Question

# Find the value of Trigonometric equation $\cos 0 + \cos \dfrac{\pi }{7} + \cos \dfrac{{2\pi }}{7} + \cos \dfrac{{3\pi }}{7} + \cos \dfrac{{4\pi }}{7} + \cos \dfrac{{5\pi }}{7} + \cos \dfrac{{6\pi }}{7} =$

Hint: To proceed with a solution we need to have basic Knowledge of trigonometry and their formulas.Here we will be using the $\cos (a + b)$ formula as a question of only cos terms.

Given:-

$\Rightarrow \cos 0 + \cos \dfrac{\pi }{7} + \cos \dfrac{{2\pi }}{7} + \cos \dfrac{{3\pi }}{7} + \cos \dfrac{{4\pi }}{7} + \cos \dfrac{{5\pi }}{7} + \cos \dfrac{{6\pi }}{7}$
Now let us rewritten the term as
$\Rightarrow \cos 0 + \left( {\dfrac{{\cos \pi }}{7} + \dfrac{{\cos 6\pi }}{7}} \right) + \left( {\dfrac{{\cos 2\pi }}{7} + \dfrac{{\cos 5\pi }}{7}} \right) + \left( {\dfrac{{\cos 3\pi }}{7} + \dfrac{{\cos 6\pi }}{7}} \right) \to (1)$
Here the term are in the form $\cos (a + b)$
Since we know the formula that $\cos (a + b) = 2\cos \left( {\dfrac{{a + b}}{2}} \right) + \cos \left( {\dfrac{{a - b}}{2}} \right)$
Now by using the above formula in equation (1) we will get
$\Rightarrow \cos 0 + 2\cos \dfrac{\pi }{2}\cos \dfrac{{5\pi }}{{14}} + 2\cos \dfrac{\pi }{2}\cos \dfrac{{3\pi }}{{14}} + 2\cos \dfrac{\pi }{2}\cos \dfrac{\pi }{{14}} \\ \Rightarrow 1 + 0 + 0 + 0 = 1 \\$
Therefore the value of the given term is 1.

NOTE: Here in this problem we have rewritten the term given, where we can use the trigonometric formula $\cos (a + b) = 2\cos \left( {\dfrac{{a + b}}{2}} \right) + \cos \left( {\dfrac{{a - b}}{2}} \right)$ .Then on applying the formula to the given term and on further simplification we get the value of the term.