I know that the equation or a sphere of radius 1 in spherical coordinates is:
$$\rho = 1$$
how to mathematically convert this above equation into one that looks like this:
$$ f(\rho, \theta, \phi) = (\rho \sin \phi \cos \theta,\ \ \rho \sin \phi sin \theta, \ \ \rho \cos \phi)$$
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$\begingroup$Assuming your question is regaurding the derivation of the spherical coordinates:
Let $\phi$ represent the angle between the z-axis and your point. Similarly let $\theta$ represent the angle between the x axis and your point.
Notice that you get:
$$cos(\phi) = \frac{z}{\rho}$$
$$z = \rho \space cos(\phi)$$
Now we still have our equations:
$$x=rcos(\theta)$$$$y=rsin(\theta)$$
So we need to find how $r$ is related to $\rho$. To do this we notice that
$$sin(\phi)=\frac{r}{\rho}$$Thus $r =\rho \space sin(\phi)$
Plugging in this result you get the equations:
$$x = \rho \space sin(\phi) cos(\theta)$$$$y = \rho \space sin(\phi) sin(\theta)$$$$z = \rho \space cos(\phi)$$
$\endgroup$ $\begingroup$$$ f(ρ,θ,ϕ) = (ρ \sin ϕ \cos θ,\ \ \ ρ \sin ϕ \sin θ,\ \ \ ρ \cos ϕ) $$
The preceding function is how you can reach any point in a 3-dimensional coordinate system, using a spherical coordinate "address" for each point in the 3-d space. The function $f(ρ,θ,ϕ)$ works by returning the (x,y,z) coordinate given a spherical coordinate $(ρ,θ,ϕ)$.
Next we have the equation for a sphere of radius 1 when specified using spherical coordinate variables of $\rho$, $\theta$, and $\phi$:
$$ρ=1$$
$$\theta = unconstrained$$
$$\phi = unconstrained$$
Thus, given the above information, the parametric function for a sphere of radius 1 is:
$$ f(θ,ϕ) = (\sin ϕ \cos θ,\ \ \ \sin ϕ \sin θ,\ \ \ \cos ϕ) $$
Allowing us to get to any point on a sphere by choosing coordinates of $f(\theta, \phi)$ and returning the (x,y,z) coordinate of this point:
Another example, just to be extra clear, the parametric function for a sphere of radius 2 is:
$$ f(θ,ϕ)=(2 \sin ϕ \cos θ,\ \ \ 2 \sin ϕ \sin θ,\ \ \ 2 \cos ϕ) $$
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