How to show that $\cos(x+y)=\cos(x)\cos(y)−\sin(x)\sin(y)$ by using power series? Original question:Proofs of $\cos(x+y) = \cos x\cos y - \sin x \sin y$
I would like to know an answer to the question linked above by using power series.
I tried to expand the $(x+y)^{2n}$ by using the binomial theorem but it didn't lead me anywhere.
 A: Let 
$$C_k = \begin{cases}
1 & k \equiv 0 \pmod 4 \\
0 & k \equiv 1 \pmod 4 \\
-1 & k \equiv 2 \pmod 4 \\
0 & k \equiv 3 \pmod 4 \\
\end{cases}$$
$$S_k = \begin{cases}
0 & k \equiv 0 \pmod 4 \\
1 & k \equiv 1 \pmod 4 \\
0 & k \equiv 2 \pmod 4 \\
-1 & k \equiv 3 \pmod 4 \\
\end{cases}$$
Then the series is
$$\cos(t) = \sum_{0 \le k} \frac{C_k}{k!} t^k$$
$$\sin(t) = \sum_{0 \le k} \frac{S_k}{k!} t^k$$
So...
$$\begin{align}
%
 & \cos(x + y)
\\
 =& \sum_{0 \le k} \frac{C_k}{k!} (x + y)^k 
\\
=& \sum_{0 \le k} \sum_{u = 0}^k \frac{C_k}{k!} {k \choose u} x^u y^{k - u}
\\
=& \sum_{0 \le u \le k \le \infty} \frac{C_k}{k!} {k \choose u} x^u y^{k - u}
\\
\\
& \{j = k - u,~ k = u + j\}
\\
\\
=& \sum_{0 \le u,~ 0 \le j} \frac{C_{u + j}}{(u + j)!} {u + j \choose u} x^u y^j
\\
\\
& \left\{ {u + j \choose u} = \dfrac{(u + j)!}{u!j!} \right\}
\\
\\
=& \sum_{0 \le u,~ 0 \le j} \frac{C_{u + j}}{u!j!} x^u y^j
\end{align}$$
And
$$\begin{align}
%
& \cos(x)\cos(y) - \sin(x)\sin(y) 
\\
= &
\left(\sum_{0 \le k} \frac{C_k}{k!} x^k \right) \left(\sum_{0 \le k} \frac{C_k}{k!} y^k \right)
- \left(\sum_{0 \le k} \frac{S_k}{k!} x^k \right) \left(\sum_{0 \le k} \frac{S_k}{k!}  y^k \right)
\\
= &
\left(\sum_{0 \le k,~ 0 \le j} \frac{C_kC_j}{k!j!} x^ky^j \right)
-\left(\sum_{0 \le k, 0 \le j} \frac{S_kS_j}{k!j!} x^ky^j \right)
\\
= &
\sum_{0 \le k, 0 \le j} \left(\frac{C_kC_j - S_kS_j}{k!j!}\right)x^ky^j
\\
\end{align}$$
Equating coefficients, it just amounts to show
$$\frac{C_{k + j}}{k!j!} = \frac{C_kC_j - S_kS_j}{k!j!}$$
which is 
$$C_{k + j} = C_kC_j - S_kS_j$$
(It isn't a coincidence this looks exactly like the original formula.)  Theres only 16 possible values for $(j, k) \pmod 4$, so just check all of them with brute force if you can't find a shortcut.

My TA asked me to, so I suppose there should be a good way.

LUL.  I think you should be careful with that TA.
A: Note:
$$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$
and:
$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
and so:
$$\cos(x+y)=\frac{e^{i(x+y)}+e^{-i(x+y)}}{2}=\frac{e^{ix}e^{iy}+e^{-ix}e^{-iy}}{2}$$
$$\cos(x)\cos(y)-\sin(x)\sin(y)=\frac{e^{ix}+e^{-ix}}{2}\frac{e^{iy}+e^{-iy}}{2}-\frac{e^{ix}-e^{-ix}}{2i}\frac{e^{iy}-e^{-iy}}{2i}=\frac{e^{ix}e^{iy}+e^{-ix}e^{iy}+e^{ix}e^{-iy}+e^{-ix}e^{-iy}+e^{ix}e^{-iy}-e^{ix}e^{iy}-e^{-ix}e^{iy}+e^{-ix}e^{-iy}}{4}=\frac{2e^{ix}e^{iy}+2e^{-ix}e^{-iy}}{4}=\cos(x+y)$$
