Proving $\sin(2x) = 2\sin(x)\cos(x)$ using Cauchy product

Question

I've stumbled upon this question. I can't understand why all even terms of the Cauchy product are $0$, since we add only positive numbers: \begin{equation*} c_n = \begin{cases} \sum\limits_{k=0}^{m} \frac{(-1)^{k}(-1)^{m-k}}{(2k+1)!(2m-2k)!} & n = 2m + 1 \\ \hspace{32 pt} 0 & n = 2m \end{cases} \end{equation*}

Answer 1

Note that $$c_n=\sum_{k=0}^n a_kb_{n-k}$$ where $a_k$ is the coefficient of $x^k$ of the series of $\sin(x)$ and $b_{n-k}$ is the coefficient of $x^{n-k}$ of the series of $\cos(x)$. Now if $n$ is even then

i) if $k$ is odd then $n-k$ is odd and $b_{n-k}=0$ which implies that $a_kb_{n-k}=0$;

ii) if $k$ is even then $a_{k}=0$ which implies that $a_kb_{n-k}=0$.

Hence $c_n=\sum_{k=0}^n 0=0$.

By using the same argument we can show that the product of an "even" series with and an "odd" series is an "odd" series.

Answer 2

Since$$\sin x=0+x+0\times x^2-\frac1{3!}x^3+\cdots$$and$$\cos x=1+0\times x-\frac1{2!}x^2+0\times x^3+\cdots$$when you compute a term of even order of the Cauchy product of these series, you get a sum with zeros everywhere; for instance, the coeffiecient of $x^2$ is$$0\times\left(-\frac1{2!}\right)+1\times0+0\times1=0.$$