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

Last updated date: 28th Mar 2023

•

Total views: 306.6k

•

Views today: 8.83k

Answer

Verified

306.6k+ views

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.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE