The value of $\cos \left( {\dfrac{{2\pi }}{7}} \right) + \cos \left( {\dfrac{{4\pi }}{7}} \right) + \cos \left( {\dfrac{{6\pi }}{7}} \right)$
A. Is equal to zero
B. Lies between $0$ and $3$
C. Is a negative number
D. Lies between $3$ and $6$

Hint: In the given problem, we need to find the value of $\cos \left( {\dfrac{{2\pi }}{7}} \right) + \cos \left( {\dfrac{{4\pi }}{7}} \right) + \cos \left( {\dfrac{{6\pi }}{7}} \right)$. At first, we will multiply and divide the above written expression by $2\sin \left( {\dfrac{\pi }{7}} \right)$. After this, we will apply the trigonometric formula $2\sin x\cos y = \sin \left( {x + y} \right) + \sin \left( {x - y} \right)$ and simplify it further to get our required answer.

Formula Used:
$2\sin x\cos y = \sin \left( {x + y} \right) + \sin \left( {x - y} \right)$

Complete step by step solution:
We have, $\cos \left( {\dfrac{{2\pi }}{7}} \right) + \cos \left( {\dfrac{{4\pi }}{7}} \right) + \cos \left( {\dfrac{{6\pi }}{7}} \right)$
We will multiply and divide the above written expression by $2\sin \left( {\dfrac{\pi }{7}} \right)$
$\dfrac{1}{{2\sin \left( {\dfrac{\pi }{7}} \right)}}\left[ {2\sin \left( {\dfrac{\pi }{7}} \right)\cos \left( {\dfrac{{2\pi }}{7}} \right) + 2\sin \left( {\dfrac{\pi }{7}} \right)\cos \left( {\dfrac{{4\pi }}{7}} \right) + 2\sin \left( {\dfrac{\pi }{7}} \right)\cos \left( {\dfrac{{6\pi }}{7}} \right)} \right]$
We know that $2\sin x\cos y = \sin \left( {x + y} \right) + \sin \left( {x - y} \right)$. Therefore, we get
$ = \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]$
Now, take L.C.M.
$ = \dfrac{1}{{2\sin \left( {\dfrac{\pi }{7}} \right)}}\left[ {\sin \left( {\dfrac{{\pi + 2\pi }}{7}} \right) + \sin \left( {\dfrac{{\pi - 2\pi }}{7}} \right) + \sin \left( {\dfrac{{\pi + 4\pi }}{7}} \right) + \sin \left( {\dfrac{{\pi - 4\pi }}{7}} \right) + \sin \left( {\dfrac{{\pi + 6\pi }}{7}} \right) + \sin \left( {\dfrac{{\pi - 6\pi }}{7}} \right)} \right]$
On simplifying the above written expression further, we get
$ = \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]$
As we know that $\sin \left( { - \theta } \right) = - \sin \theta $. Therefore, we get
$ = \dfrac{1}{{2\sin \dfrac{\pi }{7}}}\left[ {\sin \dfrac{{3\pi }}{7} - \sin \dfrac{\pi }{7} + \sin \dfrac{{5\pi }}{7} - \sin \dfrac{{3\pi }}{7} + \sin \pi - \sin \dfrac{{5\pi }}{7}} \right]$
$ = \dfrac{1}{{2\sin \dfrac{\pi }{7}}}\left[ { - \sin \dfrac{\pi }{7} + \sin \pi } \right]$
As we know that $\sin \pi = 0$. Therefore, we get
$ = \dfrac{1}{{2\sin \dfrac{\pi }{7}}}\left[ { - \sin \dfrac{\pi }{7} + 0} \right]$
$ = - \dfrac{1}{2}$
Hence, the value of $\cos \left( {\dfrac{{2\pi }}{7}} \right) + \cos \left( {\dfrac{{4\pi }}{7}} \right) + \cos \left( {\dfrac{{6\pi }}{7}} \right)$ is $ - \dfrac{1}{2}$, which is a negative number.

Option ‘C’ is correct

Note: To solve these types of problems, at first, observe the given expression carefully and try to make it in such a form in which you can apply trigonometric identities as here we multiplied and divided the expression by $2\sin \left( {\dfrac{\pi }{7}} \right)$. To solve these types of questions, one must remember all the standard formulas of trigonometric functions. Most of the trigonometric functions questions are just based on substitutions, we can only solve the problem if we know the formulas and sign convention. Students should take care of the calculations so as to be sure of their final answer.
Some trigonometric formulas are written below:
- $2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)$
- $2\cos A\sin B = \sin \left( {A + B} \right) - \sin \left( {A - B} \right)$
- $2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)$
- $2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)$
- $\sin C + \sin D = 2\sin \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$
- $\sin C - \sin D = 2\sin \dfrac{{C - D}}{2}\cos \dfrac{{C + D}}{2}$
- $\cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$
- $\cos C - \cos D = - 2\sin \dfrac{{C + D}}{2}\sin \dfrac{{C - D}}{2}$
We use them according to the given problem.