What does it mean to eliminate $\theta$ from these equations? How should I do it? 
Eliminate $\theta$ from the following pairs of equations:
A) $x=\sin \theta$, $y=\sin 2\theta$
B) $x=3\cos 2\theta +1$, $y=2\sin \theta$

My problem is I really don't understand what the question is asking? The answer in the book seems like it is asking me to put everything in terms of $x$ and $y$? And if it is, how do I go about that?
Thank you!
EDIT: My attempt:
$$y = 2\sin \theta \cos \theta $$
$${y \over {2\sin \theta }} = \cos \theta $$
so as $$x = \sin \theta $$
$$\cos \theta  = {y \over {2x}}$$
what do i do from here?
 A: You've done the hard part. Of A):

$$
\begin{align} 
\sin\theta & = x \\ \\
\cos\theta  & = {\frac y{2x}}
\end{align}$$

This is where the Pythagorean Identity (as you call it) comes in very handy, indeed:
$$\quad\quad\quad\sin^2(\theta) + \cos^2(\theta) = 1$$
$$\iff \left(\sin\theta \right)^2 + \left(\cos\theta \right)^2 = 1$$
$$x^2 + \left(\dfrac{y}{2x}\right)^2 = 1$$
Then just simplify!

Part B: 

$$x = 3 \cos 2\theta, \quad x = 2\sin \theta$$

$$x = 3(1-2\sin^2\theta) \tag{$\cos(2\theta) = 1 - 2\sin^2\theta$}$$ 
$$x= \frac 32(2 - (1 - \cos^2\theta)) \tag{$\sin^2\theta = 1 - \cos^2 \theta$}$$ 
$$x = \frac 32 (\cos^2 \theta + 1) \iff \frac 23 x - 1 = \cos^2\theta$$
$$ $$
$$ y = 2 \sin \theta \iff \sin \theta = y/2$$
$$ $$
$$\sin^2 \theta + \cos ^2 \theta = 1\tag{Pythagorean identity again}$$
$$\left(\frac y2\right)^2 + \left(\frac 23 x - 1\right) = 1$$
Simplify.
A: A) $y = \sin 2\theta$, $y = 2\sin\theta\cos\theta$, $y = 2x\sqrt{1-x^2}$
B) Similarly you will find the answer as $3-3y^2+1$
