<s>[INST] Reasoning that $ \sin2x=2 \sin x \cos x$ In mathcounts teacher told us to use the formula $ \sin2x=2 \sin x \cos x$.
What's the math behind this formula that made it true?  Can someone explain?
 [/INST] Let's start with the power series definitions:
$$\sin x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!} -+ \dots, $$
$$\cos x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-+ \dots. $$
If you expand the product $$\sin x \cos x= \left( x-\frac{x^3}{3!}+\frac{x^5}{5!} -+ \dots \right) \left( 1-\frac{x^2}{2!}+\frac{x^4}{4!}-+ \dots \right), $$
and collect terms, you will find that only odd powers of $x$ appear, and the coefficient of $x^{2n+1}$ is $$(-1)^n \left(\frac{1}{(2n+1)!0!}+\frac{1}{(2n-1)! 2!}+ \dots +\frac{1}{1!(2n)!} \right).$$
You can prove (using induction for example) that the last expression equals $$(-1)^n \frac{2^{2n}}{(2n+1)!}=(-1)^n \frac{1}{2} \frac{2^{2n+1}}{(2n+1)!}.$$
Thus we have $$\sin x \cos x=\frac{1}{2} \sum_{n=0}^\infty (-1)^n \frac{(2x)^{2n+1}}{(2n+1)!}, $$ which is precisely $\frac{1}{2} \sin 2x$.
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