Prove that cosα+cos(α+β)+cos(α+2β)+…
Question:

Prove that $\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(n-1) \beta]=\frac{\cos \left\{\alpha+\left(\frac{n-1}{2}\right) \beta\right\} \sin \left(\frac{n \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$ for all $n \in \mathbf{N}$.       [NCERT EXEMPLAR]

Solution:

Let $\mathrm{p}(n): \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(n-1) \beta]=\frac{\cos \left\{\alpha+\left(\frac{n-1}{2}\right) \beta\right\} \sin \left(\frac{n \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)} \forall n$ $\in \mathbf{N}$.

Step I : For $n=1$,

LHS $=\cos [\alpha+(1-1) \beta]=\cos \alpha$

$\mathrm{RHS}=\frac{\cos \left\{\alpha+\left(\frac{1-1}{2}\right) \beta\right\} \sin \left(\frac{\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}=\cos \alpha$

As, LHS = RHS

So, it is true for $n=1$.

Step II : For $n=k$,

Let $\mathrm{p}(k): \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(k-1) \beta]=\frac{\cos \left\{\alpha+\left(\frac{k-1}{2}\right) \beta\right\} \sin \left(\frac{k \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$ be true $\forall k \in \mathbf{N}$.

Step III : For $n=k+1$,

LHS $=\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(k-1) \beta]+\cos [\alpha+(k+1-1) \beta]$

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

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

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

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

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

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

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

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

As, $\mathrm{LHS}=\mathrm{RHS}$

So, it is also true for $n=k+1$.

Hence, $\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(n-1) \beta]=\frac{\cos \left\{\alpha+\left(\frac{n-1}{2}\right) \beta\right\} \sin \left(\frac{n \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$ for all $n$ $\in \mathbf{N}$

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