Question

Find the value of
$\cos \left( \dfrac{2\pi }{7} \right)+\cos \left( \dfrac{4\pi }{7} \right)+\cos \left( \dfrac{6\pi }{7} \right)+\dfrac{3}{2}$

Answer 1

Hint: In this case, we are given a sum of cosine values of angles differing by $\dfrac{\pi }{7}$. Therefore, we should try to express it in a form such that only the trigonometric ratio of a known angle is required and the rest of the terms get cancelled out. For this we will use the formula for addition of the sine values of the sum and difference of two angles. Thereafter, we can evaluate the expression to obtain the answer.

Complete step-by-step answer:
The expression given to us is
$\cos \left( \dfrac{2\pi }{7} \right)+\cos \left( \dfrac{4\pi }{7} \right)+\cos \left( \dfrac{6\pi }{7} \right)+\dfrac{3}{2}.............(1.1)$
However, the cosine values of the given angles are not easy to directly evaluate as they do not correspond to any standard angles. However, we can use the formula for the sine of the sum and difference of angles which is stated below
$\sin (a+b)=\sin (a)\cos (b)+\cos (a)\sin (b).........(1.2)$
and
$\sin (a-b)=\sin (a)\cos (b)-\cos (a)\sin (b).........(1.3)$
Adding equations (1.2) and (1.3), we obtain

& \sin (a+b)+\sin (a-b)=\sin (a)\cos (b)+\cos (a)\sin (b)+\left( \sin (a)\cos (b)-\cos (a)\sin (b) \right) \\\ & \Rightarrow \sin (a+b)+\sin (a-b)=2\sin (a)\cos (b).........(1.4) \\\ \end{aligned}$$ Now, to convert equation (1.1) to the form of equation(1.4), we can multiply and divide the first three terms by $2\sin \left( \dfrac{\pi }{7} \right)$ to obtain $\begin{aligned} & \cos \left( \dfrac{2\pi }{7} \right)+\cos \left( \dfrac{4\pi }{7} \right)+\cos \left( \dfrac{6\pi }{7} \right)+\dfrac{3}{2} \\\ & =\dfrac{1}{2\sin \left( \dfrac{\pi }{7} \right)}\left( 2\sin \left( \dfrac{\pi }{7} \right)\times \cos \left( \dfrac{2\pi }{7} \right)+2\sin \left( \dfrac{\pi }{7} \right)\times \cos \left( \dfrac{4\pi }{7} \right)+2\sin \left( \dfrac{\pi }{7} \right)\times \cos \left( \dfrac{6\pi }{7} \right) \right)+\dfrac{3}{2} \\\ \end{aligned}$ Now, using equation (1.4) in the first three terms, we can write the above equation as $\begin{aligned} & \cos \left( \dfrac{2\pi }{7} \right)+\cos \left( \dfrac{4\pi }{7} \right)+\cos \left( \dfrac{6\pi }{7} \right)+\dfrac{3}{2} \\\ & =\dfrac{1}{2\sin \left( \dfrac{\pi }{7} \right)}\left( \sin \left( \dfrac{\pi }{7}+\dfrac{2\pi }{7} \right)+\sin \left( \dfrac{\pi }{7}-\dfrac{2\pi }{7} \right)+\sin \left( \dfrac{\pi }{7}+\dfrac{4\pi }{7} \right)+\sin \left( \dfrac{\pi }{7}-\dfrac{4\pi }{7} \right)+\sin \left( \dfrac{\pi }{7}+\dfrac{6\pi }{7} \right)+\sin \left( \dfrac{\pi }{7}-\dfrac{6\pi }{7} \right) \right)+\dfrac{3}{2} \\\ & =\dfrac{1}{2\sin \left( \dfrac{\pi }{7} \right)}\left( \sin \left( \dfrac{3\pi }{7} \right)+\sin \left( \dfrac{-\pi }{7} \right)+\sin \left( \dfrac{5\pi }{7} \right)+\sin \left( \dfrac{-3\pi }{7} \right)+\sin \left( \dfrac{7\pi }{7} \right)+\sin \left( \dfrac{-5\pi }{7} \right) \right)+\dfrac{3}{2}................(1.5) \\\ \end{aligned}$ However, we know that sine function is an odd function. Therefore $\sin \left( \theta \right)=-\sin \left( \theta \right)..................(1.6)$ We, notice that, inside the brackets, the $\sin \left( \dfrac{3\pi }{7} \right),\sin \left( \dfrac{-3\pi }{7} \right),\sin \left( \dfrac{5\pi }{7} \right)\text{ and }\sin \left( \dfrac{-5\pi }{7} \right)$ terms cancel out and $\sin \left( \dfrac{7\pi }{7} \right)=\sin \left( \pi \right)=0$ as $\pi ={{180}^{\circ }}$ and $\sin \left( {{180}^{\circ }} \right)=0$. Thus, we can rewrite equation (1.5) as $\begin{aligned} & \cos \left( \dfrac{2\pi }{7} \right)+\cos \left( \dfrac{4\pi }{7} \right)+\cos \left( \dfrac{6\pi }{7} \right)+\dfrac{3}{2} \\\ & =\dfrac{1}{2\sin \left( \dfrac{\pi }{7} \right)}\left( \sin \left( \dfrac{-\pi }{7} \right)+0 \right)+\dfrac{3}{2} \\\ & =\dfrac{\sin \left( \dfrac{-\pi }{7} \right)}{2\sin \left( \dfrac{\pi }{7} \right)}+\dfrac{3}{2}................(1.7) \\\ \end{aligned}$ Using equation(1.6), we have $\sin \left( \dfrac{-\pi }{7} \right)=-\sin \left( \dfrac{\pi }{7} \right)$. Using it in equation (1.6), we obtain $\begin{aligned} & \cos \left( \dfrac{2\pi }{7} \right)+\cos \left( \dfrac{4\pi }{7} \right)+\cos \left( \dfrac{6\pi }{7} \right)+\dfrac{3}{2} \\\ & =\dfrac{-\sin \left( \dfrac{\pi }{7} \right)}{2\sin \left( \dfrac{\pi }{7} \right)}+\dfrac{3}{2}=\dfrac{-1}{2}+\dfrac{3}{2}=\dfrac{-1+3}{2}=\dfrac{2}{2}=1 \\\ \end{aligned}$ Thus, the answer to the given question is found to be one. Note: It is important to note that in equation (1.6), we should mention that the terms having opposite signs cancel out because the sine function is an odd function. This is not generally true of other trigonometric functions such as cosine.