Question

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

Answer 1

Do not be confused by the -1 exponent. It means the inverse function, NOT the reciprocal.

Example:

\(\tan^{-1}\theta = \arctan\theta \ne {1 \over {\tan\theta}}\)

Answer 2

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

Answer 3

\( \sin^{-1}x = \arcsin x \) 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

Answer 4

This is the unit circle, so OM = 1

Answer 5

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

Answer 6

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)?

Answer 7

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

Answer 8

Is it by pythagorean theorem?

Answer 9

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.)

Answer 10

Um, I'm not sure about that

Answer 11

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

Answer 12

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