Courses
Courses for Kids
Free study material
Free LIVE classes
More LIVE
Join Vedantu’s FREE Mastercalss

# Evaluate the equation $\cos \dfrac{2\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{6\pi }{7}$ A) Is equal to zeroB) Lies between 0 and 3C) Is a negative numberD) Lies between 3 and 6 Verified
361.8k+ views
Hint: Multiply and divide with $\sin \dfrac{\pi }{7}$ to the expression then simplify the expression using the trigonometric identities.

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)
Last updated date: 26th Sep 2023
Total views: 361.8k
Views today: 6.61k