Show sin($\bar{z}$) = $\overline{\sin(z)}$ Let $z = x + iy$ where $x$ and $y$ are real. Show that
$$\sin(\bar{z})=\overline{\sin(z)}.$$
I started by trying to take the conjugate of $z$ which was ok but I didn't know how to take the conjugate of $\sin(z)$.
 A: Remember that $$\sin(x+iy)=\cosh(y)\sin(x)+i\sinh(y)\cos(x)$$Now if we take the conjugate, we get that $$\sin(x-iy)=\cosh(-y)\sin(x)+i\sinh(-y)\cos(x)$$As $\sinh(-y)=-\sinh(y)$ and $\cosh(-y)=\cosh(y)$, we can get $$\sin(x-iy)=\cosh(y)\sin(x)-i\sinh(y)\cos(x)$$
which you can see to be the conjugate of $\sin(x+iy)$
A: Sine in the complex variable is an entire function given by the series
$$
\sin z = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}z^{2n+1}
$$
convergent everywhere in the complex plane. We can compute its complex conjugate  term by term
$$
\overline{\sin z} = \sum_{n=0}^\infty \frac{\overline{(-1)^n}}{\overline{(2n+1)!}}\overline{z^{2n+1}} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\overline{z}^{2n+1} = \sin\overline{z}.
$$
Alternatively, as suggested in the comments, the equality follows from the Schwarz reflection principle. In both cases, we exploit the fact that sine is complex analytic.
A: Facts:
$$sin(a+b) = sinacosb+cosasinb$$
$$\begin{align}
\sin{ix}&=\frac{1}{2i}\left(e^{i^2x}-e^{-i^2x}\right)\\
&=\frac{1}{2i}\left(e^{-x}-e^x\right)\\
\end{align}$$
$$\begin{align}
\cos{ix}&=\frac{1}{2}\left(e^{i^2x}-e^{-i^2x}\right)\\
&=\frac{1}{2}\left(e^{-x}-e^x\right)\\
\end{align}$$
NOW:
$sin(z) = sin(x+iy) = sin(x)cos(iy)+cos(x)sin(iy) = sin(x)\frac{1}{2}\left(e^{-y}-e^y\right)+cos(x)\frac{1}{2i}\left(e^{-y}-e^y\right)$
$sin( \bar {z}) = sin(x-iy) = sin(x)cos(iy)-cos(x)sin(iy) = sin(x)\frac{1}{2}\left(e^{-y}-e^y\right)-cos(x)\frac{1}{2i}\left(e^{-y}-e^y\right)$
you can take it from here, i.e. find $\bar{sin(z)}$
