91影视

Consider the first $D_{伪}{D}_{尾}\stackrel{鈫�}{x}$. The linear transformation$D_{尾}\stackrel{鈫�}{x}$ rotates counter clockwise by the angle$尾$ . Hence, $\stackrel{鈫�}{z}=D_{尾}\stackrel{鈫�}{x}$is a vector rotated from initial one by the angle $尾$. Then, vector$\stackrel{鈫�}{z}$ is rotated with $D_{伪}\stackrel{鈫�}{z}$by the angle $伪$. Thus, the initial vector $\stackrel{鈫�}{x}$is rotated by an angle of data-custom-editor="chemistry" $伪+尾$counter clockwise. Next, consider the first role="math" localid="1664214338414" $D_{尾}{D}_{伪}\stackrel{鈫�}{x}$. The linear transformation rotates counter clockwise by the angle$伪$ .Hence $\stackrel{鈫�}{z}=D_{伪}\stackrel{鈫�}{x}$, is a vector rotated from initial one by the angle. Then, vector$\stackrel{鈫�}{z}$ is rotated with$D_{尾}\stackrel{鈫�}{z}$ by the angle role="math" localid="1664214417501" $尾$. Thus, the initial vector $\stackrel{鈫�}{x}$is rotated by an angle of $伪+尾$counter clockwise. The conclusion is that linear transformations are the same.

The multiplication of matrices should yield the same result, that is, the following should hold.

$D_{伪}{D}_{尾}=D_{尾}{D}_{伪}$

It is easy to see that

$$\begin{gathered} D_{伪}{D}_{尾}=\left[\begin{array}{cc}\cos 伪& -\sin 伪\\ \sin 伪& \cos 伪\end{array}\right]\left[\begin{array}{cc}\cos 尾& -\sin 尾\\ \sin 尾& \cos 尾\end{array}\right] \\ =\left[\begin{array}{cc}\cos 伪\cos 尾-\sin 伪\sin 尾& --\cos 伪\sin 尾-\sin 伪\cos 尾\\ \sin 伪\cos 尾+\cos 伪\sin 尾& -\sin 伪\sin 尾+\cos 伪\cos 尾\end{array}\right] \\ =\left[\begin{array}{cc}\cos (伪+尾)& -\sin (伪+尾)\\ \sin (伪+尾)& \cos (伪+尾)\end{array}\right] \\ =D_{尾}{D}_{伪} \end{gathered}$$

Yes, it makes sense because of the identities given in the exercise.

The linear transformations are the same and$D_{尾}{D}_{伪}=D_{伪}{D}_{尾}$ .