Question

The value of $\cos \dfrac{\pi }{7}+\cos \dfrac{2\pi }{7}+\cos \dfrac{3\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{5\pi }{7}+\cos \dfrac{6\pi }{7}+\cos \dfrac{7\pi }{7}$ is.
(a) 1
(b) -1
(c) 0
(d) none of these

Answer 1

Hint:For solving this question first we will use trigonometric formula $\cos \left( \theta \right)+\cos \left( \pi -\theta \right)=0$ and trigonometric ratio $\cos \pi =-1$ for simplifying the given term. After that, we will easily solve it without making any mistakes and find the correct value of the given term and select the correct answer.

Complete step-by-step answer:
Given:
We have to find the value of $\cos \dfrac{\pi }{7}+\cos \dfrac{2\pi }{7}+\cos \dfrac{3\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{5\pi }{7}+\cos \dfrac{6\pi }{7}+\cos \dfrac{7\pi }{7}$ .
Now, before we proceed we should know the following formulas:
$\begin{aligned} & \cos \left( \theta \right)+\cos \left( \pi -\theta \right)=0..........\left( 1 \right) \\\ & \cos \pi =-1..............................\left( 2 \right) \\\ \end{aligned}$
Now, we will use the above formulas to simplify the given term.
We have the following equation:
$\begin{aligned} & \cos \dfrac{\pi }{7}+\cos \dfrac{2\pi }{7}+\cos \dfrac{3\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{5\pi }{7}+\cos \dfrac{6\pi }{7}+\cos \dfrac{7\pi }{7} \\\ & \Rightarrow \left( \cos \dfrac{\pi }{7}+\cos \dfrac{6\pi }{7} \right)+\left( \cos \dfrac{2\pi }{7}+\cos \dfrac{5\pi }{7} \right)+\left( \cos \dfrac{3\pi }{7}+\cos \dfrac{4\pi }{7} \right)+\cos \pi \\\ \end{aligned}$
Now, write $\dfrac{6\pi }{7}=\pi -\dfrac{\pi }{7}$ , $\dfrac{5\pi }{7}=\pi -\dfrac{2\pi }{7}$ and $\dfrac{4\pi }{7}=\pi -\dfrac{3\pi }{7}$ . Then,
$\begin{aligned} & \left( \cos \dfrac{\pi }{7}+\cos \dfrac{6\pi }{7} \right)+\left( \cos \dfrac{2\pi }{7}+\cos \dfrac{5\pi }{7} \right)+\left( \cos \dfrac{3\pi }{7}+\cos \dfrac{4\pi }{7} \right)+\cos \pi \\\ & \Rightarrow \left( \cos \dfrac{\pi }{7}+\cos \left( \pi -\dfrac{\pi }{7} \right) \right)+\left( \cos \dfrac{2\pi }{7}+\cos \left( \pi -\dfrac{2\pi }{7} \right) \right)+\left( \cos \dfrac{3\pi }{7}+\cos \left( \pi -\dfrac{3\pi }{7} \right) \right)+\cos \pi \\\ \end{aligned}$
Now, using the formula from the equation (1) in the above to simplify it. Then,
$\begin{aligned} & \left( \cos \dfrac{\pi }{7}+\cos \left( \pi -\dfrac{\pi }{7} \right) \right)+\left( \cos \dfrac{2\pi }{7}+\cos \left( \pi -\dfrac{2\pi }{7} \right) \right)+\left( \cos \dfrac{3\pi }{7}+\cos \left( \pi -\dfrac{3\pi }{7} \right) \right)+\cos \pi \\\ & \Rightarrow \left( \cos \dfrac{\pi }{7}-\cos \dfrac{\pi }{7} \right)+\left( \cos \dfrac{2\pi }{7}-\cos \dfrac{2\pi }{7} \right)+\left( \cos \dfrac{3\pi }{7}-\cos \dfrac{3\pi }{7} \right)+\cos \pi \\\ & \Rightarrow \cos \pi \\\ \end{aligned}$
Now, from the formula from the equation (2), we can write $\cos \pi =-1$ in the above to find the final answer. Then,
$\begin{aligned} & \cos \dfrac{\pi }{7}+\cos \dfrac{2\pi }{7}+\cos \dfrac{3\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{5\pi }{7}+\cos \dfrac{6\pi }{7}+\cos \dfrac{7\pi }{7} \\\ & \Rightarrow \cos \pi \\\ & \Rightarrow -1 \\\ \end{aligned}$
Now, from the above result, it is evident that the value of $\cos \dfrac{\pi }{7}+\cos \dfrac{2\pi }{7}+\cos \dfrac{3\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{5\pi }{7}+\cos \dfrac{6\pi }{7}+\cos \dfrac{7\pi }{7}$ will be equal to -1.
Hence, option (b) will be the correct option.

Note: Here, the student should first understand what is asked in the problem and then proceed in the right direction to get the correct answer quickly. Moreover, for objective problems, we should directly apply the formula that if $A+B=\pi$ , then $\cos A+\cos B=0$ for solving such types of problems directly without doing any long calculation.