Prove Identity: Alternatives to Derivation

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Homework Help Overview

The discussion revolves around proving a mathematical identity related to inverse trigonometric functions, specifically involving the arcsine function. Participants explore alternative methods of proof and the implications of certain variable restrictions.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the original proof methods, including integration and geometric reasoning. Questions arise regarding the domain restrictions of the variables involved, particularly concerning the values of r and a. There is also a focus on the use of known identities for inverse functions.

Discussion Status

The conversation is ongoing, with participants providing insights and questioning assumptions. Some guidance has been offered regarding the use of double angle formulas and identities for arc functions, but no consensus has been reached on a definitive proof method.

Contextual Notes

There are mentions of potential typos in the original problem statement and the need for clarity regarding the restrictions on the variables a and r. The discussion also highlights a lack of familiarity with certain identities related to inverse functions.

icystrike
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Homework Statement


[PLAIN]http://img812.imageshack.us/img812/4068/deriveidentitybydesmond.jpg

In fact i came up with this identity just wondering if there is alternative way to prove it.

Homework Equations


The Attempt at a Solution

 
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You made a slight typo, it should be [tex]\sin^{-1}\left(\frac{a}{r}\right)[/tex] at the end.

To make things clearer, you should let a/r=x and state the restriction on x for computable real values.

Ok anyway to the point... You're looking for an alternate way to proving this identity? What is the first way you proved it?
 
Without investigating further, do you require r = ±a? Because the domain of inverse sine is between ±1, but you have both a/r and r/a as arguments.
 
Tedjn said:
Without investigating further, do you require r = ±a? Because the domain of inverse sine is between ±1, but you have both a/r and r/a as arguments.

Well that's why as I said, it must've been a typo such that the inverse sine on the RHS is a/r.
 
Mentallic said:
Well that's why as I said, it must've been a typo such that the inverse sine on the RHS is a/r.

Yes sorry for my typo...
then i guess -1<=x<=1

I derived this identity when someone told me to integrate a small area bounded by the upper quadrant of circle and x=a and y-axis ... then i derive two method ... one is by substitution and integrate with repect to theta and next is by geometric means (area of part of circle minus a small triangle) Ignore my proof.. I more interested in a more generalized identity or theorem that can explain this
 
Last edited:
That's a little complicated, but fair enough :smile:

Let [tex]a/r=x[/tex] and [tex]\sin^{-1}x=A[/tex], now look at the LHS - you should obviously know your double angle formulae, mainly [tex]\sin(2A)=2 \sin A \cos A[/tex]

Do you know how to find [tex]\cos\left(\sin^{-1}x\right)[/tex]?

After doing this you'll notice that your identity can be simplified greatly.
 
Ok! I've proved... i wasnt quite aware of the existence of the list of identities for arc function ... ( I am wondering at which level you will be more exposed to the identities of the inverse function?)
 
Identities for arc functions? If you mean [tex]\cos(\sin^{-1}x)=\sqrt{1-x^2}[/tex] and such as being identities, well, frankly I'm quite surprised you haven't learned them already.
 

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