Question

The sum $s=\sin \theta +\sin 2\theta +..............+\sin n\theta $ equal to?
A. $\dfrac{\sin \dfrac{1}{2}(n+1)\theta \sin \dfrac{1}{2}n\theta }{\sin \dfrac{\theta }{2}}$
B. $\dfrac{\cos \dfrac{1}{2}(n+1)\theta \sin \dfrac{1}{2}n\theta }{\sin \dfrac{\theta }{2}}$
C. $\dfrac{\sin \dfrac{1}{2}(n+1)\theta \sin \dfrac{1}{2}n\theta }{\cos \dfrac{\theta }{2}}$
D. $\dfrac{\cos \dfrac{1}{2}(n+1)\theta \sin \dfrac{1}{2}n\theta }{\cos \dfrac{\theta }{2}}$

Answer 1

Hint: In this question, we have to find the sum of $s=\sin \theta +\sin 2\theta +..............+\sin n\theta $. To solve this question, we consider only first term of the series and multiply it by $2\sin \dfrac{\theta }{2}$. As the equation becomes in the form of $2\sin C\sin D$, so we use the formula of $2\sin C\sin D$to expand the equation. Then we use the second term of the series and do the same process till the last term of the series. After that, we add all the equations and adding and subtracting the terms and more simplifying it, we are able to get the sum of the series.

Formula Used:
We will use the formula
$2\sin C\sin D=\cos (C-D)-\cos (C+D)$

Complete Step- by- step Solution:
Given that sum $s=\sin \theta +\sin 2\theta +..............+\sin n\theta $………………………………. (1)
We know that
$2\sin C\sin D=\cos (C-D)-\cos (C+D)$
Let us consider first term of sequence (1)
i.e. $\sin \theta $, multiply it by $2\sin \dfrac{\theta }{2}$
we get $2\sin \theta \sin \dfrac{\theta }{2}=\cos (\theta -\dfrac{\theta }{2})-\cos (\theta +\dfrac{\theta }{2})$
that is $2\sin \theta \sin \dfrac{\theta }{2}=\cos \dfrac{\theta }{2}-\cos \dfrac{3\theta }{2}$………………………………………………………...(2)
Same process we do with the second term i.e. $\sin 2\theta $
Multiply $\sin 2\theta $ with $2\sin \dfrac{\theta }{2}$, we get
$2\sin 2\theta \sin \dfrac{\theta }{2}=\cos (2\theta -\dfrac{\theta }{2})-\cos (2\theta +\dfrac{\theta }{2})$
 That is $2\sin \theta \sin \dfrac{\theta }{2}=\cos \dfrac{3\theta }{2}-\cos \dfrac{5\theta }{2}$…………………………………………………… (3)
Similarly for $\sin n\theta $, we get
$2\sin n\theta \sin \dfrac{\theta }{2}=\cos (n\theta -\dfrac{\theta }{2})-\cos (n\theta +\dfrac{\theta }{2})$
$2\sin n\theta \sin \dfrac{\theta }{2}=\cos \left( \left( \dfrac{2n-1}{2} \right)\theta \right)-\cos \left( \left( \dfrac{2n+1}{2} \right)\theta \right)$ …………………………(n)
Now we will add (2),(3)…………………..,(n), we get
$\begin{align}
  & 2\sin \theta \sin \dfrac{\theta }{2}+2\sin 2\theta \sin \dfrac{\theta }{2}+...........+2\sin n\theta \sin \dfrac{\theta }{2}=\cos \dfrac{\theta }{2}-\cos \dfrac{3\theta }{2}+\cos \dfrac{3\theta }{2}-\cos \dfrac{5\theta }{2}+ \\
 & ...........+\cos \left( \left( \dfrac{2n-1}{2} \right)\theta \right)-\cos \left( \left( \dfrac{2n+1}{2} \right)\theta \right) \\
\end{align}$
Therefore we observe the terms similar in magnitude but opposite in the sign cancel each other. Hence all the terms cancel each other except the starting and the last term as they have no term opposite to it. now we solve the left terms and take $2\sin \dfrac{\theta }{2}$common from the above equation, we get
$2\sin \dfrac{\theta }{2}(\sin \theta +\sin 2\theta +...........+\sin n\theta )=\cos \dfrac{\theta }{2}-\cos \left( \left( \dfrac{2n+1}{2} \right)\theta \right)$
Now we use the formula $2\sin C\sin D=\cos (C-D)-\cos (C+D)$ in the above equation, we get
$2\sin \dfrac{\theta }{2}(\sin \theta +\sin 2\theta +...........+\sin n\theta )=2\sin \dfrac{\left( \dfrac{\theta }{2}+\left( \dfrac{2n+1}{2} \right)\theta \right)}{2}\times \sin \dfrac{\left( \left( \dfrac{2n+1}{2} \right)\theta -\dfrac{\theta }{2} \right)}{2}$
\[(\sin \theta +\sin 2\theta +...........+\sin n\theta )=\dfrac{\sin \dfrac{\left( \dfrac{\theta }{2}+n\theta +\dfrac{\theta }{2} \right)}{2}\times \sin \dfrac{\left( n\theta +\dfrac{\theta }{2}-\dfrac{\theta }{2} \right)}{2}}{\sin \dfrac{\theta }{2}}\]
Then $s=\sin \theta +\sin 2\theta +..............+\sin n\theta $= $\dfrac{\sin \left( \dfrac{n\theta +\theta }{2} \right)\sin \left( \dfrac{n\theta }{2} \right)}{\sin \dfrac{\theta }{2}}$
That is $s=\dfrac{\sin \left( \dfrac{n\theta +\theta }{2} \right)\sin \left( \dfrac{n\theta }{2} \right)}{\sin \dfrac{\theta }{2}}$

Thus, Option ( A ) is correct.

Note: In these types of questions, students made mistakes that on expanding the equation by using trigonometric identity, they use the identity on all the terms at one time and they get confused in solving the series. We always try to apply the formula one by one on each term. In this way we cannot ger confused and easily solve the question.