Question

The value of $\cos \dfrac{\pi }{{11}} + \cos \dfrac{{3\pi }}{{11}} + \cos \dfrac{{5\pi }}{{11}} + \cos \dfrac{{7\pi }}{{11}} + \cos \dfrac{{9\pi }}{{11}}$ is:-
${\text{A}}{\text{.0}} \\\ {\text{B}}{\text{. - }}\dfrac{1}{2} \\\ {\text{C}}{\text{.}}\dfrac{1}{2} \\\ {\text{D}}{\text{. }}None{\text{ }}of{\text{ }}these \\\$

Answer 1

- Hint- First, we will multiply a term with this given expression to form this expression as in a state, where we can use trigonometry identities. Then, we will change the inappropriate angles according to their range and eventually we will divide the same term from the calculated result.

Complete step-by-step solution -

Multiply the given expression by $\sin \dfrac{{5\pi }}{{11}}$
$= \sin \left( {\dfrac{{5\pi }}{{11}}} \right)\cos \left( {\dfrac{\pi }{{11}}} \right) + \sin \left( {\dfrac{{5\pi }}{{11}}} \right)\cos \left( {\dfrac{{3\pi }}{{11}}} \right) + \sin \left( {\dfrac{{5\pi }}{{11}}} \right)\cos \left( {\dfrac{{5\pi }}{{11}}} \right) + \sin \left( {\dfrac{{5\pi }}{{11}}} \right)\cos \left( {\dfrac{{7\pi }}{{11}}} \right) + \sin \left( {\dfrac{{5\pi }}{{11}}} \right)\cos \left( {\dfrac{{9\pi }}{{11}}} \right)$
Using the formula $\sin {\text{C cosD = }}\dfrac{1}{2}\left[ {\sin \left( {{\text{C + D}}} \right) + \sin \left( {{\text{C - D}}} \right)} \right]$
We get,
= \dfrac{1}{2}\left[ \begin{gathered} \left\\{ {\sin \left( {\dfrac{{6\pi }}{{11}}} \right) + \sin \left( {\dfrac{{4\pi }}{{11}}} \right)} \right\\} + \left\\{ {\sin \left( {\dfrac{{8\pi }}{{11}}} \right) + \sin \left( {\dfrac{{2\pi }}{{11}}} \right)} \right\\} + \left\\{ {\sin \left( {\dfrac{{10\pi }}{{11}}} \right) + \sin \left( 0 \right)} \right\\} + \left\\{ {\sin \left( {\dfrac{{12\pi }}{{11}}} \right) + \sin \left( {\dfrac{{ - 2\pi }}{{11}}} \right)} \right\\} \\\ \+ \left\\{ {\sin \left( {\dfrac{{14\pi }}{{11}}} \right) + \sin \left( {\dfrac{{ - 4\pi }}{{11}}} \right)} \right\\} \\\ \end{gathered} \right]
Now. To find the exact quadrant of all the angles given we subtract $2\pi$ from the angles whose values are increasing beyond $\pi$.
$= \dfrac{1}{2}\left[ \begin{gathered} \sin \left( {\dfrac{{6\pi }}{{11}}} \right) + \sin \left( {\dfrac{{4\pi }}{{11}}} \right) + \sin \left( {\dfrac{{8\pi }}{{11}}} \right) + \sin \left( {\dfrac{{2\pi }}{{11}}} \right) + \sin \left( {\dfrac{{10\pi }}{{11}}} \right) + \sin \left( {\dfrac{{12\pi }}{{11}} - 2\pi } \right) + \sin \left( {\dfrac{{ - 2\pi }}{{11}}} \right) + \\\ \sin \left( {\dfrac{{14\pi }}{{11}} - 2\pi } \right) + \sin \left( {\dfrac{{ - 4\pi }}{{11}}} \right) \\\ \end{gathered} \right] \\\ = \dfrac{1}{2}\left[ \begin{gathered} \sin \left( {\dfrac{{6\pi }}{{11}}} \right) + \sin \left( {\dfrac{{4\pi }}{{11}}} \right) + \sin \left( {\dfrac{{8\pi }}{{11}}} \right) + \sin \left( {\dfrac{{2\pi }}{{11}}} \right) + \sin \left( {\dfrac{{10\pi }}{{11}}} \right) + \sin \left( {\dfrac{{ - 10\pi }}{{11}}} \right) + \sin \left( {\dfrac{{ - 2\pi }}{{11}}} \right) + \sin \left( {\dfrac{{ - 8\pi }}{{11}}} \right) + \\\ \sin \left( {\dfrac{{ - 4\pi }}{{11}}} \right) \\\ \end{gathered} \right] \\\$
Now as we know $\sin \left( {{\text{ - }}\theta } \right) = - \sin \theta$
Using this formula we get,$= \dfrac{1}{2}\left[ {\sin \left( {\dfrac{{6\pi }}{{11}}} \right) + \sin \left( {\dfrac{{4\pi }}{{11}}} \right) - \sin \left( {\dfrac{{4\pi }}{{11}}} \right) + \sin \left( {\dfrac{{8\pi }}{{11}}} \right) - \sin \left( {\dfrac{{8\pi }}{{11}}} \right) + \sin \left( {\dfrac{{2\pi }}{{11}}} \right) - \sin \left( {\dfrac{{2\pi }}{{11}}} \right) + \sin \left( {\dfrac{{10\pi }}{{11}}} \right) - \sin \left( {\dfrac{{10\pi }}{{11}}} \right)} \right] \\\ = \dfrac{1}{2} \times \left[ {\sin \left( {\dfrac{{6\pi }}{{11}}} \right)} \right] \\\ {\text{As sin}}\theta {\text{ = sin}}\left( {\pi {\text{ - }}\theta } \right), \\\ = \dfrac{1}{2} \times \left[ {\sin \left( {\dfrac{{5\pi }}{{11}}} \right)} \right] \\\$
Since we had multiplied the expression by $\sin \left( {\dfrac{{5\pi }}{{11}}} \right)$ .we divide by the same now
The expression then becomes=\dfrac{{\left[ {\dfrac{1}{2} \times \left\\{ {\sin \left( {\dfrac{{5\pi }}{{11}}} \right)} \right\\}} \right]}}{{\left\\{ {\sin \left( {\dfrac{{5\pi }}{{11}}} \right)} \right\\}}} = \dfrac{1}{2}

Note - Given, the trigonometric product identities are important for these type of questions and also one should know the range of angles and how to make an angle within range as only this approach will lead us to an answer and to avoid mistakes do remember to divide the term in the end which is multiplied in the beginning of the solution as it is one of the typical step to remember.