Question

${\sum }_{r=0}^{10}co{s}^3\frac{\pi r}{3}=$

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

Answer 1

The correct option is D

$-\frac{1}{8}$

$LetI={\sum }_{r=0}^{10}co{s}^3\left(\frac{\pi r}{3}\right)\Rightarrow \frac{1}{4}{\sum }_{r=0}^{10}(cos\pi r+3cos\frac{\pi r}{3})I_1=\frac{1}{4}{\sum }_{r=0}^{10}cos\pi r{I}_1=\frac{1}{4}[cos0+cos\pi +cos2\pi +cos3\pi +..........+cos10\pi ]=\frac{1}{4}[1-1+1-1+\mathrm{1........1}+1]=\frac{1}{4}.$
$I_2=\left\{\frac{cos(\frac{10}{2}.\frac{\pi }{3})sin\frac{11\pi }{6}}{sin\frac{\pi }{6}}\right\}3.\frac{1}{4}\therefore I_2=-\frac{3}{8}\therefore I_1+I_2=\frac{1}{4}-\frac{3}{8}=-\frac{1}{8}$