Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

${\sum }_{r=0}^{10}{\cos }^3\frac{r\pi }{3}=\frac{1}{4}{\sum }_{r=0}^{10}\left(3\cos \frac{r\pi }{3}+\cos r\pi \right)$. Now ${\sum }_{r=0}^{10}\cos r\pi =1-1+1-\cdots +1=1$ and ${\sum }_{r=0}^{10}\cos \frac{r\pi }{3}=\frac{\cos \frac{10\pi }{6}\sin \frac{11\pi }{6}}{\sin \frac{\pi }{6}}=\frac{\cos \frac{5\pi }{3}\cdot \left(-\frac{1}{2}\right)}{\frac{1}{2}}=-\frac{1}{2}$. Hence the sum $=\frac{1}{4}\left[3\left(-\frac{1}{2}\right)+1\right]=-\frac{1}{8}=\frac{-a}{b}$, so $b=8$. \end{document}