mathematical induction

**grgrsanjay** · November 23rd 2010, 06:43 AM

P.T sinx +sin3x+ . . . +sin(2n-1)x = [sin^(2) (nx)] / sinx for any natural number n using mathematical induction

**wonderboy1953** · November 23rd 2010, 09:49 AM

Originally Posted by grgrsanjay

P.T sinx +sin3x+ . . . +sin(2n-1)x = [sin^(2) (nx)] / sinx for any natural number n using mathematical induction

How far did you get?

**tonio** · November 23rd 2010, 09:54 AM

Originally Posted by grgrsanjay

P.T sinx +sin3x+ . . . +sin(2n-1)x = [sin^(2) (nx)] / sinx for any natural number n using mathematical induction

So you must prove that $\displaystyle{\sum\limits^{n+1}_{k=1}\sin(2k-1)x=\frac{\sin^2(n+1)x}{\sin x}}$ , as the case $n=1$ is trivially true and assuming true for n:

$\displaystyle{\sum\limits^{n+1}_{k=1}\sin(2k-1)x=\sum\limits^n_{k=1}\sin(2k-1)x+\sin(2n+1)x=\frac{\sin^2nx}{\sin x}+\sin(2n+1)x}}=$

$=\displaystyle{\frac{\sin^2nx+\sin x\sin(2n+1)x}{\sin x}}$ . Thus, it's enough to show that $\sin^2(n+1)x=\sin^2nx+\sin x\sin(2n+1)x$ .

For this you'll need identities: in the LHS use the well known $\sin(\alpha+\beta)=\sin\alpha\cos\beta+sin\beta\co s\alpha$ (watch

that square!), and on the RHS, second term, use the same identity.

After the above's done pass all the terms to the LHS, use the trigonometric Pythagoras identity

and also $\cos(2\beta)=1-2\sin^2\beta$ ...

Tonio

**tonio** · November 23rd 2010, 09:56 AM

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