# Question involving trigonometric identities and inverse functions

1. Mar 16, 2008

### AussieDave

[SOLVED] Question involving trigonometric identities and inverse functions

1. The problem statement, all variables and given/known data

2. Relevant equations
I've tried to combine the following known equations to come up with a solution:

$$\frac{d}{dx}$$(sin$$^{-1}$$x) = $$\frac{1}{\sqrt{1-x^{2}}}$$

sin x = y
sin $$^{-1}$$ y = x

$$\frac{d}{dx}$$(sin x) = cos x

cos$$^{2}$$x + sin$$^{2}$$x = 1

3. The attempt at a solution
I've been writing down a whole bunch of different equations on a sheet of paper to try and come up with something to connect the two equations. I feel like I'm kind of shooting in the dark though as I'm not sure where to begin and how to use this knowledge of the derivatives (if that's needed) and the relationship between cos and sin. I've tried fiddling around with the pythagorean identity but I end up with things like:

x = sin$$^{-1}$$x$$\sqrt{1-cos^{2}x}$$

and I'm not sure where to go from there.

Your help will be much appreciated.

Kind regards,

David

2. Mar 16, 2008

### Snazzy

Draw a triangle. sin(y) = x = opp/hyp

3. Mar 16, 2008

### AussieDave

I'm still running into trouble here.

I have:

cos(y) = adj/hyp = sin^-1(opp/hyp) = x

but I can't find a way to move on. Do I have to use hyp = SqrRoot(opp^2 + adj^2) ?

4. Mar 16, 2008

### Snazzy

$$sinY=x=\frac{opp}{hyp}$$ Yes? So what's the hypotenuse then? Can you figure out the length of the adjacent side by knowing the opposite side length and the hypotenuse length? Can you then determine the ratio of the adjacent side to the hypotenuse, which gives $$cosY$$? Remember that cos relates the angle between the adjacent side and hypotenuse to the ratio of the adjacent side to the hypotenuse.

Last edited: Mar 16, 2008
5. Mar 16, 2008

### AussieDave

Well Adj^2 + Opp^2 = Hyp^2 so that's the relationship there. If I rearrange and sub that into the equation you give me and that doesn't seem to make things any less complicated.

6. Mar 16, 2008

### tiny-tim

… turn the equation round! …

Hi David!

When you have inverse functions, just turn the equation round:

$$y = sin^{-1}x$$ , so x = siny , so cosy = … ?

7. Mar 16, 2008

### AussieDave

I do understand the basic principles of inverse functions but I'm struggling to relate that to the basic triangle and produce numbers similar to those given as the possible answers to the question.

8. Mar 16, 2008

### tiny-tim

… on a triangle …

Ah!

Well, y is one of the angles of the right-angled triangle, and x is the opposite side, and the hypotenuse is 1.

So cosy is the adjacent side.

So just use good ol' Pythagoras …

9. Mar 16, 2008

### Snazzy

10. Mar 16, 2008

### AussieDave

I never knew that the hypotenuse equalled 1. Can you please tell me why that is the case? Given that, I was able to calculate cos y = $$\sqrt{1-x^{2}}$$ which is answer (c). Is this correct?

EDIT: Actually, after just looking at Snazzy's diagram, I understand why the hypotenuse = 1 because y = sin^-1(opp/hyp) and y = sin^-1(x) and x = opp so hyp = 1.

Thank you very much for your help. I'm guessing (c) is therefore correct?

Last edited: Mar 16, 2008
11. Mar 16, 2008

### tiny-tim

Yes: always put hypotenuse = 1.

It's because "sin = opposite over hypotneuse" - so if you put hypotenuse = 1, then the formula is simply "sin = opposite"!

(Same for cos, of course.)
Yes!

12. Mar 16, 2008

### Snazzy

$$sinY=\frac{opp}{hyp}=\frac{x}{hyp}=x$$
$$\frac{x}{x}=hyp=1$$

So yes, once you figure out the hypotenuse = 1, you can find the length of the adjacent side using the Pythagorean theorem and it will give you $$adj=\sqrt{1-x^2}$$

$$cosY=\frac{adj}{hyp}=\frac{\sqrt{1-x^2}}{1}=\sqrt{1-x^2}$$

Last edited: Mar 16, 2008
13. Mar 16, 2008

### AussieDave

Well thank you to both of you. It's good to get that little guy out of the way. Now I have to do that [SOLVED] thing. Hmmm.

14. Dec 30, 2009

### mohini patil

Re: [SOLVED] Question involving trigonometric identities and inverse functions