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

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)}}$