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

Last updated date: 12th Aug 2024
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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.

$\Rightarrow \cos 0 + \cos \dfrac{\pi }{7} + \cos \dfrac{{2\pi }}{7} + \cos \dfrac{{3\pi }}{7} + \cos \dfrac{{4\pi }}{7} + \cos \dfrac{{5\pi }}{7} + \cos \dfrac{{6\pi }}{7}$
$\Rightarrow \cos 0 + \left( {\dfrac{{\cos \pi }}{7} + \dfrac{{\cos 6\pi }}{7}} \right) + \left( {\cos\dfrac{{ 2\pi }}{7} + \cos\dfrac{{ 5\pi }}{7}} \right) + \left( {\cos\dfrac{{ 3\pi }}{7} + \cos\dfrac{{ 6\pi }}{7}} \right) \to (1)$
Here the term are in the form $\cos (a + b)$
Since we know the formula that $\cos (a + b) = 2\cos \left( {\dfrac{{a + b}}{2}} \right) + \cos \left( {\dfrac{{a - b}}{2}} \right)$
$\Rightarrow \cos 0 + 2\cos \dfrac{\pi }{2}\cos \dfrac{{5\pi }}{{14}} + 2\cos \dfrac{\pi }{2}\cos \dfrac{{3\pi }}{{14}} + 2\cos \dfrac{\pi }{2}\cos \dfrac{\pi }{{14}}$
$\Rightarrow 1 + 0 + 0 + 0 = 1$
Note: 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.