Parametrization of $x^2+y^2=z^2$ How can we show that any point on $x^2+y^2=z^2$ can be written in the form (z $\cos(\theta)$, z $\sin(\theta)$, z) for some $\theta$? Here is how I tried to approach it: 
$$(z \cos(\theta))^2+(z \sin(\theta))^2)^2=(z)^2$$ 
$$z^2 \cos^2(\theta)+z^2 \sin^2(\theta)=z^2$$
$$z^2 (\cos^2(\theta)+\sin^2(\theta))=z^2$$
$$z^2 (\cos^2(\theta)+\sin^2(\theta))=z^2$$
$$z^2 (1)=z^2 = t^2$$
Is my approach correct or am I way off? If not, any hints would be wonderful with some detailed explanation. The way I attempted definitely does not look right in someway.
 A: You have shown that if $x=z\cos\theta$ and $y=z\sin\theta$, then $x^2+y^2=z^2$. 
The proof could be shortened a bit. Suppose that $x=z\cos\theta$ and $y=z\sin\theta$. Then
$$x^2+y^2=z^2(\cos^2\theta+\sin^2\theta)=(z^2)(1)=z^2.$$
However, that is not what you are being asked to show. You are asked to show that if $x^2+y^2=z^2$, then there exists a $\theta$ such that $x=z\cos\theta$ and $y=z\sin\theta$. 
First deal with the special case $z=0$. Then if $x^2+y^2=z^2$, we must have $x=y=0$. Then taking $\theta$ to be any angle (number) we get that $x=z\cos \theta$ and $y=z\sin\theta$. The case $z=0$ is exceptional, in that any $\theta$ will work.
Suppose now that $z\ne 0$. Then if $x^2+y^2=z^2$, we must have $|x|\le |z|$ and $|y|\le |z|$. It follows that $\frac{x}{z}$ and $\frac{y}{z}$ are both between $-1$ and $1$. 
Since $\frac{x}{z}$ is between $-1$ and $1$, it is the cosine of something. There is a $\theta$ in the interval $[0,2\pi)$ such that $\frac{x}{z}=\cos\theta$. Actually, this $\theta$ is almost determined by $\frac{x}{z}$. 
Typically there are two such $\theta$. For example, if $\frac{x}{z}$ is positive and less than $1$, there is one such $\theta$ in the first quadrant and one in the fourth quadrant.  So we almost know $\sin\theta$. 
The sign of $\frac{y}{z}$ determines, in this case, whether $\theta$ is in the first or fourth quadrant.  
The geometry: We took a relentlessly algebraic approach. However, the geometry provides much more insight. Let $z$ be a fixed non-zero number. Then $x^2+y^2=z^2$ is the equation of a circle of radius $|z|$. For simplicity take $z$ positive. Take a point $P$ on that circle. Then  $\theta$ be the angle we must rotate starting at the positive $x$-axis to get to $P$. And now drawing an appropriate triangle, and paying attention to sign, we see that $P=(z\cos\theta,z\sin\theta)$. 
A: You are going in the wrong direction: the first line assumes what you want to prove.  Assuming you are working in $\Bbb R$, you should say that as $y^2 \ge 0, x^2 \le z^2$, then define $\theta= \cos^{-1}\frac xz$, etc
