Question

The expression $\cos \dfrac{{10\pi }}{{13}} + \cos \dfrac{{8\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}$ is
A. $- 1$
B. $0$
C. $1$
D.None of these

Answer 1

Hint : We can see that in the given question, we have a trigonometric function, as cosine or $\cos$ is a trigonometric ratio. So we will apply the trigonometric identity to solve this question, such as
$\cos \left( {\pi - \theta } \right) = - \cos \theta$ .
We will try to bring the first and second term similar to the other two terms by using this formula.

Complete step-by-step answer :
Here we have been given
$\cos \dfrac{{10\pi }}{{13}} + \cos \dfrac{{8\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}$ .
Let us take the first term i.e.
$\cos \dfrac{{10\pi }}{{13}}$ .
We can write this also as
$\cos \left( {\pi - \dfrac{{3\pi }}{{13}}} \right)$ , as on simplifying it gives us the original value i.e.
$\cos \left( {\dfrac{{13\pi - 3\pi }}{{13}}} \right) = \cos \left( {\dfrac{{10\pi }}{{13}}} \right)$
Now we will take the second term
$\cos \dfrac{{8\pi }}{{13}}$ .
Again we can write this also as
$\cos \left( {\pi - \dfrac{{5\pi }}{{13}}} \right)$ , as on simplifying it gives us the original value i.e.
$\cos \left( {\dfrac{{13\pi - 5\pi }}{{13}}} \right) = \cos \left( {\dfrac{{8\pi }}{{13}}} \right)$ .
Now by applying the trigonometric identity, we can write
$\cos \left( {\pi - \dfrac{{3\pi }}{{13}}} \right) = - \cos \dfrac{{3\pi }}{{13}}$
Similarly , we can write
$\cos \left( {\pi - \dfrac{{5\pi }}{{13}}} \right) = - \cos \dfrac{{5\pi }}{{13}}$
By putting the values back in the expression we can write :
$- \cos \dfrac{{3\pi }}{{13}} + \left( { - \cos \dfrac{{5\pi }}{{13}}} \right) + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}}$ .
By arranging the terms, we have :
$\cos \dfrac{{3\pi }}{{13}} - \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}} - \cos \dfrac{{5\pi }}{{13}}$ .
It gives us the value $0$ .
Hence the correct option is (b) $0$ .
So, the correct answer is “Option b”.

Note : We should note that we can also solve this question, with an alternate method. We will take the first and the term together and the second and fourth term together .
So we have
$\cos \dfrac{{10\pi }}{{13}} + \cos \dfrac{{3\pi }}{{13}} + \cos \dfrac{{5\pi }}{{13}} + \cos \dfrac{{8\pi }}{{13}}$ .
We will use the formula
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
By applying the formula we can write :
$2\cos \left( {\dfrac{{\dfrac{{10\pi }}{{13}} + \dfrac{{3\pi }}{{13}}}}{2}} \right) + \cos \left( {\dfrac{{\dfrac{{8\pi }}{{13}} - \dfrac{{5\pi }}{{13}}}}{2}} \right)$ .
On simplifying the value we have:
$2\cos \left( {\dfrac{{\dfrac{{10\pi + 3\pi }}{{13}}}}{2}} \right) + \cos \left( {\dfrac{{\dfrac{{8\pi - 5\pi }}{{13}}}}{2}} \right) = 2\cos \left( {\dfrac{{\dfrac{{13\pi }}{{13}}}}{2}} \right) + \cos \left( {\dfrac{{\dfrac{{3\pi }}{{13}}}}{2}} \right)$
We can cancel the same factors from numerator and denominator and we have:
$2\cos \left( {\dfrac{\pi }{2}} \right)\cos \left( {\dfrac{{3\pi }}{{26}}} \right)$
We know that the value of
$\cos \left( {\dfrac{\pi }{2}} \right) = 0$ , so anything multiplied by this will also give the result as zero.
Hence our answer is
$2 \times 0 \times \cos \left( {\dfrac{{3\pi }}{{26}}} \right) = 0$ .