Question

Find the value of \[\cos \theta + \cos 2\theta + \cos 3\theta + .. + \cos \{ \left( {n - 1} \right)\theta \} + \cos n\theta \]
a. $\sin \left( {\dfrac{\theta }{2}} \right)$
b. $\dfrac{{\cos \left( {n + 1} \right)\theta }}{{\sin \left( {\dfrac{\theta }{2}} \right)}}$
c. $\dfrac{{\cos \left( {\theta + \dfrac{{\left( {n - 1} \right)\theta }}{2}} \right)\sin \left( {\dfrac{{n\theta }}{2}} \right)}}{{\sin \left( {\dfrac{\theta }{2}} \right)}}$
d. $\sin \left( {\dfrac{{n\theta }}{2}} \right).\cos \{ \left( {n + 1} \right)\theta \} $

Answer 1

Hint: First, we shall analyze the given information so that we are able to solve this problem. Here, we are given\[\cos \theta + \cos 2\theta + \cos 3\theta + .. + \cos \{ \left( {n - 1} \right)\theta \} + \cos n\theta \]and we are asked to calculate the value of\[\cos \theta + \cos 2\theta + \cos 3\theta + .. + \cos \{ \left( {n - 1} \right)\theta \} + \cos n\theta \].
We need to consider\[\cos \theta + \cos 2\theta + \cos 3\theta + .. + \cos \{ \left( {n - 1} \right)\theta \} + \cos n\theta \] this in terms of summation.
That is, \[\cos \theta + \cos 2\theta + \cos 3\theta + .. + \cos \{ \left( {n - 1} \right)\theta \} + \cos n\theta = \sum\limits_{k = 1}^n {cos(k\theta )} \]
Since we all know${e^{i\theta }} = \cos \theta + i\sin \theta $, we need to substitute it in the obtained equation.

Formula to be used:
a) The formula to calculate the sum of the geometric series is as follows.
$1 + x + {x^2} + .... + {x^n} = \dfrac{{{x^{n + 1}} - 1}}{{x - 1}}$ where$x \geqslant 1$
b) $\sin \theta = \dfrac{{{e^{i\theta }} - {e^{ - i\theta }}}}{{2i}}$

Complete step by step answer:
We know that${e^{i\theta }} = \cos \theta + i\sin \theta $
We are asked to calculate\[\cos \theta + \cos 2\theta + \cos 3\theta + .. + \cos \{ \left( {n - 1} \right)\theta \} + \cos n\theta \]
Thus, \[\cos \theta + \cos 2\theta + \cos 3\theta + .. + \cos \{ \left( {n - 1} \right)\theta \} + \cos n\theta \]can be written in terms of summation as follows.
\[\cos \theta + \cos 2\theta + \cos 3\theta + .. + \cos \{ \left( {n - 1} \right)\theta \} + \cos n\theta = \sum\limits_{k = 1}^n {cos(k\theta )} \]
\[\sum\limits_{k = 1}^n {cos(k\theta )} = \operatorname{Re} \sum\limits_{k = 1}^n {{e^{ik\theta }}} \] ($\cos \theta $is the real part of${e^{i\theta }}$)
                       $ = \operatorname{Re} \left( {{e^{i\theta }} + {e^{i2\theta }} + {e^{i3\theta }} + ..... + {e^{in\theta }}} \right)$ (We have expanded the above expression)
Now, we shall take ${e^{i\theta }}$as a common term.
\[\sum\limits_{k = 1}^n {cos(k\theta )} = \operatorname{Re} \left( {{e^{i\theta }}\left( {1 + {e^{i\theta }} + {e^{i2\theta }} + {e^{i3\theta }} + ..... + {e^{i\left( {n - 1} \right)\theta }}} \right)} \right)\]
                      \[ = \operatorname{Re} \left( {{e^{i\theta }}\left( {1 + {e^{i\theta }} + {e^{i2\theta }} + {e^{i3\theta }} + ..... + {e^{i\left( {n - 1} \right)\theta }}} \right)} \right)\]
Now, using the formula to calculate the sum of the geometric series$1 + x + {x^2} + .... + {x^n} = \dfrac{{{x^{n + 1}} - 1}}{{x - 1}}$, we get
\[\sum\limits_{k = 1}^n {cos(k\theta )} = \operatorname{Re} \left( {{e^{i\theta }}\left( {\dfrac{{{e^{in\theta }} - 1}}{{{e^{i\theta }} - 1}}} \right)} \right)\]
Here, we shall apply\[{e^{i\theta }} = {e^{\dfrac{{i\theta }}{2}}}.{e^{\dfrac{{i\theta }}{2}}}\] ,\[{e^{in\theta }} = {e^{\dfrac{{in\theta }}{2}}}.{e^{\dfrac{{in\theta }}{2}}}\] and\[1 = {e^{\dfrac{{in\theta }}{2}}}.{e^{\dfrac{{ - in\theta }}{2}}}\] in the above equation.
\[\sum\limits_{k = 1}^n {cos(k\theta )} = \operatorname{Re} \left( {{e^{i\theta }}\left( {\dfrac{{{e^{\dfrac{{in\theta }}{2}}}.{e^{\dfrac{{in\theta }}{2}}} - {e^{\dfrac{{in\theta }}{2}}}.{e^{\dfrac{{ - in\theta }}{2}}}}}{{{e^{\dfrac{{i\theta }}{2}}}.{e^{\dfrac{{i\theta }}{2}}} - {e^{\dfrac{{i\theta }}{2}}}.{e^{\dfrac{{ - i\theta }}{2}}}}}} \right)} \right)\]
Now, we shall pick the common terms inside the brackets.
\[\sum\limits_{k = 1}^n {cos(k\theta )} = \operatorname{Re} \left( {{e^{i\theta }}\dfrac{{{e^{\dfrac{{in\theta }}{2}}}}}{{{e^{\dfrac{{i\theta }}{2}}}}}\left( {\dfrac{{{e^{\dfrac{{in\theta }}{2}}} - {e^{\dfrac{{ - in\theta }}{2}}}}}{{{e^{\dfrac{{i\theta }}{2}}} - {e^{\dfrac{{ - i\theta }}{2}}}}}} \right)} \right)\] ……………..$\left( 1 \right)$
Here, \[{e^{i\theta }}\dfrac{{{e^{\dfrac{{in\theta }}{2}}}}}{{{e^{\dfrac{{i\theta }}{2}}}}} = {e^{i\theta }}{e^{\dfrac{{in\theta }}{2}}}{e^{\dfrac{{ - i\theta }}{2}}}\]
                         \[ = {e^{i\theta + \dfrac{{in\theta }}{2}\dfrac{{ - i\theta }}{2}}}\]
                         \[ = {e^{i\left( {\theta + \dfrac{{n\theta }}{2}\dfrac{{ - \theta }}{2}} \right)}}\]
                          \[ = {e^{i\left( {\dfrac{{2\theta + n\theta - \theta }}{2}} \right)}}\]
                          \[ = {e^{i\theta \left( {\dfrac{{n + 1}}{2}} \right)}}\]
 We need to substitute \[{e^{i\theta }}\dfrac{{{e^{\dfrac{{in\theta }}{2}}}}}{{{e^{\dfrac{{i\theta }}{2}}}}} = {e^{i\left( {n + 1} \right)\dfrac{\theta }{2}}}\] in the equation$\left( 1 \right)$.
\[\sum\limits_{k = 1}^n {cos(k\theta )} = \operatorname{Re} \left( {{e^{i\left( {n + 1} \right)\dfrac{\theta }{2}}}\left( {\dfrac{{{e^{\dfrac{{in\theta }}{2}}} - {e^{\dfrac{{ - in\theta }}{2}}}}}{{{e^{\dfrac{{i\theta }}{2}}} - {e^{\dfrac{{ - i\theta }}{2}}}}}} \right)} \right)\] ……….$\left( 2 \right)$
Using the formula$\sin \theta = \dfrac{{{e^{i\theta }} - {e^{ - i\theta }}}}{{2i}}$, we have${e^{i\theta }} - {e^{ - i\theta }} = 2i\sin \theta $
Similarly, \[{e^{\dfrac{{in\theta }}{2}}} - {e^{\dfrac{{ - in\theta }}{2}}} = 2i\sin \dfrac{{n\theta }}{2}\] and\[{e^{\dfrac{{i\theta }}{2}}} - {e^{\dfrac{{ - i\theta }}{2}}} = 2i\sin \dfrac{\theta }{2}\] .
We shall substitute the above results in the equation$\left( 2 \right)$
\[\sum\limits_{k = 1}^n {cos(k\theta )} = \operatorname{Re} \left( {{e^{i\left( {n + 1} \right)\dfrac{\theta }{2}}}\left( {\dfrac{{2i\sin \dfrac{{n\theta }}{2}}}{{2i\sin \dfrac{\theta }{2}}}} \right)} \right)\]
Since${e^{i\theta }} = \cos \theta + i\sin \theta $we have
\[\sum\limits_{k = 1}^n {cos(k\theta )} = \operatorname{Re} \left( {\cos \left( {\left( {n + 1} \right)\dfrac{\theta }{2}} \right) + \sin \left( {i\left( {n + 1} \right)\dfrac{\theta }{2}} \right)\left( {\dfrac{{\sin \dfrac{{n\theta }}{2}}}{{\sin \dfrac{\theta }{2}}}} \right)} \right)\]
\[ \Rightarrow \sum\limits_{k = 1}^n {cos(k\theta )} = \operatorname{Re} \left( {\cos \left( {\left( {n + 1} \right)\dfrac{\theta }{2}} \right) + i\sin \left( {\left( {n + 1} \right)\dfrac{\theta }{2}} \right)\left( {\dfrac{{\sin \dfrac{{n\theta }}{2}}}{{\sin \dfrac{\theta }{2}}}} \right)} \right)\]
\[ \Rightarrow \sum\limits_{k = 1}^n {cos(k\theta )} = \operatorname{Re} \left( {\cos \left( {\left( {n + 1} \right)\dfrac{\theta }{2}} \right)\left( {\dfrac{{\sin \dfrac{{n\theta }}{2}}}{{\sin \dfrac{\theta }{2}}}} \right) + i\sin \left( {\left( {n + 1} \right)\dfrac{\theta }{2}} \right)\left( {\dfrac{{\sin \dfrac{{n\theta }}{2}}}{{\sin \dfrac{\theta }{2}}}} \right)} \right)\]
Now, we need to consider the real part alone.
\[ \Rightarrow \sum\limits_{k = 1}^n {cos(k\theta )} = \cos \left( {\left( {n + 1} \right)\dfrac{\theta }{2}} \right)\left( {\dfrac{{\sin \dfrac{{n\theta }}{2}}}{{\sin \dfrac{\theta }{2}}}} \right)\]
                             $ = \dfrac{{\cos \left( {\dfrac{{n\theta + \theta }}{2}} \right)\sin \left( {\dfrac{{n\theta }}{2}} \right)}}{{\sin \left( {\dfrac{\theta }{2}} \right)}}$
                            $ = \dfrac{{\cos \left( {\dfrac{{2\theta + n\theta - \theta }}{2}} \right)\sin \left( {\dfrac{{n\theta }}{2}} \right)}}{{\sin \left( {\dfrac{\theta }{2}} \right)}}$
                          $ = \dfrac{{\cos \left( {\dfrac{{2\theta }}{2} + \dfrac{{n\theta - \theta }}{2}} \right)\sin \left( {\dfrac{{n\theta }}{2}} \right)}}{{\sin \left( {\dfrac{\theta }{2}} \right)}}$
                           $ = \dfrac{{\cos \left( {\theta + \dfrac{{\left( {n - 1} \right)\theta }}{2}} \right)\sin \left( {\dfrac{{n\theta }}{2}} \right)}}{{\sin \left( {\dfrac{\theta }{2}} \right)}}$
Therefore, \[\cos \theta + \cos 2\theta + \cos 3\theta + .. + \cos \{ \left( {n - 1} \right)\theta \} + \cos n\theta \]$ = \dfrac{{\cos \left( {\theta + \dfrac{{\left( {n - 1} \right)\theta }}{2}} \right)\sin \left( {\dfrac{{n\theta }}{2}} \right)}}{{\sin \left( {\dfrac{\theta }{2}} \right)}}$

So, the correct answer is “Option c”.

Note: We know that${e^{i\theta }} = \cos \theta + i\sin \theta $. Generally, a complex number consists of a real part and an imaginary part. For example, let us consider a complex number$z = 5 + 7i$. In this example, the real part of$z$ is$5$and the imaginary part of$z$ is$7$.
Similarly, for${e^{i\theta }} = \cos \theta + i\sin \theta $, the real part of${e^{i\theta }}$ is$\cos \theta $ and the imaginary part of${e^{i\theta }}$ is$\sin \theta $
Therefore, \[\cos \theta + \cos 2\theta + \cos 3\theta + .. + \cos \{ \left( {n - 1} \right)\theta \} + \cos n\theta \]$ = \dfrac{{\cos \left( {\theta + \dfrac{{\left( {n - 1} \right)\theta }}{2}} \right)\sin \left( {\dfrac{{n\theta }}{2}} \right)}}{{\sin \left( {\dfrac{\theta }{2}} \right)}}$