-x) is sin(x). Hence, we get:

\[LHS = \dfrac{{\sqrt 3 }}{8}(\sin 60^\circ - \sin 20^\circ + \sin 20^\circ )\]

Cancelling the two sine terms, we get:

\[LHS = \dfrac{{\sqrt 3 }}{8}\sin 60^\circ \]

We know that the value of sin(60) is \[\dfrac{{\sqrt 3 }}{2}\]. Hence, we have:

\[LHS = \dfrac{{\sqrt 3 }}{8}\dfrac{{\sqrt 3 }}{2}\]

\[LHS = \dfrac{3}{{16}}\]

\[LHS = RHS\]

Hence, we proved that the left hand side is equal to the right hand side of the equation.

Note: You can also proceed by combining the sine terms containing 20 degrees and 40 degrees and use product rule to simplify, the final answer will be the same. Regrouping the terms of trigonometric functions is essential for such a type of question.

Complete step-by-step answer:

Let us assign the left hand side of the equation to LHS and solve to get the right hand side, hence proving it.

\[LHS = \sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ \]

We know that the value of sin(60) is \[\dfrac{{\sqrt 3 }}{2}\]. Substituting it in the above equation, we get:

\[LHS = \dfrac{{\sqrt 3 }}{2}\sin 20^\circ \sin 40^\circ \sin 80^\circ \]

We now group the terms sin(40) and sin(80) to get:

\[LHS = \dfrac{{\sqrt 3 }}{2}\sin 20^\circ (\sin 40^\circ \sin 80^\circ ).............(1)\]

We know the formula for the product of sine terms, that is, as follows:

2sin(A) sin(B) = cos(A-B) - cos(A+B) ……………. (2)

Using formula (2) in equation (1), we get:

\[LHS = \dfrac{{\sqrt 3 }}{4}\sin 20^\circ (\cos (40^\circ - 80^\circ ) - \cos (40^\circ + 80^\circ ))\]

Simplifying the expression, we get:

\[LHS = \dfrac{{\sqrt 3 }}{4}\sin 20^\circ (\cos ( - 40^\circ ) - \cos (120^\circ ))\]

We know that the value of cos(-x) is cos(x), hence we have:

\[LHS = \dfrac{{\sqrt 3 }}{4}\sin 20^\circ (\cos 40^\circ - \cos 120^\circ )\]

We know that the value of cos(120) is \[ - \dfrac{1}{2}\], using this in the above equation, we get:

\[LHS = \dfrac{{\sqrt 3 }}{4}\sin 20^\circ \left( {\cos 40^\circ + \dfrac{1}{2}} \right)\]

Simplifying it further, we get:

\[LHS = \dfrac{{\sqrt 3 }}{8}(2\sin 20^\circ \cos 40^\circ + \sin 20^\circ )........(3)\]

We know the formula for the product of sine and cosine terms.

2sinAcosB = sin(A+B) + sin(A-B)

Using this formula in equation (3), we get:

\[LHS = \dfrac{{\sqrt 3 }}{8}((\sin (20^\circ + 40^\circ )\sin (20^\circ - 40^\circ )) + \sin 20^\circ )\]

Simplifying further, we get:

\[LHS = \dfrac{{\sqrt 3 }}{8}(\sin 60^\circ + \sin ( - 20^\circ ) + \sin 20^\circ )\]

We know that the value of sin(-x) is sin(x). Hence, we get:

\[LHS = \dfrac{{\sqrt 3 }}{8}(\sin 60^\circ - \sin 20^\circ + \sin 20^\circ )\]

Cancelling the two sine terms, we get:

\[LHS = \dfrac{{\sqrt 3 }}{8}\sin 60^\circ \]

We know that the value of sin(60) is \[\dfrac{{\sqrt 3 }}{2}\]. Hence, we have:

\[LHS = \dfrac{{\sqrt 3 }}{8}\dfrac{{\sqrt 3 }}{2}\]

\[LHS = \dfrac{3}{{16}}\]

\[LHS = RHS\]

Hence, we proved that the left hand side is equal to the right hand side of the equation.

Note: You can also proceed by combining the sine terms containing 20 degrees and 40 degrees and use product rule to simplify, the final answer will be the same. Regrouping the terms of trigonometric functions is essential for such a type of question.

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