Prove that if $T$ is a rotation then $-T$ is also a rotation 
Let $V$ be a $2$ dimensional real vector space with inner product,and let $T:V \rightarrow V$ be a linear operator on $V$. Prove that if $T$ is a rotation then -T is also a rotation. 

Rotation Definition:Let T be a linear operator on a finite-dimensional real inner product space V. The operator T is called a rotation if T is the identity on V or if there exists a two-dimensional subspace W of V, an orthonormal basis $\beta={x_1,x_2}$ for W, and a real number $\theta$ such that $$T(x_1)=(cos \theta)x_1+(sin \theta)x_2$$, $$T(x_2)=(-sin \theta)x_1+(cos \theta)x_2$$, and$ T(y)=y$ for all $y \in W^{\perp}$. 
Theorem: Let T be an orthogonal operator on a two-dimensional real inner product space V. T is a rotation if and only if det(T) = 1, and T is a reflection if and only if det(T) = −1.
Let $T$ be a $2$ by $2$ matrix, then since it is a rotation, we have det($T$)=1. So it is easy to see that det($-T$) also equals $1$. Does this work?
 A: The proof you currently have does not work unless you have already proven that a $2 \times 2$ matrix is a rotation if and only if it is orthogonal with determinant $1$.
Hint: Note that $-\cos(\theta) = \cos(\theta + 180^\circ)$ and $-\sin \theta = \sin(\theta + 180^\circ)$.
Alternatively, note that the map $R(x) = -x$ is a rotation, and use the fact that a composition of rotations is also a rotation. 
A: Assuming you are taking the defintion that coordinate rotations in any dimension are represented by orthogonal matrices i.e. such that $Q^{T}Q = Id$, defining $P := -Q$ and taking the determinant both sides will work as you suggested thanks to $Q$ beeing a rotation.
A: If $T$ is a rotation, then $T$ can be expressed as 
$$[T] = \begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta 
\end{bmatrix} $$
for some $\theta \in \mathbb{R}$, where $[T]$ denotes the matrix representation of $T$. Therefore
$$[-T] = \begin{bmatrix}
-\cos\theta & -\sin\theta \\
\sin\theta & -\cos\theta 
\end{bmatrix}
= \begin{bmatrix}
\cos(\theta + \pi) & \sin(\theta + \pi) \\
-\sin(\theta + \pi) & \cos(\theta + \pi) 
\end{bmatrix}. $$
That is, if $T$ rotates a vector by $\theta$ radius, then $-T$ would rotate it by $(\theta + \pi)$, which is of course also a rotation.
A: A rotation is something of the form $\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$.  Thus you can check that $-r$ is a rotation through $\theta+\pi$.  That is, $-r=\begin{bmatrix}-\cos\theta&\sin\theta\\-\sin\theta&-\cos\theta\end{bmatrix}=\begin{bmatrix}\cos(\theta+π)&-\sin(\theta+\pi)\\\sin(\theta+π)&\cos(\theta+\pi)\end{bmatrix}$.
Alternatively, multiplication by $-1$ is just a rotation through $π$.  
