Question

Find which of the following options equals the value of $4\cos 6\theta \cos 4\theta \cos 2\theta $?
(a) $\cos 12\theta +\cos 8\theta +\cos 4\theta +1$,
(b) $\cos 12\theta +\cos 8\theta -\cos 4\theta +1$,
(c) $\cos 12\theta -\cos 8\theta +\cos 4\theta +1$,
(d) $\cos 12\theta -\cos 8\theta -\cos 4\theta +1$.

Answer 1

Hint: We start solving the problem by writing the given product rearranging the terms in product. We then use the identity \[2\cos A\cos B\ =\ \cos \left( A+B \right)+\cos \left( A-B \right)\] in the rearranged product by assuming the values for A and B to proceed further into the problem. We then make necessary calculations and make use of identities $1+\cos 2A=2{{\cos }^{2}}A$ and \[2\cos A\cos B\ =\ \cos \left( A+B \right)+\cos \left( A-B \right)\] to proceed further in the problem. We then make necessary calculations to get the required result.

Complete step-by-step answer:
According to the problem, we need to compute the given function $4\cos 6\theta \cos 4\theta \cos 2\theta $ to the given options.
We are given a function as $4\cos 6\theta \cos 4\theta \cos 2\theta $. Let us rewrite the given product as shown below:
$\Rightarrow 4\cos 6\theta \cos 4\theta \cos 2\theta =\left( 2\cos 4\theta \right)\times \left( 2\cos 6\theta \cos 2\theta \right)$ ---(1).
Let us apply the trigonometric identity \[2\cos A\cos B\ =\ \cos \left( A+B \right)+\cos \left( A-B \right)\] for the terms $2\cos 6\theta \cos 2\theta $ in equation (1), assuming A as \[6\theta \] and B as \[2\theta \].
Now, let us solve further.
$\Rightarrow 4\cos 6\theta \cos 4\theta \cos 2\theta =2\cos 4\theta \times \left[ \cos \left( 6\theta +2\theta \right)+\cos \left( 6\theta -2\theta \right) \right]$.
$\Rightarrow 4\cos 6\theta \cos 4\theta \cos 2\theta =2\cos 4\theta \times \left[ \cos \left( 8\theta \right)+\cos \left( 4\theta \right) \right]$.
Let us now perform the multiplication operation on the Right-Hand Side (R.H.S).
$\Rightarrow 4\cos 6\theta \cos 4\theta \cos 2\theta =2\cos 4\theta \cos 8\theta +2\cos 4\theta \cos 4\theta $.
$\Rightarrow 4\cos 6\theta \cos 4\theta \cos 2\theta =2\cos 4\theta \cos 8\theta +2{{\cos }^{2}}4\theta $ ---(2).
We know that $1+\cos 2A=2{{\cos }^{2}}A$. We use this result in equation (2) to solve the problem further.
$\Rightarrow 4\cos 6\theta \cos 4\theta \cos 2\theta =2\cos 4\theta \cos 8\theta +\left( 1+\cos 8\theta \right)$.
\[\Rightarrow 4\cos 6\theta \cos 4\theta \cos 2\theta =2\cos 4\theta \cos 8\theta +\cos 8\theta +1\] ---(3).
Now, let us use the identity \[2\cos A\cos B\ =\ \cos \left( A+B \right)+\cos \left( A-B \right)\] in equation (1), assuming A as \[8\theta \] and B as \[4\theta \].
Now, let us solve the problem further.
\[\Rightarrow 4\cos 6\theta \cos 4\theta \cos 2\theta =\cos \left( 8\theta +4\theta \right)+\cos \left( 8\theta -4\theta \right)+\cos 8\theta +1\].
\[\Rightarrow 4\cos 6\theta \cos 4\theta \cos 2\theta =\cos \left( 12\theta \right)+\cos \left( 4\theta \right)+\cos 8\theta +1\].
\[\Rightarrow 4\cos 6\theta \cos 4\theta \cos 2\theta =\cos 12\theta +\cos 4\theta +\cos 8\theta +1\].
So, we have found that the value of \[4\cos 6\theta \cos 4\theta \cos 2\theta \] as \[\cos 12\theta +\cos 4\theta +\cos 8\theta +1\].

So, the correct answer is “Option A”.

Note: We can solve this problem easily by assuming an angle for $\theta $ and checking all the options which is equal to the given product. We should not confuse the trigonometric identities present in the problem. We should not make calculation mistakes while solving this problem. Whenever we get problems involving multiplication of two or more trigonometric functions, we should start solving by using the trigonometric as we just used in the problem.