what is the connection between the orientation of a parametric curve and the sin(t) and cos(t) functions?

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Problem I need to find a piecewise smooth parametrization for the semicircle in the yz plane that begins at (0,0,4), passes through (0,-4,0), and terminates at (0,0,-4).

My thought process Since we are in the yz plane that must mean x=0. Also, from the information above we can figure out the center of the circle is at the orgin (0,0,0). Now I need to figure out an equation for y and z so that I get the points (0,4), (-4,0) and (0,-4). I determined this to be

$y=4cos(t)$

$z = 4sin(t)$

for $\frac{\pi}{2} < t < \frac{3\pi}{2}$

However, the answer in the textbook is

$y=-4sin(t)$

$z = 4cos(t)$

for $0 < t < \pi$

When I plugin values of t (like $\pi/2, \pi, 3\pi/2$ for my equations I get the correct points.

But I also get the correct points when I plugin values of t (like 0, $\pi/2, \pi$) for the textbook's equations.

So are we both right? Or is there something I"m missing with the orientation (or direction) of the curve?