Matrix Algebra, Proof of some Trigonometric Identities Please Refer to the image to see the problem.  This was the easiest way to input the question as it has some difficult symbols to input from a keyboard.
[Edit: Image with task replaced by $\LaTeX$:]

Consider the column vectors:
$$x=\begin{bmatrix}x_1\\x_2\end{bmatrix} \text{ and } y=\begin{bmatrix}y_1\\y_2\end{bmatrix}$$
Treat these as vectors in a plane with x-coordinate being the first
  element and the y-coordinate being the second element. Suppose x and y
  make angles $\theta_1$ and $\theta_2$ with the x-axis.
(a) Show that:
$$x=\Vert x\Vert_2\begin{bmatrix}\cos(\theta_1)\\\sin(\theta_1)\end{bmatrix}\text{ and } y=\Vert y\Vert_2\begin{bmatrix}\cos(\theta_2)\\\sin(\theta_2)\end{bmatrix}.$$
(b) Show that:
$$x^Ty=\Vert x\Vert ~\Vert y\Vert\cos(\theta_1-\theta_2)$$
(c) What is the value of $x^Ty$ when the two vectors are perpendicular
  to each other?

 A: a) For a nonzero $x$, put
$$
x=\begin{bmatrix}x_1\\x_2\end{bmatrix}=\sqrt{x_1^2+x_2^2}\begin{bmatrix}\frac{x_1}{\sqrt{x_1^2+x_2^2}}\\\frac{x_2}{\sqrt{x_1^2+x_2^2}}\end{bmatrix}=\|x\|_2\begin{bmatrix}c(x)\\s(x)\end{bmatrix}.
$$
Since $c(x)^2+s(x)^2=1$ and $0\leq|c(x)|\leq 1$ and $0\leq|s(x)|\leq 1$, we can set
$$
c(x) = \frac{x_1}{\sqrt{x_1^2+x_2^2}}=\cos\theta, \quad
s(x) = \frac{x_2}{\sqrt{x_1^2+x_2^2}}=\sin\theta
$$
for some $\theta\in[0,2\pi)$.
b) If 
$$
x=\|x\|_2\begin{bmatrix}\cos\theta_1\\\sin\theta_1\end{bmatrix},
\quad
y=\|y\|_2\begin{bmatrix}\cos\theta_2\\\sin\theta_2\end{bmatrix},
$$
then
$$\begin{split}
x^Ty&=\|x\|_2\|y\|_2\begin{bmatrix}\cos\theta_1\\\sin\theta_1\end{bmatrix}^T\begin{bmatrix}\cos\theta_2\\\sin\theta_2\end{bmatrix}
=\|x\|_2\|y\|_2(\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2)\\
&=\|x\|_2\|y\|_2\cos(\theta_1-\theta_2)
\end{split}$$
using a handy trigonometric identity.
c) $x$ and $y$ are perpendicular if $\theta_1-\theta_2=(2k+1)\pi/2$ for some $k\in\mathbb{Z}$ for which $\cos(\theta_1-\theta_2)=\cos((2k+1)\pi/2)=0$. Hence $x^Ty=0$.
