The sum $S=\sin \theta+\sin 2 \theta+\ldots .+\sin n \theta$, equals

Options:
A $\sin \frac{1}{2}(n+1) \theta \sin \frac{1}{2} n \theta / \sin \frac{\theta}{2}$
B $\cos \frac{1}{2}(n+1) \theta \sin \frac{1}{2} n \theta / \sin \frac{\theta}{2}$
C $\sin \frac{1}{2}(n+1) \theta \cos \frac{1}{2} n \theta / \sin \frac{\theta}{2}$
D $\cos \frac{1}{2}(n+1) \theta \cos \frac{1}{2} n \theta / \sin \frac{\theta}{2}$

The correct answer is:
$\sin \frac{1}{2}(n+1) \theta \sin \frac{1}{2} n \theta / \sin \frac{\theta}{2}$

Solution:
we have,
\(\operatorname{sum}(S)=\sin \theta+\sin 2 \theta+\ldots+\sin n \theta \text {. }\) ...(1)
we know that,
\(\begin{aligned}
& 2 \sin C \cdot \sin \theta=\cos (C-D)-\cos (C+\theta) \\
\Rightarrow & \sin C \cdot \sin \theta=
\end{aligned}\)
we have.
\(S=\operatorname{Sin} \theta+\sin 2 \theta+\ldots+\sin n \theta .\)
we will use the formula
\(\begin{aligned}
2 \sin C \cdot \sin \theta=\cos (C-D)-\cos (C+D) . \rightarrow(1) \\
\cos C-\cos D=2 \sin \left(\frac{c+D}{2}\right) \times \sin \left(\frac{D-C}{2}\right) \rightarrow(2) .
\end{aligned}\)
Let us consider first term of sequence (1)
ie. \(\sin \theta\) now, multiply it by \(2 \cdot \sin \frac{\theta}{2}\)
\(\begin{aligned}
& \Rightarrow 2 \times \sin \theta \cdot \sin \frac{\theta}{2}=\cos \left(\theta-\frac{\theta}{2}\right)-\cos \left(\theta+\frac{\theta}{2}\right) \text { (from (1)) } \\
& \Rightarrow 2 \sin \theta \cdot \sin \frac{\theta}{2}=\cos \frac{\theta}{2}-\frac{\cos 3 \theta}{2} \rightarrow \text { (1) }
\end{aligned}\)
Now. Similarly for 2nd term
\(\begin{aligned}
& 2 \sin 2 \theta \cdot \sin \frac{\theta}{2}=\cos \left(2 \theta-\frac{\theta}{2}\right)-\cos \left(2 \theta+\frac{\theta}{2}\right) \\
& \Rightarrow 2 \sin 2 \theta \cdot \sin \frac{\theta}{2}=\cos \frac{3 \theta}{2}-\frac{\cos 5 \theta}{2} \rightarrow(2) \\
& \qquad \vdots \qquad \qquad \qquad \quad \vdots \\
& 2 \sin n \theta \cdot \sin \frac{\theta}{2}=\cos \left(n \theta-\frac{\theta}{2}\right)-\cos \left(n \theta+\frac{\theta}{2}\right) \\
& \Rightarrow 2 \sin n \theta \cdot \sin \frac{\theta}{2}=\cos \left(\left(\frac{2 n-1}{2}\right) \theta\right)-\cos \left(\left(\frac{2 n+1}{2}\right) \theta\right) \rightarrow(n)
\end{aligned}\)
Now,
we will add (1), (2)... so on till (n).
\(\Rightarrow 2 \sin \theta \cdot \sin \frac{\theta}{2}+2 \sin 2 \theta \cdot \sin \frac{\theta}{2}+\ldots+2 \sin n \theta \cdot \sin \frac{\theta}{2}=\)
\(\cos \frac{\theta}{2}-\cos \frac{3 \theta}{2}+\frac{\cos 3 \theta}{2}-\frac{\cos 5 \theta}{2}+\ldots+\cos \left(\left(\frac{2 n-1}{2}\right) \theta\right)-\cos \left(\frac{2 n+1}{2}\right) \theta\)
\(\begin{aligned}
& \Rightarrow 2 \sin \frac{\theta}{2}(\sin \theta+\sin 2 \theta+\ldots+\sin n \theta)=\cos \frac{\theta}{2}-\left(\cos \left(\frac{2 n+1}{2}\right) \theta\right) \\
& \Rightarrow 2 \sin \frac{\theta}{2}(\sin \theta+\sin 2 \theta+\ldots+\sin n \theta)=2 \times \sin \frac{\left(\frac{\theta}{2}+\left(\frac{2 n+1}{2}\right) \theta\right)}{2} \cdot \frac{\sin \left(\left(\frac{2 n+1}{2}\right) \theta-\frac{\theta}{2}\right)}{2} \quad \text { (using formula (2)) }
\end{aligned}\)
\(\Rightarrow \sin \theta+\sin 2 \theta+\ldots+\sin n \theta=\frac{\sin \left(\frac{\frac{\theta}{2}+n \theta+\frac{\theta}{2}}{2}\right) \cdot \sin \left(\frac{n \theta+\frac{\theta}{2}-\frac{\theta}{2}}{2}\right)}{\sin \frac{\theta}{2}}\)
\(\Rightarrow s=\sin \theta+\sin 2 \theta+\ldots+\sin n \theta=\frac{\sin \left(\frac{n \theta+\theta}{2}\right) \cdot \sin \left(\frac{n \theta}{2}\right)}{\sin \frac{\theta}{2}}\)
\(\Rightarrow s=\frac{\sin \left(\frac{n+1}{2}\right) \theta \cdot \sin \left(\frac{n \theta}{2}\right)}{\sin \frac{\theta}{2}}\)
\(\Rightarrow\) option A. \(\frac{\sin \left(\frac{n+1}{2}\right) \theta \cdot \sin \left(\frac{n \theta}{2}\right)}{\sin \frac{\theta}{2}}\) is correct Answer.

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main