Sine of a matrix

the trick is to write $A=S\Lambda S^{-1}\to \sin(A)=S\sin(\Lambda) S^{-1} \ and\ \cos(A)=S\cos(\Lambda)S^{-1}$ where $S$ is the eigenvector matrix and $\Lambda$ is the eigenvalue matrix. this can be shown by taking power series for sine and cosine and factoring out $S$ and $S^{-1}$ from both sides. notice that since $\Lambda$ is diagonal, $\cos(\Lambda) =\cos \begin{bmatrix} 2\pi &0&\cdots&\cdots&0\\0&4\pi&\cdots&\cdots&0\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ \cdots& \cdots&\cdots&\cdots&\cdots\\0&0&\cdots&\cdots&20\pi\end{bmatrix}=\begin{bmatrix} \cos(2\pi) &0&\cdots&\cdots&0\\0&\cos(4\pi)&\cdots&\cdots&0\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ \cdots& \cdots&\cdots&\cdots&\cdots\\0&0&\cdots&\cdots&\cos(20\pi)\end{bmatrix}=I$ so $\cos(A)=S\cos(\Lambda)S^{-1}=SIS^{-1}=SS^{-1}=I$ similarly $\sin(\Lambda) =\sin \begin{bmatrix} 2\pi &0&\cdots&\cdots&0\\0&4\pi&\cdots&\cdots&0\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ \cdots& \cdots&\cdots&\cdots&\cdots\\0&0&\cdots&\cdots&20\pi\end{bmatrix}=\begin{bmatrix} \sin(2\pi) &0&\cdots&\cdots&0\\0&\sin(4\pi)&\cdots&\cdots&0\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ \cdots& \cdots&\cdots&\cdots&\cdots\\0&0&\cdots&\cdots&\sin(20\pi)\end{bmatrix}=\textbf{0}$ so $\sin(A)=S\sin(\Lambda)S^{-1}=\textbf{0}$ we have $\sin(A)+\cos(A)=\textbf{0}+I=\boxed{I}$