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Prove that sin x + sin 3x + … + sin (2n – 1) x = \(\frac{sin^2nx}{sin\,x}\) for all n ϵ N

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Let, P(n) be the given statement,

now, P(n); sin x + sin 3x + ....+ sin(2n-1)x = \(\frac{sin^2nx}{sin\,x}\)

step1: P(n): sin x = \(\frac{sin^2nx}{sin\,x}\)

Thus, P(1) is true. 

Step2: Let, P(m) be true.

then,  sin x + sin 3x + ....+ sin(2m-1)x = \(\frac{sin^2mx}{sin\,x}\)

Now, we need to show that P(m+1) is true when P(m) is true. 

As P(m) is true

Thus, P(m+1) is divisible by x+y. So, by the principle of mathematical induction P(n) is true for all n.

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