Show if $A^TA = I$ and $\det A = 1$ , then $A$ is a rotational matrix Show if $A^TA = I$ and $\det A = 1$ where 
$ A =
  \begin{bmatrix}
    a & b \\
    c & d
  \end{bmatrix}
$, then $A =\begin{bmatrix}
    \cos\theta & -\sin\theta \\
    \sin\theta & \cos\theta
  \end{bmatrix}$.
attempt:
Suppose $ A^TA =\begin{bmatrix}
    a & c \\
    b & d
  \end{bmatrix}
  \begin{bmatrix}
    a & b \\
    c & d
  \end{bmatrix}$ = $\begin{bmatrix}
    a^2 + c^2 & ab + cd \\
    ab + cd & b^2 + d^2
  \end{bmatrix}$ = $\begin{bmatrix}
    1 & 0 \\
    0 & 1
  \end{bmatrix}$
. Then $a^2 + c^2 = 1$ implies $a = \cos\theta$, and $c = \sin\theta$ or $c = - \sin\theta$ using the unit circle . 
Similarly $ d = \cos\theta$, and $b = \sin\theta$ or $b = -\sin\theta$.
So know I am stuck in showing how $b = -\sin\theta$ has to be chosen and $c = \sin\theta$.
Can someone please help? Thank you!

If you had $A =\begin{bmatrix} \cos\theta & \sin\theta \\  \sin\theta & \cos\theta \end{bmatrix}$ or $A =\begin{bmatrix} \cos\theta & -\sin\theta \\  -\sin\theta & \cos\theta \end{bmatrix}$ instead of $A =\begin{bmatrix} \cos\theta & -\sin\theta \\  \sin\theta & \cos\theta \end{bmatrix}$, would it then be true that $A^T A = I \vphantom{\frac{\int^\int}\int} $?
