Prove that + 3x+...+ (2n-1)x= ^2nx for all n . - Vidyadip

Question

Prove that $\sin\text{x}+\sin3\text{x}+...+\sin(\text{2n-1})\text{x}=\frac{\sin ^2\text{nx}}{\sin\text{x}}$ for all $\text{n}\in\text{N}.$

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Answer

Let P(n): $\sin\text{x}+\sin3\text{x}+...+\sin(\text{2n-1})\text{x}=\frac{\sin ^2\text{nx}}{\sin\text{x}}$
For n = 1
$\sin \text{x}=\frac{\sin^2\text{x}}{\sin\text{x}}$
$\sin\text{x}=\sin\text{x}$
⇒ p(n) is true for n = 1
Let p(n) is true for n = k, So
$\sin\text{x}+\sin3\text{x}+...+\sin(\text{2k-1})\text{x}=\frac{\sin ^2\text{kx}}{\sin\text{x}}$
We have to show that,
$\sin\text{x}+\sin3\text{x}+...+\sin(\text{2k-1})\text{x}+\sin(\text{2k+1})\text{x}=\frac{\sin ^2(\text{k+1})\text{x}}{\sin\text{x}}$
Now,
$\Big\{\sin\text{x}+\sin3\text{x}+...+\sin(\text{2k}-1)\text{x}\Big\}+\sin(2\text{k}+1)\text{x}$
$=\frac{\sin^2\text{kx}}{\sin\text{x}}+\frac{\sin(2\text{k}+1)\text{x}}{1}$
$=\frac{\sin^2\text{kx}+{\sin(2\text{k}+1)\text{x}}\sin\text{x}}{\sin\text{x}}$ [Using equation (1)]
$=\frac{2\sin^2\text{kx}+\cos\big[(2\text{k}+1)\text{x}-\text{x}-\big]\big[(2\text{k}+1)\text{x}+\text{x}\big]}{2\sin\text{x}}$
$=\frac{2\sin^2\text{kx}+\cos2\text{kx}-\cos(2\text{kx}+2\text{x})}{2\sin\text{x}}$
$=\frac{1-\cos2\text{kx}+\cos2\text{kx}-\cos2\text{x}(\text{k}+1)}{2\sin\text{x}}$
$=\frac{1-\cos2\text{x}(\text{k}+1)}{2\sin\text{x}}$
$=\frac{2\sin^2\text{x}(\text{k}+1)}{2\sin\text{x}}$
$=\frac{\sin^2\text{x}(\text{k}+1)}{\sin\text{x}}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for $\text{n}\in\text{N}$ by PMI

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