What do they mean exactly by evaluate cos(sin^-1(t))?
Draw a triangle diagram, or cheat by consulting
Compositions of Trig and Inverse Trig Functions
... assuming evaluate is meant to mean simplify.
Hmmm we haven't done much of this in class. Usually when a sin or cos pops up it's to calculate the rate of change or find the derivative. Any clues on how to simplify this? I looked at the link and don't really recognize doing anything similar in class.
Yes, but this particular task isn't actually calculus, just trig and pythag.
Draw a right triangle. Label one of the two 'legs' as t = sin a. Label one of the corners (which one?) as a = arcsin t. Label the remaining leg as cos a = cos(arcsin(t)) = ... using pythag, sqrt(1 - t^2).
Let $\displaystyle y=\arcsin(x)$ then $\displaystyle y'=\frac{1}{\sqrt{1-x^2}}$. But, we could have also done that $\displaystyle y=\arcsin(x)\implies \sin(y)=x\implies y'\cos(y)=1\implies y'=\frac{1}{\cos(y)}=\frac{1}{\cos(\arcsin(x))}$. Comparing these two gives $\displaystyle \cos(\arcsin(x))=\sqrt{1-x^2}$
$\displaystyle f(t)=\cos(\arcsin t)$ makes sense as long as $\displaystyle |t|\le1,$ now $\displaystyle f(t)\ge0,$ so by using $\displaystyle \cos t=\pm\sqrt{1-\sin^2t}$ and putting $\displaystyle t\mapsto\arcsin t$ we get $\displaystyle f(t)=\pm\sqrt{1-t^2},$ but $\displaystyle f\ge0,$ thus $\displaystyle f(t)=\sqrt{1-t^2}$ and we're done.