Find parametrization of the curve given by the intersection of the unit sphere $x^2+y^2+z^2=1$ and $x+y=1$
So look on the intersection:
$$x^2+(1-x)^2+z^2=1\iff 2x^2-2x+z^2=0\iff 4(x-\frac{1}{2})^2+2z^2=1$$
Plug in just $x=\cos t,y=\sin t$ would not be useful?
we need to chose functions of $\cos t$ and $\sin t$ that will be based on $\cos^2x+\sin^2x=1$? or $x=\frac{1}{2}\cos t+\frac{1}{2}$ and $z=\frac{1}{\sqrt{2}} \sin t$?
This yields the same result but gives insight on the intersection shape.
Define $u=\frac{1}{\sqrt{2}}(x+y)$ and $v=\frac{1}{\sqrt{2}}(x-y)$. We get $u^{2}+v^{2}+z^{2}=1$, $u=\frac{1}{\sqrt{2}}$, and consequently $v^{2}+z^{2}=\frac{1}{2}$.
This is a circle on the $yz$ plane, centered on origin with radius $\frac{1}{\sqrt{2}}$, translated $\frac{1}{\sqrt{2}}\hat{i}$ and rotated $\frac{\pi}{4}\hat{k}$.
The parametrization is $u=\frac{1}{\sqrt{2}}$, $v=\frac{1}{\sqrt{2}}\cos{t}$, $z=\frac{1}{\sqrt{2}}\sin{t}$ or $x=\frac{1}{2}(1+\cos{t})$, $y=\frac{1}{2}(1-\cos{t})$, $z=\frac{1}{\sqrt{2}}\sin{t}$