Chapter 2 - Matrices - 2.3 The Inverse of a Matrix - 2.3 Exercises - Page 72: 59

Yes, the matrix $A$ is invertible and its inverse is $A^{-1}=\left[\begin{array}{cc} sin\theta & -cos \theta\\ cos \theta& sin \theta \end{array}\right]$

Given $A=\left[\begin{array}{cc} sin\theta & cos \theta\\ -cos \theta& sin \theta \end{array}\right]$ then $|A|=(sin\theta)^{2}+(cos\theta)^{2}=1\ne0 $ thus the matrix $A$ is nonsingular and invertible therefore $A^{-1}=\frac{1}{|A|} *\left[\begin{array}{cc} sin\theta & -cos \theta\\ cos \theta& sin \theta \end{array}\right]=\left[\begin{array}{cc} sin\theta & -cos \theta\\ cos \theta& sin \theta \end{array}\right]$