Question

The value of ${\sum }_{r=0}^{10}{\cos }^3\left(\frac{\pi r}{3}\right)$ is equal to:

• A. $\frac{-9}{2}$
• B. $\frac{-7}{2}$
• C. $\frac{-9}{8}$
• D. $\frac{-1}{8}$

Answer 1

The correct option is D $\frac{-1}{8}$

$\cos 3\theta =4{\cos }^3\theta -3\cos \theta$

$\frac{\cos 3\theta +3\cos \theta }{4}={\cos }^3\theta$

$\theta =\frac{\pi r}{3}$

$\sum _{r=0}^{10}{\cos }^3\frac{\pi r}{3}=\sum _{r=0}^{10}\frac{\cos \pi r+3\cos \left(\frac{\pi r}{3}\right)}{4}$

$=\frac{1}{4}+\frac{3}{4}\sum _{r=0}^{10}\cos \left(\frac{\pi r}{3}\right)$

$=\frac{1}{4}+\frac{3}{4}(\cos \theta +\cos \frac{\pi }{3}+\cos \frac{2\pi }{3}+......\cos \frac{10\pi }{3})$

$=\frac{1}{4}+\frac{3}{4}\frac{\sin \left(\frac{11}{2}\frac{\pi }{3}\right)}{\sin \left(\frac{1}{2}\frac{\pi }{3}\right)}\cos \left(\frac{0+\frac{10\pi }{3}}{2}\right)$

$=\frac{1}{4}+\frac{3}{4}\left(\frac{\frac{-1}{2}}{\frac{1}{2}}\right)\left(\frac{1}{2}\right)$

$=\frac{1}{4}-\frac{3}{8}=\frac{-1}{8}$