Problem calculating the sine of a matrix Given the matrix $A=\begin{pmatrix}-\frac{3\pi}{4} & \frac{\pi}{2}\\\frac{\pi}{2}&0\end{pmatrix}$, I want to calculate the sine $\sin(A)$.
I do so by diagonalizing A and plugging it in the power series of the sine:
\begin{align}
\sin (A) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} A^{2k+1}.
\end{align}
The diagonalization leads to:
\begin{align}
A = \frac{1}{5}
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}
\begin{pmatrix}-\pi & 0\\0&\frac{\pi}{4}\end{pmatrix}
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}
\end{align}
and thus:
\begin{align}
A^n = \frac{1}{5}
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}
\begin{pmatrix}-\pi & 0\\0&\frac{\pi}{4}\end{pmatrix}^n
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}.
\end{align}
Hence:
\begin{align}
\sin (A) &= \begin{pmatrix}-2 & 1\\1&2\end{pmatrix}
\begin{pmatrix}\sin(-\pi) & 0\\0&\sin(\frac{\pi}{4})\end{pmatrix}
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}\\
&= \begin{pmatrix}-2 & 1\\1&2\end{pmatrix}
\begin{pmatrix}0 & 0\\0&\frac{1}{\sqrt{2}})\end{pmatrix}
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}\\
&= \frac{1}{5}\begin{pmatrix}\frac{1}{\sqrt{2}} & \sqrt{2}\\\sqrt{2}&2\sqrt{2}\end{pmatrix},
\end{align}
which differs from "Wolfram Alpha's result"
\begin{align}
\sin(A) &= \begin{pmatrix}-\frac{1}{\sqrt{2}} & 1\\ 1 & 0 \end{pmatrix} .
\end{align}
How can this happen?

The problem is that Wolfram Alpha interprets "sin(A)" for a matrix A (or array of however many dimensions, or list of list of lists, or what have you) as meaning simply the result of applying sin component-wise.
This is not what you intended, and you did your intended calculation perfectly fine.
