# Parametric Equation of circle parallel to equator on a sphere

Here they say that a circle on a unit sphere obtained by fixing $$\theta=\frac{\pi}{4}$$ and letting $$\phi$$ vary can be described parametrically as follows $$\left(\cos\frac{\pi}{4}\cos\pi t, \cos\frac{\pi}{4}\sin\pi t, \sin\frac{\pi}{4}\right) \qquad \text{ where } t\in[-1, 1]$$ How does one obtain this?

# Idea 1

I thought maybe they simply computed the radius of the circle $$r_{\text{circle}} = r_{\text{sphere}} \cdot \sin\theta = \sin \frac{\pi}{4}$$ and then used the parametric form of a circle \begin{align} x &= r_{\text{circle}} \cos \phi \\ y &= r_{\text{circle}} \sin \phi \end{align}

which would give

\begin{align} x &= \sin \frac{\pi}{4}\cos\phi \\ y &= \sin\frac{\pi}{4}\sin\phi \\ z &= r_{\text{sphere}}\cos\theta = \cos\frac{\pi}{4} \end{align}

and setting $$\phi = \pi t$$ for $$t\in[-1, 1]$$. However this is different from what they have..

# Idea 2

I also tried simply using $$\theta = \frac{\pi}{4}$$ and $$\phi=\pi t$$ for $$t\in[-1, 1]$$ in the parametric equation for a sphere but I get the same result as in "Idea 1".

They do seem to have $$\sin$$ and $$\cos$$ mixed up, and your Idea 1 does work. But for what it's worth, $$\cos \pi / 4 = \sin \pi / 4$$ so technically they're not wrong!