Question

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

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

Answer 1

The correct option is B $\frac{3}{4}$

$cos3x=4co{s}^{3}x-3cosx$

$⟹co{s}^{3}x=\frac{cos3x+3cosx}{4}$

$co{s}^3\theta +co{s}^3(\frac{2\pi }{3}+\theta )+co{s}^3(\frac{4\pi }{3}+\theta )$

$=\frac{cos3\theta +3cos\theta }{4}+\frac{cos3(\frac{2\pi }{3}+\theta )+3cos(\frac{2\pi }{3}+\theta )}{4}+\frac{cos3(\frac{4\pi }{3}+\theta )+3cos(\frac{4\pi }{3}+\theta )}{4}$

$=\frac{cos3\theta +3cos\theta }{4}+\frac{cos(2\pi +3\theta )+3cos(\frac{2\pi }{3}+\theta )}{4}+\frac{cos(4\pi +3\theta )+3cos(\frac{4\pi }{3}+\theta )}{4}$

$=\frac{cos3\theta +3cos\theta }{4}+\frac{cos\left(3\theta \right)+3cos(\frac{2\pi }{3}+\theta )}{4}+\frac{cos\left(3\theta \right)+3cos(\frac{4\pi }{3}+\theta )}{4}$

$=\frac{3cos3\theta +3cos\theta }{4}+\frac{3\left(cos\right(\frac{2\pi }{3}+\theta )+cos(\frac{4\pi }{3}+\theta \left)\right)}{4}$

$=\frac{3cos3\theta +3cos\theta }{4}+\frac{3\left(2cos\right(\pi +\theta \left)cos\right(\frac{-\pi }{3}\left)\right)}{4}$

$=\frac{3cos3\theta +3cos\theta }{4}+\frac{3cos(\pi +\theta )}{4}$

$=\frac{3cos3\theta +3cos\theta }{4}-\frac{3cos\left(\theta \right)}{4}$

$=\frac{3cos3\theta }{4}$

$⟹a=\frac{3}{4}$

Hence, the answer is option(B).