Question

Evaluate $\cos \dfrac{\pi }{{11}} + \cos \dfrac{{3\pi }}{{11}} + \cos \dfrac{{5\pi }}{{11}} + \cos \dfrac{{7\pi }}{{11}} + \cos \dfrac{{9\pi }}{{11}} = $
1) $ - \dfrac{1}{2}$
2) $\dfrac{1}{2}$
3) $1$
4) $ - 1$

Answer 1

Hint: To find the sum of the given cosine functions, first we have to introduce a sine function, namely $2\sin \dfrac{\pi }{{11}}$ by multiplying and dividing at the same time. Then we are to operate the functions using the required trigonometric formulas. Finally we will get a form of equation that can be operated or cancelled to get the required solution.

Complete step-by-step solution:
To find, $\cos \dfrac{\pi }{{11}} + \cos \dfrac{{3\pi }}{{11}} + \cos \dfrac{{5\pi }}{{11}} + \cos \dfrac{{7\pi }}{{11}} + \cos \dfrac{{9\pi }}{{11}}$.
Now, multiplying and dividing the terms with $2\sin \dfrac{\pi }{{11}}$, we get,
$ = \dfrac{{2\sin \dfrac{\pi }{{11}}\left( {\cos \dfrac{\pi }{{11}} + \cos \dfrac{{3\pi }}{{11}} + \cos \dfrac{{5\pi }}{{11}} + \cos \dfrac{{7\pi }}{{11}} + \cos \dfrac{{9\pi }}{{11}}} \right)}}{{2\sin \dfrac{\pi }{{11}}}}$
Opening the brackets, we get,
$ = \dfrac{{2\sin \dfrac{\pi }{{11}}\cos \dfrac{\pi }{{11}} + 2\sin \dfrac{\pi }{{11}}\cos \dfrac{{3\pi }}{{11}} + 2\sin \dfrac{\pi }{{11}}\cos \dfrac{{5\pi }}{{11}} + 2\sin \dfrac{\pi }{{11}}\cos \dfrac{{7\pi }}{{11}} + 2\sin \dfrac{\pi }{{11}}\cos \dfrac{{9\pi }}{{11}}}}{{2\sin \dfrac{\pi }{{11}}}}$
Now, we know, $2\sin \theta \cos \theta = \sin 2\theta $ and $2\sin \theta \cos \phi = \sin \left( {\theta + \phi } \right) - \sin \left( {\theta - \phi } \right)$.
Using these formulas in the terms of the above equation, gives us,
$2\sin \dfrac{\pi }{{11}}\cos \dfrac{\pi }{{11}} = \sin \dfrac{{2\pi }}{{11}}$
$2\sin \dfrac{\pi }{{11}}\cos \dfrac{{3\pi }}{{11}} = \sin \dfrac{{4\pi }}{{11}} - \sin \dfrac{{2\pi }}{{11}}$
$2\sin \dfrac{\pi }{{11}}\cos \dfrac{{5\pi }}{{11}} = \sin \dfrac{{6\pi }}{{11}} - \sin \dfrac{{4\pi }}{{11}}$
$2\sin \dfrac{\pi }{{11}}\cos \dfrac{{7\pi }}{{11}} = \sin \dfrac{{8\pi }}{{11}} - \sin \dfrac{{6\pi }}{{11}}$
$2\sin \dfrac{\pi }{{11}}\cos \dfrac{{9\pi }}{{11}} = \sin \dfrac{{10\pi }}{{11}} - \sin \dfrac{{8\pi }}{{11}}$
Replacing, these terms in the given series, gives us,
$ = \dfrac{{\sin \dfrac{{2\pi }}{{11}} + \left( {\sin \dfrac{{4\pi }}{{11}} - \sin \dfrac{{2\pi }}{{11}}} \right) + \left( {\sin \dfrac{{6\pi }}{{11}} - \sin \dfrac{{4\pi }}{{11}}} \right) + \left( {\sin \dfrac{{8\pi }}{{11}} - \sin \dfrac{{6\pi }}{{11}}} \right) + \left( {\sin \dfrac{{10\pi }}{{11}} - \sin \dfrac{{8\pi }}{{11}}} \right)}}{{2\sin \dfrac{\pi }{{11}}}}$
Opening the brackets and simplifying, we get,
$ = \dfrac{{\sin \dfrac{{2\pi }}{{11}} + \sin \dfrac{{4\pi }}{{11}} - \sin \dfrac{{2\pi }}{{11}} + \sin \dfrac{{6\pi }}{{11}} - \sin \dfrac{{4\pi }}{{11}} + \sin \dfrac{{8\pi }}{{11}} - \sin \dfrac{{6\pi }}{{11}} + \sin \dfrac{{10\pi }}{{11}} - \sin \dfrac{{8\pi }}{{11}}}}{{2\sin \dfrac{\pi }{{11}}}}$
We can see clearly that in the numerator all the terms get cancelled, except $\sin \dfrac{{10\pi }}{{11}}$.
So,
\[ = \dfrac{{\sin \dfrac{{10\pi }}{{11}}}}{{2\sin \dfrac{\pi }{{11}}}}\]
Now, we know, $\sin \dfrac{{10\pi }}{{11}} = \sin \left( {\pi - \dfrac{\pi }{{11}}} \right)$
Using this property, we get,
$ = \dfrac{{\sin \left( {\pi - \dfrac{\pi }{{11}}} \right)}}{{2\sin \dfrac{\pi }{{11}}}}$
Now, we know, $\sin \left( {2\dfrac{\pi }{2} - \theta } \right) = \sin \theta $.
So, using this property, we can clearly say that, $\sin \left( {\pi - \dfrac{\pi }{{11}}} \right) = \sin \left( {2\dfrac{\pi }{2} - \dfrac{\pi }{{11}}} \right) = \sin \dfrac{\pi }{{11}}$.
Therefore, we can write as,
$ = \dfrac{{\sin \dfrac{\pi }{{11}}}}{{2\sin \dfrac{\pi }{{11}}}}$
Now, cancelling the $\sin \dfrac{\pi }{{11}}$ in the numerator and denominator, we get,
$\cos \dfrac{\pi }{{11}} + \cos \dfrac{{3\pi }}{{11}} + \cos \dfrac{{5\pi }}{{11}} + \cos \dfrac{{7\pi }}{{11}} + \cos \dfrac{{9\pi }}{{11}} = \dfrac{1}{2}$
Therefore, the correct option is 2.

Note: Many a times, we may get confused and panic due to the complex angles in the cosine functions and feel that the solutions would be complex, but we had to just use simple and commonly used trigonometric properties to get the answer. Sometimes, we can also make calculation mistakes in solving the equations and using the trigonometric properties properly.