Verify that $$ \left[\begin{matrix}\cos32 & -\sin32\\ \sin32 & \cos32 \end{matrix}\right]\cdot\left[\begin{matrix}\cos40 & -\sin40\\ \sin40 & \cos40 \end{matrix}\right]=\left[\begin{matrix}\cos72 & -\sin72\\ \sin72 & \cos72 \end{matrix}\right] $$ and explain why this result could have been expected.
I have verified that the product matrix is correct using the matrix operation techniques, but I am not sure as to why the result is expected.
The matrix
$$R\left(\theta\right)=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$$
when acted on a vector, rotates the vector by an angle $\theta$. Thus it is expected that a rotation by $\theta_{1}$, followed by a rotation by $\theta_{2}$, will be equivalent to a rotation by $\theta_{1}+\theta_{2}$. In matrix notation this is translated into
$$R\left(\theta_{1}\right)R\left(\theta_{2}\right)=R\left(\theta_{1}+\theta_{2}\right)$$
In addition to the comment above, I think it deserves a little explaination why: $R\left(\theta_{1}\right)R\left(\theta_{2}\right)=R\left(\theta_{1}+\theta_{2}\right)$
By using matrix multiplication we get:
$$R\left(\theta_{1}\right)R\left(\theta_{2}\right)=\begin{pmatrix}\cos\theta_{1}&-\sin\theta_{1}\\\sin\theta_{1}&\cos\theta_{1}\end{pmatrix}\cdot\begin{pmatrix}\cos\theta_{2}&-\sin\theta_{2}\\\sin\theta_{2}&\cos\theta_{2}\end{pmatrix} = \begin{pmatrix}\cos\theta_{1}\cos\theta_{2}-\sin\theta_{1}\sin\theta_{2} & -\left(\sin\theta_{1}\cos\theta_{2}+\cos\theta_{1}\sin\theta_{2}\right)\\ \sin\theta_{1}\cos\theta_{2}+\cos\theta_{1}\sin\theta_{2} & -\sin\theta_{1}\sin\theta_{2}+\cos\theta_{1}\cos\theta_{2} \end{pmatrix}$$
We will use the trigonometric identities:
$$\left(1\right) \cos\left(\theta_{1}+\theta_{2}\right)=\cos\theta_{1}\cos\theta_{2}-\sin\theta_{1}\sin\theta_{2}\\ \left(2\right) \sin\left(\theta_{1}+\theta_{2}\right)=\sin\theta_{1}\cos\theta_{2}+\cos\theta_{1}\sin\theta_{2}$$ And we will recieve: $$R\left(\theta_{1}\right)R\left(\theta_{2}\right)=\begin{pmatrix}\cos\theta_{1}\cos\theta_{2}-\sin\theta_{1}\sin\theta_{2} & -\left(\sin\theta_{1}\cos\theta_{2}+\cos\theta_{1}\sin\theta_{2}\right)\\ \sin\theta_{1}\cos\theta_{2}+\cos\theta_{1}\sin\theta_{2} & -\sin\theta_{1}\sin\theta_{2}+\cos\theta_{1}\cos\theta_{2} \end{pmatrix}=\begin{pmatrix}\cos\left(\theta_{1}+\theta_{2}\right) & \sin\left(\theta_{1}+\theta_{2}\right)\\ \sin\left(\theta_{1}+\theta_{2}\right) & \cos\left(\theta_{1}+\theta_{2}\right) \end{pmatrix} = R\left(\theta_{1}+\theta_{2}\right)$$ As said in the comment above.
