Question

If ${\cos }^3\theta +{\cos }^3\left(\frac{2\pi }{3}+\theta \right)+{\cos }^3\left(\frac{4\pi }{3}+\theta \right)=a\cos 3\theta$, then $a$ is equal to

• A. $\frac{1}{4}$
• B. $\frac{3}{4}$
• C. $\frac{5}{4}$
• D. $\frac{7}{4}$

Answer 1

Answer: (B)

We have,
${\cos }^3\theta +{\cos }^3\left(\frac{2\pi }{3}+\theta \right)+{\cos }^3\left(\frac{4\pi }{3}+\theta \right)=a\cos 3\theta$
$\Rightarrow \frac{\cos 3\theta +3\cos \theta }{4}+\frac{\cos 3\left(\frac{2\pi }{3}+\theta \right)+3\cos \left(\frac{2\pi }{3}+\theta \right)}{4}$
$+\frac{\cos 3\left(\frac{4\pi }{3}+\theta \right)+3\cos \left(\frac{4\pi }{3}+\theta \right)}{4}=a\cos 3\theta$
$\Rightarrow \cos 3\theta +3\cos \theta +\cos (2\pi +3\theta )+3\cos \left(\frac{2\pi }{3}+\theta \right)$
$+\cos (4\pi +3\theta )+3\cos \left(\frac{4\pi }{3}+\theta \right)=4a\cos 3\theta$
$\Rightarrow \cos 3\theta +3\cos \theta +\cos 3\theta +\cos 3\theta +3[\cos \left(\frac{2\pi }{3}+\theta \right)$
$+\cos \left(\frac{4\pi }{3}+\theta \right)]=4a\cos 3\theta$
$\Rightarrow 3\cos 3\theta +3\cos \theta +3\left[2\cos (\pi +\theta )\cos \frac{\pi }{3}\right]$
$=4a\cos 3\theta$
$\Rightarrow 3\cos 3\theta +3\cos \theta +3\times 2\times (-\cos \theta )\times \frac{1}{2}$
$=4a\cos 3\theta$
$\Rightarrow 3\cos 3\theta +3\cos \theta -3\cos \theta =4a\cos 3\theta$
$\Rightarrow 3\cos 3\theta =4a\cos 3\theta$
$\Rightarrow a=\frac{3}{4}$