Prove that $ϕ$ is a homomorphism and describe its kernel. $$ ϕ:ℝ→GL(2,ℝ),\qquad ϕ(x)= \begin{bmatrix}\cos(2x)&\sin(2x)\\-\sin(2x)&\cos(2x)\end{bmatrix} $$ where $x∈ℝ$
I have begun by saying $$ y=\begin{bmatrix}\cos(2y)&\sin(2y)\\-\sin(2y)&\cos(2y)\end{bmatrix} $$ $ℝ$ denote the set of real numbers under addition so I need to show that $ϕ(x+y)=ϕ(x)ϕ(y)$
So I have calculated $$ ϕ(x+y)\begin{bmatrix}\cos(2y)+\cos(2y)&\sin(2y)+\sin(2y)\\-\sin(2y)-\sin(2y)&\cos(2y)+\cos(2y)\end{bmatrix} $$ Is this right? And I don't know how to go from there. any help would be appreciated.
Note that by definition, $$ \phi(x+y)= \begin{bmatrix} \cos(2(x+y)) & \sin(2(x+y)) \\ -\sin(2(x+y)) & \cos(2(x+y)) \end{bmatrix} $$ Now note that \begin{align} \cos(2x+2y)&=\cos(2x)\cos(2y)-\sin(2x)\sin(2y) \\ \sin(2x+2y)&=\sin(2x)\cos(2y)+\cos(2x)\sin(2y) \end{align} and compute $$ \phi(x)\phi(y)= \begin{bmatrix} \cos(2x) & \sin(2x) \\ -\sin(2x) & \cos(2x) \end{bmatrix} \begin{bmatrix} \cos(2y) & \sin(2y) \\ -\sin(2y) & \cos(2y) \end{bmatrix} $$ as a matrix product.