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  • Prove that, sinθ plus sin 2θ plus sin 3θ plus …. plus sin nθ for all n ϵ N.

Prove that, \sin \theta+\sin 2\theta+\sin 3\theta+...+\sin n\theta\begin{aligned} &=\frac{\frac{\sin \mathrm{n} \theta}{2} \sin \frac{(\mathrm{n}+1)}{2} \theta}{\sin \frac{\theta}{2}} \end{aligned} for all n \in N

Answers (1)

Given:

 

P(n) : \sin \theta+\sin 2\theta+\sin 3\theta+...+\sin n\theta\begin{aligned} &=\frac{\frac{\sin \mathrm{n} \theta}{2} \sin \frac{(\mathrm{n}+1)}{2} \theta}{\sin \frac{\theta}{2}} \end{aligned}

Now, we’ll substitute different values for n=1,

\\\sin \theta=[\sin (1 \theta / 2) \sin (1+1) \theta / 2] / \sin \theta / 2

Thus, it is true.Now,

let us assume that, for n=k is true

\sin \theta+\sin 2 \theta+\sin 3 \theta+\ldots \ldots+\sin k \theta=[\sin (k \theta / 2) \sin (k+1) \theta / 2] / \sin \theta / 2

At n = k+1

\sin \theta+\sin 2 \theta+\sin 3 \theta+\ldots \ldots+\sin k \theta+\sin (k+1) \theta\\

\\= \sin \mathrm{k} \theta / 2 \sin (\mathrm{k}+1) \theta / 2+\sin \theta / 2 \sin (\mathrm{k}+1) \Theta] / \sin \theta / 2$ $\\ =[\cos (\theta / 2)-\cos (2 k+1) \theta / 2+\cos (2 k+1) \theta / 2-\cos (2 k+3) \theta / 2] / 2 \sin \theta / 2$ $\\ =[\cos (\theta / 2)-\cos (2 k+3) \Theta / 2] / 2 \sin \theta / 2$ $\\ =[\sin \mathrm{k} \theta / 2 \sin (\mathrm{k}+1) \ominus / 2+\sin \theta / 2 \sin (\mathrm{k}+1) \Theta] / \sin \theta / 2

\\ =\sin \mathrm{k} \theta / 2 \sin (\mathrm{k}+1) \theta / 2+\sin \theta / 2 \sin (\mathrm{k}+1) \Theta] / \sin \theta / 2$ $\\ =[\cos (\theta / 2)-\cos (2 k+1) \theta / 2+\cos (2 k+1) \theta / 2-\cos (2 k+3) \theta / 2] / 2 \sin \theta / 2$ $\\ =[\cos (\theta / 2)-\cos (2 k+3) \Theta / 2] / 2 \sin \theta / 2$ $\\ =[\sin \mathrm{k} \theta / 2 \sin (\mathrm{k}+1) \ominus / 2+\sin \theta / 2 \sin (\mathrm{k}+1) \Theta] / \sin \theta / 2

Thus, n=k+1 is true

Thus, by mathematical Induction,

For each natural no. n it is true that,

\sin \theta+\sin 2\theta+\sin 3\theta+...+\sin n\theta\begin{aligned} &=\frac{\frac{\sin \mathrm{n} \theta}{2} \sin \frac{(\mathrm{n}+1)}{2} \theta}{\sin \frac{\theta}{2}} \end{aligned}

 

Posted by

infoexpert21

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