# If $A = {30^0}$, then $4\cos A\cos \left( {{{60}^0} - A} \right)\cos \left( {{{60}^0} + A} \right) = \cos 3A$.

If the above statement is true enter 1 else 0.

Last updated date: 24th Mar 2023

•

Total views: 306.6k

•

Views today: 3.86k

Answer

Verified

306.6k+ views

Hint – In this question first simplify the L.H.S by putting direct values of standard cosine angles later on simplify the R.H.S and check whether it is equal or not so use these concepts to reach the solution of the question.

Given equation is

$4\cos A\cos \left( {{{60}^0} - A} \right)\cos \left( {{{60}^0} + A} \right) = \cos 3A$

Consider L.H.S

$ \Rightarrow 4\cos A\cos \left( {{{60}^0} - A} \right)\cos \left( {{{60}^0} + A} \right)$

Now it is given that $A = {30^0}$

So, substitute the value of A in above equation, we have

\[

\Rightarrow 4\cos {30^0}\cos \left( {{{60}^0} - {{30}^0}} \right)\cos \left( {{{60}^0} + {{30}^0}} \right) \\

\Rightarrow 4\cos {30^0}\cos {30^0}\cos {90^0} \\

\]

As we know that the value of $\cos {30^0} = \dfrac{{\sqrt 3 }}{2},{\text{ }}\cos {90^0} = 0$, so substitute this value in above equation we have,

$ \Rightarrow 4\cos {30^0}\cos {30^0}\cos {90^0} = 4\left( {\dfrac{{\sqrt 3 }}{2}} \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right)\left( 0 \right)$

Now as we all know multiplication with zero is always zero

Therefore

$ \Rightarrow 4\cos A\cos \left( {{{60}^0} - A} \right)\cos \left( {{{60}^0} + A} \right) = 0$……………………………… (1)

Now consider R.H.S of the given equation we have,

$ \Rightarrow \cos 3A$

Now it is given that $A = {30^0}$

So, substitute the value of A in above equation, we have

$\therefore \cos \left( {3 \times {{30}^0}} \right) = \cos {90^0}$

As we know that the value of $\cos {90^0} = 0$, so substitute this value in above equation we have,

R.H.S $ = \cos {90^0} = 0$ ……………………………………….. (2)

Now from equation (1) and (2)

L.H.S = R.H.S

So, the given statement is true.

Hence enter 1.

So, this is the required answer.

Note – In such types of questions it is advised to simplify the LHS or the RHS according to their complexity of trigonometric functions . Sometimes proving LHS = RHS needs simplification on both sides of the equation. Remember to convert dissimilar trigonometric functions to get to the final result, and check whether R.H.S is equal to L.H.S or not if yes then enter 1, which is the required answer.

__Complete step-by-step solution -__Given equation is

$4\cos A\cos \left( {{{60}^0} - A} \right)\cos \left( {{{60}^0} + A} \right) = \cos 3A$

Consider L.H.S

$ \Rightarrow 4\cos A\cos \left( {{{60}^0} - A} \right)\cos \left( {{{60}^0} + A} \right)$

Now it is given that $A = {30^0}$

So, substitute the value of A in above equation, we have

\[

\Rightarrow 4\cos {30^0}\cos \left( {{{60}^0} - {{30}^0}} \right)\cos \left( {{{60}^0} + {{30}^0}} \right) \\

\Rightarrow 4\cos {30^0}\cos {30^0}\cos {90^0} \\

\]

As we know that the value of $\cos {30^0} = \dfrac{{\sqrt 3 }}{2},{\text{ }}\cos {90^0} = 0$, so substitute this value in above equation we have,

$ \Rightarrow 4\cos {30^0}\cos {30^0}\cos {90^0} = 4\left( {\dfrac{{\sqrt 3 }}{2}} \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right)\left( 0 \right)$

Now as we all know multiplication with zero is always zero

Therefore

$ \Rightarrow 4\cos A\cos \left( {{{60}^0} - A} \right)\cos \left( {{{60}^0} + A} \right) = 0$……………………………… (1)

Now consider R.H.S of the given equation we have,

$ \Rightarrow \cos 3A$

Now it is given that $A = {30^0}$

So, substitute the value of A in above equation, we have

$\therefore \cos \left( {3 \times {{30}^0}} \right) = \cos {90^0}$

As we know that the value of $\cos {90^0} = 0$, so substitute this value in above equation we have,

R.H.S $ = \cos {90^0} = 0$ ……………………………………….. (2)

Now from equation (1) and (2)

L.H.S = R.H.S

So, the given statement is true.

Hence enter 1.

So, this is the required answer.

Note – In such types of questions it is advised to simplify the LHS or the RHS according to their complexity of trigonometric functions . Sometimes proving LHS = RHS needs simplification on both sides of the equation. Remember to convert dissimilar trigonometric functions to get to the final result, and check whether R.H.S is equal to L.H.S or not if yes then enter 1, which is the required answer.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths 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

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE