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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.

Rico
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  • Do you have to use power series? These sorts of things are always easier if you use the exponential formula for cosine. – user413766 Mar 13 '19 at 05:16
  • @JonHales My TA asked me to, so I suppose there should be a good way. – Rico Mar 13 '19 at 05:17
  • I would advise showing that $e^{x + y} = e^x e^y$ for complex $x$, $y$. It's not too hard to do this with binomial theorem, if I remember correctly. – Theo Bendit Mar 13 '19 at 05:31

2 Answers2

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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.

DanielV
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  • Btw $C_n$ is https://oeis.org/A056594 which has among its closed forms $$\frac 12 \left(i^n + (-i)^n\right)$$ – DanielV Mar 13 '19 at 09:26
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Note: $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$ and: $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ and so: $$\cos(x+y)=\frac{e^{i(x+y)}+e^{-i(x+y)}}{2}=\frac{e^{ix}e^{iy}+e^{-ix}e^{-iy}}{2}$$ $$\cos(x)\cos(y)-\sin(x)\sin(y)=\frac{e^{ix}+e^{-ix}}{2}\frac{e^{iy}+e^{-iy}}{2}-\frac{e^{ix}-e^{-ix}}{2i}\frac{e^{iy}-e^{-iy}}{2i}=\frac{e^{ix}e^{iy}+e^{-ix}e^{iy}+e^{ix}e^{-iy}+e^{-ix}e^{-iy}+e^{ix}e^{-iy}-e^{ix}e^{iy}-e^{-ix}e^{iy}+e^{-ix}e^{-iy}}{4}=\frac{2e^{ix}e^{iy}+2e^{-ix}e^{-iy}}{4}=\cos(x+y)$$

Henry Lee
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