# Fallacious moving of powers resulting with a correct trigonometric series identity.

Prove that $$\\ \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1} ( \sin^{2(n+r+1)}x + \cos^{2(n+r+1)}x )\right) = \sum_{r=0}^n \frac{ (-1)^r {n \choose r} } {n+r+1}$$ for all values of $$x$$.

I came across this joke which said that you could just bring the powers out of the brackets and everything works out right.

$$\require{cancel} \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1} ( \sin^{2(\textbf{n+r+1})}x + \cos^{2(\textbf{n+r+1})}x )\right) \\ = \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1} (\cancelto1{\sin^2x + \cos^2x })^{\textbf{n+r+1}}\right) \\ = \sum_{r=0}^n \frac{ (-1)^r {n \choose r} } {n+r+1}$$

But how would you actually go about proving this? I noticed that this value is also equal to $$B(n+1,n+1)$$, which may or may not be relevant.

Let's prove $$\\ \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1} ( \sin^{2(n+r+1)}x + \cos^{2(n+r+1)}x )\right) = \sum_{r=0}^n \frac{ (-1)^r {n \choose r} } {n+r+1} \tag1$$ for all values of $$x$$.
By differentiating both sides with respect to $$x$$, on the right hand side one gets $$0$$, on the left hand side one gets $$\sum_{r=0}^n (-1)^r {n \choose r} \left( 2\cos x\cdot\sin^{2(n+r+1)-1}x - 2\sin x\cdot\cos^{2(n+r+1)-1}x \right)$$ $$2\cos x\cdot\sin^{2n+1}x \sum_{r=0}^n (-1)^r {n \choose r} \sin^{2r}x - 2\sin x\cdot\cos^{2n+1}x \sum_{r=0}^n (-1)^r {n \choose r} \cos^{2r}x$$ $$2\cos x\cdot\sin^{2n+1}x\cdot \left(1-\sin^2x \right)^n-2\sin x\cdot\cos^{2n+1}x\cdot \left(1-\cos^2x \right)^n=0.$$ Since both sides of $$(1)$$ clearly agree at $$x=0$$, then $$(1)$$ is true for all values of $$x$$.