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!

You have that there is some $\theta$ for which you can write
$$a = \cos\theta \qquad\text{and}\qquad c = \sin\theta$$
Now, note that the vanishing elements of the matrix product give the equation 
$$a b + c d = 0 \tag{1}$$
and that the assumed relation $\det A = 1$ gives
$$a d - b c = 1 \tag{2}$$
Simply solve $(1)$ and $(2)$ for $b$ and $d$ in terms of $a$ and $c$.
