Question

The value of $\sum\limits_{r = 0}^{10} {{{\cos }^3}} \left( {\dfrac{{\pi r}}{3}} \right)$ is equal to:
A. $ - \dfrac{9}{2}$
B. $ - \dfrac{7}{2}$
C. $ - \dfrac{9}{8}$
D. $ - \dfrac{1}{8}$

Answer 1

Hint: Start by dissolving ${\cos ^3}\left( {\dfrac{{\pi r}}{3}} \right)$ by using the formula of $\cos 3\theta $, and then apply the summation.
$\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta $
$\dfrac{{\cos 3\theta + 3\cos \theta }}{4} = {\cos ^3}\theta $

Complete Step-by-Step solution:
Since in our question instead of $\theta $ we have$\dfrac{{\pi r}}{3}$, therefore, put $\theta = \dfrac{{\pi r}}{3}$,
$ \Rightarrow \sum\limits_{r = 0}^{10} {{{\cos }^3}\dfrac{{\pi r}}{3}} = \sum\limits_{r = 0}^{10} {\dfrac{{\cos \pi r + 3\cos \left( {\dfrac{{\pi r}}{3}} \right)}}{4}} $
Next step is to take the constants out of the summation, therefore,
$ \Rightarrow \dfrac{1}{4} + \dfrac{3}{4}\sum\limits_{r = 0}^{10} {\cos \left( {\dfrac{{\pi r}}{3}} \right)} $
Next step is to solve the summation, therefore,
$ \Rightarrow \dfrac{1}{4} + \dfrac{3}{4}\left( {\cos \theta + \cos \dfrac{\pi }{3} + \cos \dfrac{{2\pi }}{3} + ... + \cos \dfrac{{10\pi }}{3}} \right)$
We can simplify the bracket by writing it as,
$ \Rightarrow \dfrac{1}{4} + \dfrac{3}{4}\dfrac{{\sin \left( {\dfrac{{11}}{2}\dfrac{\pi }{3}} \right)}}{{\sin \left( {\dfrac{1}{2}\dfrac{\pi }{3}} \right)}}\cos \left( {\dfrac{{0 + \dfrac{{10\pi }}{3}}}{2}} \right)$
Solving the above equation, we get,
$ \Rightarrow \dfrac{1}{4} + \dfrac{3}{4}\left( {\dfrac{{ - \dfrac{1}{2}}}{{\dfrac{1}{2}}}} \right)\left( {\dfrac{1}{2}} \right)$
On simplifying the above, we get,
$ \Rightarrow \dfrac{1}{4} + \dfrac{3}{8} = - \dfrac{1}{8}$
Therefore, the answer is option D.

Note: Make sure you take note of the quadrants in which the trigonometric ratios are lying (sin and cos in this case) so that the appropriate signs are added in accordance to the quadrants in which they are lying.