Solveeit Logo
Login

Question

Question: Evaluate the equation \(\cos \dfrac{2\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{6\pi }{7}\) A) Is...

Evaluate the equation cos⁡2π7+cos⁡4π7+cos⁡6π7\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

Explanation

Solution

Hint: Multiply and divide with sin⁡π7\sin \dfrac{\pi }{7} to the expression then simplify the expression using the trigonometric identities.

Complete step-by-step answer:
We have expression
cos⁡2π7+cos⁡4π7+cos⁡6π7\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⁡2π7+cos⁡4π7+cos⁡6π7............(1)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 2sin⁡π72\sin \dfrac{\pi }{7} to both sides and then apply trigonometric identity as follows:
2Msin⁡π7=2sin⁡π7cos⁡2π7+2sin⁡π7cos⁡4π7+2sin⁡π7cos⁡6π7..........(2)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
2Msin⁡π7=sin⁡(π7+2π7)+sin⁡(π7−2π7)+sin⁡(π7+4π7)+sin⁡(π7−4π7)+sin⁡(π7+6π7)+sin⁡(π7−6π7)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;
2Msin⁡π7=sin⁡3π7+sin⁡(−π7)+sin⁡5π7+sin⁡(−3π7)+sin⁡(7π7)+sin⁡(−5π7)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⁡(−θ)=−sin⁡θ................(4)\sin \left( -\theta \right)=-\sin \theta ................\left( 4 \right)
Therefore, we can write the value of 2Msin⁡π72M\sin \dfrac{\pi }{7} by using the relation (4), we get;
2Msin⁡π7=sin⁡3π7−sin⁡π7+sin⁡5π7−sin⁡3π7+sin⁡7π7−sin⁡5π72M\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;
2Msin⁡π7=−sin⁡π7+sin⁡(7π7)2M\sin \dfrac{\pi }{7}=-\sin \dfrac{\pi }{7}+\sin \left( \dfrac{7\pi }{7} \right)
Or
2Msin⁡π7=−sin⁡π7+sin⁡π2M\sin \dfrac{\pi }{7}=-\sin \dfrac{\pi }{7}+\sin \pi
We have value if sin⁡π\sin \pi is zero. Hence, above equation can be written as
2Msin⁡π7=−sin⁡π7 M=−sin⁡π72sin⁡π7 \begin{aligned} & 2M\sin \dfrac{\pi }{7}=-\sin \dfrac{\pi }{7} \\\ & M=\dfrac{-\sin \dfrac{\pi }{7}}{2\sin \dfrac{\pi }{7}} \\\ \end{aligned}
Therefore M=−12M=\dfrac{-1}{2}
Hence, value of cos⁡2π7+cos⁡4π7+cos⁡6π7\cos \dfrac{2\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{6\pi }{7} is −12\dfrac{-1}{2} .
Hence, option C is correct from the given options.
Note: Key point of the question is multiplication by sin⁡π7\sin \dfrac{\pi }{7} to the expression cos⁡2π7+cos⁡4π7+cos⁡6π7.\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)