If F(x)= \(\begin{bmatrix}\cos x&\sin x&0\\\sin x&cos x&0\\0&0&1\end{bmatrix}\)and F(y)=\(\begin{bmatrix}\cos y&-\sin y&0\\\sin y&cos y&0\\0&0&1\end{bmatrix}\),show that F(x)+F(y)=F(x+y)
F(x)=\(\begin{bmatrix}\cos x&-\sin x&0\\\sin x&cos x&0\\0&0&1\end{bmatrix}\),F(y)=\(\begin{bmatrix}\cos y&-\sin y&0\\\sin y&cos y&0\\0&0&1\end{bmatrix}\)
F (x+y)=\(\begin{bmatrix}\cos (x+y)&-\sin (x+y)&0\\\sin (x+y)&cos (x+y)&0\\0&0&1\end{bmatrix}\)
F(x)F(y)=\(\begin{bmatrix}\cos x&-\sin x&0\\\sin x&cos x&0\\0&0&1\end{bmatrix}\)\(\begin{bmatrix}\cos y&-\sin y&0\\\sin y&cos y&0\\0&0&1\end{bmatrix}\)
=\(\begin{bmatrix}\cos x\cos y-\sin x\sin y+0&-\cos x\sin y-\sin x\cos y+0&0\\\sin x\cos y+\cos x\sin y&-\sin x\sin y+\cos x\cos y+0&0\\0&0&0\end{bmatrix}\)
=\(\begin{bmatrix}\cos (x+y)&-\sin(x+y)&0\\\sin (x+y)&\cos(x+y)&0\\0&0&1\end{bmatrix}\)
=F(x+y)
\(\therefore\) F(x)+F(y)=F(x+y)