Find the inverse of a Trig Matrix I don't have a clue of what's going on. We haven't learn this in class so I need all the help possible. The more detailed of an explanation, the better. Thanks in advance. The only info I have is that this matrix is Orthogonal. Which means I know the answer, just don't know how to get it.
\begin{bmatrix}
       \cos\theta & -\sin\theta \\
       \sin\theta & \cos\theta 
     \end{bmatrix}
 A: This is the matrix of rotation. Hence substituting $\theta$ by $-\theta$ should give you the inverse matrix,what you can verify by multiplying.
A: Let $A(\theta)=\begin{bmatrix}
       \cos\theta & -\sin\theta \\
       \sin\theta & \cos\theta 
     \end{bmatrix}$. 
By the definition of an inverse matrix, $A^{-1}=\text{adj}(A)/\det(A)$. Notice that $\det(A)$ can be calculated easily as follows:$$\det(A)=\left|\begin{matrix}\cos\theta &-\sin\theta \\ \sin\theta & \cos\theta \end{matrix}\right|=\sin^2\theta+\cos^2\theta=1$$
$$A^{-1}=A(-\theta)=\begin{bmatrix}\cos\theta & \sin\theta \\-\sin\theta & \cos\theta\end{bmatrix}$$
A: There is a very nice relation between matrix multiplication, vectors, inverse and transpose. Consider a coordinate frame which has 2 Axes X and Y. If there is a point P which makes an angle $\theta$ with respect to the positive X Axis then the coordinates of the point will be $P(\cos \theta, \sin \theta)$ in vector notation it is written as $$\overrightarrow{OP}= \cos \theta \widehat{i} + \sin \theta \widehat{j}=\left(\begin{array}{c}\cos \theta\\ \sin \theta\end{array}\right)=\left(\begin{array}{c}\text{Projection of P on X axis}\\ \text{Projection of P on Y axis}\end{array}\right)$$
You can see that this vector is of unit magnitude having two components $\cos \theta$ on X Axis and $\sin \theta$ on Y Axis.

Now consider a second frame {2} which is at an angle $\theta$ with respect to frame {1}. Or simply we rotated the first frame {1} and the original X Y Axes to a second coordinate frame {2} with new X' Y' Axes. 
Every point on the second frame {2} can be written in terms of the original frame {1}. 

$$\text{Old coordinates in 1st frame} = \text{(Some transformation from the old frame to new frame)(New coordinates in 2nd frame)}$$
$$ X = X' \cos \theta - Y' \sin \theta$$
$$ Y = X' \sin \theta + Y' \cos \theta$$
which can be written as 
$$\begin{bmatrix}  X \\ Y \end{bmatrix}=\begin{bmatrix}\cos \theta & - \sin \theta \\\sin \theta & \cos \theta \end{bmatrix}\begin{bmatrix}  X' \\ Y' \end{bmatrix}$$
So if you want to go from frame {1} to frame {2} you use this matrix given by:
$$A(\theta)=\begin{bmatrix}\cos \theta & -\sin \theta \\\sin \theta & \cos \theta \end{bmatrix}$$
Another way of seeing this matrix is :
$$A(\theta) =\begin{bmatrix}\cos \theta & -\sin \theta \\\sin \theta & \cos \theta \end{bmatrix}$$as$$\begin{bmatrix}\text{Projection of New X' axis on Old X Axis}& \text{Projection of New Y' axis on Old X Axis}\\\text{Projection of New X' axis on Old Y Axis} &\text{Projection of New Y' axis on Old Y Axis} \end{bmatrix}$$
or it can be viewed as a collection of 2 column vectors
$$\left(\begin{array}{c}\cos \theta\\ \sin \theta\end{array}\right),\left(\begin{array}{c}-\sin \theta\\ \cos \theta\end{array}\right)$$
in which the first column vector is the X' vector wrt X and Y and the second vector is the Y' vector wrt X and Y axes.
$$ X' = X \cos \theta + Y \sin \theta$$
$$ Y' = -X \sin \theta + Y \cos \theta$$
Or
$$\begin{bmatrix}  X \\ Y \end{bmatrix}=\begin{bmatrix}\cos \theta &  \sin \theta \\-\sin \theta & \cos \theta \end{bmatrix}\begin{bmatrix}  X' \\ Y' \end{bmatrix}$$
Now the determinant of this matrix is $1$. And if you rotate the frame {2} back to {1} you see that the matrix obtained is nothing but the transpose of the matrix. But wait isn't the transformation from {1} to {2] and the transformation from {2} to {1} inverse operations? And you find out all of a sudden that the matrix 
$$A(\theta)^{-1}=\begin{bmatrix}\cos \theta & \sin \theta \\-\sin \theta & \cos \theta \end{bmatrix}$$
And 
$$A(\theta)^{T}=A(\theta)^{-1}$$
Now just think again what is the operation actually from {1} to {2}. It is a rotation by an angle $\theta$ and the transformation from {2} to {1} is rotation by angle $-\theta$ which is the inverse of the first transformation. So you can say,
$$A(\theta)^T=A(\theta)^{-1}=A(-\theta)$$ 
Also the determinant $|A(\theta)|=1$
So the matrix is orthogonal. Hope you got the physical significance of the inverse of an orthogonal matrix through this example.
