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Prove that $\sin(z_{1}+z_{2})=\sin z_{1} \cos z_{2} + \cos z_{1} \sin z_{2}$

My solution

Let $z_{2}$ be a fixed real number. Then, $f(z)=\sin(z + z_{2})$ and $g(z)=\sin z\cos z_{2}+\cos z\sin z_{2}$ are two entire functions (of $z$) which agree for all real values $z=z_{1}$ and, hence, for all complex values $z=z_{1}$, as well.

Let $z=z_{1}$ be any such complex number. Then, $f(z)=\sin(z_{1}+z)$ and $g(z)=\sin z_{1}\cos z+\cos z_{1}\sin z$ agree for all real values $z=z_{2}$ and, hence,for all complex values $z=z_{2}$ as well.

Is my method correct?

  • 0
    This looks fine, and can be streamlined if you know a little about holomorphic functions of several complex variables. In particular, $\mathbb{R}^2$ is a set of uniqueness for entire function of two variables, just as $\mathbb{R}$ is a set of uniqueness in the one variable setting.2012-05-02

1 Answers 1

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An alternative:

$\begin{align*} \sin z&= \dfrac{e^{iz}-e^{-iz}}{2i}\\ \cos z&= \dfrac{e^{iz}+e^{-iz}}{2}. \end{align*}$

Now substitute into $\sin(z_1)\cos(z_2)+\sin(z_2)\cos(z_1)$, expand, and simplify.