[SOLVED] Evaluate cos(arcsin(t))?

**thekrown** · March 8th 2010, 01:55 PM

What do they mean exactly by evaluate cos(sin^-1(t))?

**tom@ballooncalculus** · March 8th 2010, 02:05 PM

Draw a triangle diagram, or cheat by consulting

Compositions of Trig and Inverse Trig Functions

... assuming evaluate is meant to mean simplify.

**thekrown** · March 8th 2010, 02:21 PM

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.

**tom@ballooncalculus** · March 8th 2010, 02:39 PM

Originally Posted by thekrown

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).

**Drexel28** · March 8th 2010, 02:42 PM

Originally Posted by thekrown

What do they mean exactly by evaluate cos(sin^-1(t))?

Let $y=\arcsin(x)$ then $y'=\frac{1}{\sqrt{1-x^2}}$ . But, we could have also done that $y=\arcsin(x)\implies \sin(y)=x\implies y'\cos(y)=1\implies y'=\frac{1}{\cos(y)}=\frac{1}{\cos(\arcsin(x))}$ . Comparing these two gives $\cos(\arcsin(x))=\sqrt{1-x^2}$

**Krizalid** · March 8th 2010, 02:59 PM

$f(t)=\cos(\arcsin t)$ makes sense as long as $|t|\le1,$ now $f(t)\ge0,$ so by using $\cos t=\pm\sqrt{1-\sin^2t}$ and putting $t\mapsto\arcsin t$ we get $f(t)=\pm\sqrt{1-t^2},$ but $f\ge0,$ thus $f(t)=\sqrt{1-t^2}$ and we're done.

**thekrown** · March 8th 2010, 05:02 PM

Hmm I don't really get this to much. Thanks for your help guys but I think I'm going to need some tutor assistance with this.

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