The projection of a vector value function onto the xz-plane. Okay, so I missed CalcIII today and I'm struggling a bit here.
$r(t) = (\sin t,\cos t,7\sin t + 4\cos 2t)$
Find the projection of $r(t)$ onto the xz-plane for $−1 \leq x \leq 1$
Answer as an equation using the variables $x$, $y$, and $z$.
P.S. I clepped CalcI and II.... in 2010. Haven't done much since then so I'm just a little out of practice, need to kick my brain back into gear. 
 A: The question asks you to describe a curve without the parameter $t$ too. That means after you project (by setting $y = 0$), you have to relate $x$ and $z$ directly instead of via $t$. Now note that after setting $y = 0$, you have
\begin{align}
x & = \sin t \\
z & = 7 \sin t + 4 \cos 2t.
\end{align}
The problem is expressing $z$ in terms of $x$, which you can do with a bit of trigonometry:
$$
z = 7x + 4(1 - 2x^2).
$$
The complete answer would have to include the constraint $-1 \le x \le 1$ (because we lost this condition, which was automatic when $x$ was expressed in terms of $\sin$ of something), and the equation $y = 0$.
A: Looks kinda like a parabola:

And with good reason.  It just so happens that the $z$-component can be rewritten:
$$7 \sin (t)+4 \cos (2 t)=4-8 \sin ^2(t)+7 \sin (t).$$
Thus, the $x$ and $z$ components satisfy the relationship $z=4+7x-8x^2$.
A: The idea is to get rid of the y-component of the vector so that it lies in the xz-plane. I.e. write the vector as
$$(\sin t,0,7 \sin t+ 4\cos 2t)$$
