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Question involving trigonometric identities and inverse functions

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[SOLVED] Question involving trigonometric identities and inverse functions

1. Homework Statement
http://img141.imageshack.us/img141/2651/quiz1question5zi8.jpg [Broken]


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

[tex]\frac{d}{dx}[/tex](sin[tex]^{-1}[/tex]x) = [tex]\frac{1}{\sqrt{1-x^{2}}}[/tex]

sin x = y
sin [tex]^{-1}[/tex] y = x

[tex]\frac{d}{dx}[/tex](sin x) = cos x

cos[tex]^{2}[/tex]x + sin[tex]^{2}[/tex]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[tex]^{-1}[/tex]x[tex]\sqrt{1-cos^{2}x}[/tex]

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

Your help will be much appreciated.

Kind regards,

David
 
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Answers and Replies

458
0
Draw a triangle. sin(y) = x = opp/hyp
 
47
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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) ?
 
458
0
[tex]sinY=x=\frac{opp}{hyp}[/tex] 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 [tex]cosY[/tex]? Remember that cos relates the angle between the adjacent side and hypotenuse to the ratio of the adjacent side to the hypotenuse.
 
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47
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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.
 
tiny-tim
Science Advisor
Homework Helper
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… turn the equation round! …

Hi David! :smile:

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

[tex]y = sin^{-1}x[/tex] , so x = siny , so cosy = … ? :smile:
 
47
0
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.
 
tiny-tim
Science Advisor
Homework Helper
25,789
249
… 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 … :smile:
 
458
0
160wzsn.jpg
 
47
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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 = [tex]\sqrt{1-x^{2}}[/tex] 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?
 
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tiny-tim
Science Advisor
Homework Helper
25,789
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I never knew that the hypotenuse equalled 1. Can you please tell me why that is the case?
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.)
Given that, I was able to calculate cos y = [tex]\sqrt{1-x^{2}}[/tex] which is answer (c). Is this correct?
Yes! :smile: :smile:
 
458
0
[tex]sinY=\frac{opp}{hyp}=\frac{x}{hyp}=x[/tex]
[tex]\frac{x}{x}=hyp=1[/tex]

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 [tex]adj=\sqrt{1-x^2}[/tex]

[tex]cosY=\frac{adj}{hyp}=\frac{\sqrt{1-x^2}}{1}=\sqrt{1-x^2}[/tex]
 
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47
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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.
 
Re: [SOLVED] Question involving trigonometric identities and inverse functions

1. Homework Statement
http://img141.imageshack.us/img141/2651/quiz1question5zi8.jpg [Broken]


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

[tex]\frac{d}{dx}[/tex](sin[tex]^{-1}[/tex]x) = [tex]\frac{1}{\sqrt{1-x^{2}}}[/tex]

sin x = y
sin [tex]^{-1}[/tex] y = x

[tex]\frac{d}{dx}[/tex](sin x) = cos x

cos[tex]^{2}[/tex]x + sin[tex]^{2}[/tex]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[tex]^{-1}[/tex]x[tex]\sqrt{1-cos^{2}x}[/tex]

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

Your help will be much appreciated.

Kind regards,

David
160wzsn.jpg
with reference to your problem
to make sin y=x, hypotenuse=1
therefore adjacent side is=sqrt{1-x^{2}}
therefore cos y=sqrt{1-x^{2}}.
 
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