Question

Find the value of ${{\cos }^{3}}\theta +{{\cos }^{3}}\left( {{120}^{\circ }}+\theta \right)+{{\cos }^{3}}\left( {{120}^{\circ }}-\theta \right)$
A) $ \dfrac{3}{4}\sin 3\theta $
B) $ \dfrac{3}{4}\cos 3\theta $
C) $ \dfrac{3}{4}\tan 3\theta $
D) $ \dfrac{3}{4}\cot 3\theta $

Answer 1

Hint:
We have to solve the given trigonometric expression. For that, we will use the trigonometric identities for each term of the expression and then we will simplify the terms one by one and then add and subtract the like terms to get the simplified and final value of the given trigonometric expression.

Complete step by step solution:
We have
${{\cos }^{3}}\theta +{{\cos }^{3}}\left( {{120}^{\circ }}+\theta \right)+{{\cos }^{3}}\left( {{120}^{\circ }}-\theta \right)$ ………. $\left( 1 \right)$
We know the trigonometric identity that
$\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta $
We can write this identity as
${{\cos }^{3}}\theta =\dfrac{\cos 3\theta +3\cos \theta }{4}$
Now, we will use this identity for all three terms in equation 1.
 $=\dfrac{\cos 3\theta +3\cos \theta }{4}+\dfrac{\cos 3\left( {{120}^{\circ }}+\theta \right)+3\cos \left( {{120}^{\circ }}+\theta \right)}{4}+\dfrac{\cos 3\left( {{120}^{\circ }}-\theta \right)+3\cos \left( {{120}^{\circ }}-\theta \right)}{4}$
On simplifying the terms, we get
$=\dfrac{\cos 3\theta +3\cos \theta }{4}+\dfrac{\cos \left( {{360}^{\circ }}+3\theta \right)+3\cos \left( {{120}^{\circ }}+\theta \right)}{4}+\dfrac{\cos \left( {{360}^{\circ }}-3\theta \right)+3\cos \left( {{120}^{\circ }}-\theta \right)}{4}$
Taking $\dfrac{1}{4}$ common from all the terms, we get
$=\dfrac{1}{4}\left[ \cos 3\theta +3\cos \theta +\cos \left( {{360}^{\circ }}+3\theta \right)+3\cos \left( {{120}^{\circ }}+\theta \right)+\cos \left( {{360}^{\circ }}-3\theta \right)+3\cos \left( {{120}^{\circ }}-\theta \right) \right]$
Using periodic identities for the terms, we get
$=\dfrac{1}{4}\left[ \cos 3\theta +3\cos \theta +\cos 3\theta +3\cos \left( {{120}^{\circ }}+\theta \right)+\cos 3\theta +3\cos \left( {{120}^{\circ }}-\theta \right) \right]$
We know from the sum to product formulas of trigonometry that
\[\cos A+\cos B\text{ }=2\cos \frac{A+B}{2}.~\cos \frac{A-B}{2}~\].
We will use the trigonometric formula for the terms $\cos \left( {{120}^{\circ }}-\theta \right)$ and $\cos \left( {{120}^{\circ }}+\theta \right)$.
$=\dfrac{1}{4}\left[ \cos 3\theta +3\cos \theta +\cos 3\theta +\cos 3\theta +3\times 2\times \cos \left( \dfrac{{{120}^{\circ }}+\theta +{{120}^{\circ }}-\theta }{2} \right)\times \cos \left( \dfrac{{{120}^{\circ }}+\theta -{{120}^{\circ }}+\theta }{2} \right) \right]$
On simplifying the terms inside the brackets, we get
$=\dfrac{1}{4}\left[ \cos 3\theta +3\cos \theta +\cos 3\theta +\cos 3\theta +3\cos {{120}^{\circ }}\times \cos \theta \right]$ ……… $\left( 2 \right)$
We know the value of $\cos {{120}^{\circ }}$ is $-\dfrac{1}{2}$ .
Therefore, substituting the values in equation 2, we get
$=\dfrac{1}{4}\left[ \cos 3\theta +3\cos \theta +\cos 3\theta +\cos 3\theta +3\times 2\times -\dfrac{1}{2}\times \cos \theta \right]$
On further simplification, we get
$=\dfrac{1}{4}\left[ \cos 3\theta +3\cos \theta +\cos 3\theta +\cos 3\theta -3\cos \theta \right]$
Adding and subtracting the like terms inside the bracket, we get
$=\dfrac{1}{4}\left[ \cos 3\theta +\cos 3\theta +\cos 3\theta \right]$
Adding the terms inside the bracket, we get
$=\dfrac{3}{4}\cos 3\theta $

Thus, the correct option is option B.

Note:
We need to know the meaning of the trigonometric identities as we have used the trigonometric identities in this question. Trigonometric identities are defined as the equalities which involve the trigonometric functions and they are true for every value of the occurring variables for which both sides of the equality are defined. Remember that all the trigonometric identities are periodic in nature. They repeat their values after a certain interval.