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

Observe the value of $\cos \theta +\cos \left( {{120}^{\circ }}+\theta \right)+\cos \left( {{120}^{\circ }}-\theta \right)$ .
If $a+b+c=0$ , then ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$ .
The following identities are useful:
${{\sin }^{2}}\theta +{{\cos}^{2}}\theta =1$
$\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B$
$\cos (A+B)\cos (A-B)={{\cos }^{2}}A-{{\sin }^{2}}B$
Use the fact that $\cos (\pi -\theta )=-\cos \theta$ and $\cos 60^{\circ } =\dfrac{1}{2}$ .
Triple angle formula: $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos\theta$ .

Complete step-by-step answer:
Let us look at the value of $\cos \theta +\cos \left( {{120}^{\circ }}+\theta \right)+\cos \left( {{120}^{\circ }}-\theta \right)$ .
We know that $\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B$ .
= $\cos \theta +\left( \cos {{120}^{\circ }}\cos \theta -\sin {{120}^{\circ }}\sin \theta \right)+\left( \cos {{120}^{\circ }}\cos \theta +\sin {{120}^{\circ }}\sin \theta \right)$
= $\cos \theta +2\cos {{120}^{\circ }}\cos \theta$
= $\cos \theta +2\cos \left( {{180}^{\circ }}-{{60}^{\circ }} \right)\cos \theta$
Using $\cos (\pi -\theta )=-\cos \theta$ , we get:
= $\cos \theta +2\left( -\cos {{60}^{\circ }} \right)\cos \theta$
Substituting the value $\cos 60^{\circ } =\dfrac{1}{2}$ , we get:
= $\cos \theta +2\left( -\dfrac{1}{2} \right)\cos \theta$
= $\cos \theta -\cos \theta$
= 0
Now, we also know that if $a+b+c=0$ , then ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$ .
∴ ${{\cos }^{3}}\theta +{{\cos }^{3}}\left( {{120}^{\circ }}+\theta \right)+{{\cos }^{3}}\left( {{120}^{\circ }}-\theta \right)$
= $3\cos \theta \cos \left( {{120}^{\circ }}+\theta \right)\cos \left( {{120}^{\circ }}-\theta \right)$
Using $\cos (A+B)\cos (A-B)={{\cos }^{2}}A-{{\sin }^{2}}B$ , we get:
= $3\cos \theta \left( {{\cos }^{2}}{{120}^{\circ }}-{{\sin }^{2}}\theta \right)$
Substituting the value of $\cos 120^{\circ }=-\dfrac{1}{2}$ calculated earlier and squaring it, we get:
= $3\cos \theta \left( \dfrac{1}{4}-{{\sin }^{2}}\theta \right)$
Using ${{\sin }^{2}}\theta + {{\cos}^{2}}\theta =1$ , we can write:
= $3\cos \theta \left( \dfrac{1}{4}+{{\cos }^{2}}\theta -1 \right)$
= $3\cos \theta \left( {{\cos }^{2}}\theta -\dfrac{3}{4} \right)$
Taking out $\dfrac{1}{4}$ as a common multiple, this can be written as:
= $\dfrac{3}{4}\cos \theta \left( 4{{\cos }^{2}}\theta -3 \right)$
= $\dfrac{3}{4}\left( 4{{\cos }^{3}}\theta -3\cos \theta \right)$
On using the triple angle formula $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos\theta$ , we get:
= $\dfrac{3}{4}\cos 3\theta$
Hence, the correct answer is B. $\dfrac{3}{4}\cos 3\theta$ .

Note: (i)${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=(a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)$
If $a+b+c=0$ , then ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$ .
(ii)Using the identity for $\sin (A+B)$ and $\cos (A+B)$ , we can find the expressions of $\sin 3A$ and $\cos 3A$ by using the fact that $3A=2A+A$ , and then $\sin 5A$ and $\cos 5A$ as $5A=3A+2A$ , and so on.
(iii)Since ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ , we can always calculate $\cos A$ from $\sin A$ , and then the other ratios follow.