Prove the following statement(Binomial Theorem)?

**fardeen_gen** · February 10th 2009, 05:56 AM

Prove that:
{∑(r = 0 to n) nCr cos^(n - r) x sin ^(r) x}{∑(r = 0 to n) (-1)^(r) nCr cos^(n - r) x sin^(r) x} = cos ^(n) 2x

NOTE: nCr is n choose r.

**ThePerfectHacker** · February 13th 2009, 09:22 AM

Originally Posted by fardeen_gen

Prove that:
{∑(r = 0 to n) nCr cos^(n - r) x sin ^(r) x}{∑(r = 0 to n) (-1)^(r) nCr cos^(n - r) x sin^(r) x} = cos ^(n) 2x

NOTE: nCr is n choose r.

Remember that $(x+y)^n = \sum_{r=0}^n {n\choose r}x^ry^{n-r}$ .
Therefore, $\sum_{r=0}^n {n\choose r}\cos^{n-r}x\sin^r x = (\cos x + \sin x) \text{ and }$ $\sum_{r=0}^n (-1)^r{n\choose r}\cos^{n-r}x\sin^r x = (\cos x - \sin x)^n$

If you multiply them you get $(\cos^2 x - \sin^2 x)^n = \cos^n (2x)$ .

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