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  • Prove that for all n ϵ N Cos α + cos (α + β) + cos (α + 2β) + .... + cos (α + (n -1)β)

Prove that for all n ϵ N

\begin{array}{c} \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(n-1) \beta) \\\\ =\frac{\cos \left(\alpha+\left(\frac{n-1}{2}\right) \beta ) \sin \left(\frac{n \beta}{2}\right)\right.}{\sin \frac{\beta}{2}} \end{array}

 

 

Answers (1)

Given data:

\begin{array}{c} \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(n-1) \beta) \\\\ =\frac{\cos \left(\alpha+\left(\frac{n-1}{2}\right) \beta ) \sin \left(\frac{n \beta}{2}\right)\right.}{\sin \frac{\beta}{2}} \end{array}

Now, we’ll substitute different values for n,

n = 1,

\begin{aligned} \cos (\alpha+(1-1) \beta) &=\cos \alpha \\ &=\frac{\cos \left(\alpha+\left(\frac{1-1}{2}\right) \beta\right) \sin \left(\frac{\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)} \end{aligned}

Thus, it’s true

Now, n = 2,

\begin{aligned} \cos \alpha+\cos (\alpha+(2-1) \beta)=& \cos \alpha+\cos (\alpha+\beta) \\ =& \frac{\cos \left(\alpha+\left(\frac{2-1}{2}\right) \beta\right) \sin \left(\frac{2 \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)} \\ 2 \cos (\alpha+\beta / 2) \cos (\beta / 2) \end{aligned}

 

Thus, it’s true.

Now, let n = k

\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(k-1) \beta)=\frac{\cos \left(\alpha+\left(\frac{k-1}{2}\right) \beta\right) \sin \left(\frac{k \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)} be true

Thus, n=k+1\\ \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(k-1) \beta)+\cos (\alpha+(k+1-1) \beta)= \frac{\cos \left(\alpha+\left(\frac{k-1}{2}\right) \beta\right) \sin \left(\frac{k \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}+\cos (\alpha+k \beta) \\=\frac{\cos \left(\alpha+\left(\frac{k-1}{2}\right) \beta\right) \sin \left(\frac{k \beta}{2}\right)+\cos (\alpha+k \beta) \sin \left(\frac{\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}

=\frac{\sin \left(\alpha+\left(\frac{2 k-1}{2}\right) \beta\right) \sin \left(\alpha-\frac{\beta}{2}\right)+\sin (\alpha+(2 k-1) \beta / 2) \sin \left(\alpha+(2 k-1) \frac{\beta}{2}\right)}{2 \sin \left(\frac{\beta}{2}\right)}=\frac{\sin \left(\alpha+\left(\frac{2 k-1}{2}\right) \beta\right) \sin \left(\alpha-\frac{\beta}{2}\right)}{2 \sin \left(\frac{\beta}{2}\right)}\\=\frac{\cos \left(\alpha+\left(\frac{k}{2}\right) \beta\right) \sin \left(\frac{(k-1) \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}

Thus, n = k + 1 is true

Thus, by mathematical Induction,

For each natural no. n it is true that,

 

\begin{array}{c} \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(n-1) \beta) \\\\ =\frac{\cos \left(\alpha+\left(\frac{n-1}{2}\right) \beta ) \sin \left(\frac{n \beta}{2}\right)\right.}{\sin \frac{\beta}{2}} \end{array}

Posted by

infoexpert21

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