Question

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.

Answer 1

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.

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.