OpenStudy (anonymous):

Rewrite the expression as an algebraic function of x and y. cos(sin^−1 (x) − tan^−1 (y))

3 years ago
OpenStudy (larseighner):

Do not be confused by the -1 exponent. It means the inverse function, NOT the reciprocal. Example: \(\tan^{-1}\theta = \arctan\theta \ne {1 \over {\tan\theta}}\)

3 years ago
OpenStudy (anonymous):

Could you explain how to rewrite the expression? I have no idea how to start.

3 years ago
OpenStudy (larseighner):

\( \sin^{-1}x = \arcsin x \) means the angle whose sine is x. Can you figure out how to find the cosine of the angle whose sine is x?) Here is a sketch of the unit circle, you can copy it to label parts if you need to. http://larswiki.larseighner.com/uploads/Main/pinchsinthetaovertheta.svg

3 years ago
OpenStudy (larseighner):

This is the unit circle, so OM = 1

3 years ago
OpenStudy (anonymous):

How do I find the cosine of the angle the sine is on? Is it given in the problem?

3 years ago
OpenStudy (larseighner):

No you figure it from the basic relation between sine and cosine. In the diagram \(\theta\) is the angle whose sine is MP (=x in this example). How do you find the cosine (OP)?

3 years ago
OpenStudy (larseighner):

The triangle OPM has one leg OP \((=x=sin\theta)\) and the hypotenuse is 1. How do you find the cosine (OP)?

3 years ago
OpenStudy (anonymous):

Is it by pythagorean theorem?

3 years ago
OpenStudy (larseighner):

Yes. So sin x is one leg. Cos x is the other leg. Can you find an algebraic expression for cos x? (That means without trig terms.)

3 years ago
OpenStudy (anonymous):

Um, I'm not sure about that

3 years ago
OpenStudy (larseighner):

Here is the Pythagorean identity which might help here: \(\sin^2\theta + \cos^2\theta = 1 \)

3 years ago
OpenStudy (larseighner):

Now you know \(\sin\theta = x \) so \(\sin^2\theta = x^2 \)

3 years ago