Question

Evaluate the equation $\cos \dfrac{2\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{6\pi }{7}$
A) Is equal to zero
B) Lies between 0 and 3
C) Is a negative number
D) Lies between 3 and 6

Answer 1

Hint: Multiply and divide with $\sin \dfrac{\pi }{7}$ to the expression then simplify the expression using the trigonometric identities.

Complete step-by-step answer:
We have expression
$\cos \dfrac{2\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{6\pi }{7}$
Let us suppose the given expression is M.
$M=\cos \dfrac{2\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{6\pi }{7}............\left( 1 \right)$
Now, multiply the equation (1) by using $2\sin \dfrac{\pi }{7}$ to both sides and then apply trigonometric identity as follows:
\[2M\sin \dfrac{\pi }{7}=2\sin \dfrac{\pi }{7}\cos \dfrac{2\pi }{7}+2\sin \dfrac{\pi }{7}\cos \dfrac{4\pi }{7}+2\sin \dfrac{\pi }{7}\cos \dfrac{6\pi }{7}..........\left( 2 \right)\]
Here, we have to apply relation;
2 sin A cos B = sin (A + B) + sin (A – B)............................(3)
Now apply the above trigonometric identity in equation (2), we get
$2M\sin \dfrac{\pi }{7}=\sin \left( \dfrac{\pi }{7}+\dfrac{2\pi }{7} \right)+\sin \left( \dfrac{\pi }{7}-\dfrac{2\pi }{7} \right)+\sin \left( \dfrac{\pi }{7}+\dfrac{4\pi }{7} \right)+\sin \left( \dfrac{\pi }{7}-\dfrac{4\pi }{7} \right)+\sin \left( \dfrac{\pi }{7}+\dfrac{6\pi }{7} \right)+\sin \left( \dfrac{\pi }{7}-\dfrac{6\pi }{7} \right)$ One simplifying the above equation, we get;
$2M\sin \dfrac{\pi }{7}=\sin \dfrac{3\pi }{7}+\sin \left( \dfrac{-\pi }{7} \right)+\sin \dfrac{5\pi }{7}+\sin \left( \dfrac{-3\pi }{7} \right)+\sin \left( \dfrac{7\pi }{7} \right)+\sin \left( \dfrac{-5\pi }{7} \right)$
As, we know that;
$\sin \left( -\theta \right)=-\sin \theta ................\left( 4 \right)$
Therefore, we can write the value of $2M\sin \dfrac{\pi }{7}$ by using the relation (4), we get;
\[2M\sin \dfrac{\pi }{7}=\sin \dfrac{3\pi }{7}-\sin \dfrac{\pi }{7}+\sin \dfrac{5\pi }{7}-\sin \dfrac{3\pi }{7}+\sin \dfrac{7\pi }{7}-\sin \dfrac{5\pi }{7}\]
Cancelling out same terms with positive and negative signs, we get;
$2M\sin \dfrac{\pi }{7}=-\sin \dfrac{\pi }{7}+\sin \left( \dfrac{7\pi }{7} \right)$
Or
$2M\sin \dfrac{\pi }{7}=-\sin \dfrac{\pi }{7}+\sin \pi $
We have value if $\sin \pi $ is zero. Hence, above equation can be written as
$\begin{align}
  & 2M\sin \dfrac{\pi }{7}=-\sin \dfrac{\pi }{7} \\
 & M=\dfrac{-\sin \dfrac{\pi }{7}}{2\sin \dfrac{\pi }{7}} \\
\end{align}$
Therefore $M=\dfrac{-1}{2}$
Hence, value of $\cos \dfrac{2\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{6\pi }{7}$ is $\dfrac{-1}{2}$ .
Hence, option C is correct from the given options.
Note: Key point of the question is multiplication by $\sin \dfrac{\pi }{7}$ to the expression $\cos \dfrac{2\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{6\pi }{7}.$ As, we know after multiplying, we will get expression of type 2sinAcosB which cancel out all terms.
One can go wrong while applying the formula of 2sinAcosB. Confusion of plus or minus sign between sin(A+B) and sin (A-B) may occur. So, we can verify that by just expanding sin(A+B) and sin(A-B). And, we get to know identity as;
2sinAcosB=sin(A+B)+sin(A-B)