Solve Algebra Expression: \frac{1}{\sqrt{\left( \frac{y}{x}\right)^2 +1}}

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Homework Help Overview

The discussion revolves around simplifying the algebraic expression \(\frac{1}{\sqrt{\left( \frac{y}{x}\right)^2 +1}}\), which is related to trigonometric identities involving \(\cos \tan^{-1} \frac{y}{x}\).

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the simplification process of the expression and questions how their calculator arrived at a different form. Participants suggest various algebraic manipulations and properties of fractions to clarify the simplification.

Discussion Status

Participants are exploring different algebraic techniques to simplify the expression. Some guidance has been offered regarding properties of fractions and manipulation of the expression, but there is no explicit consensus on a single method or understanding.

Contextual Notes

The original poster indicates that this is not a homework problem, suggesting a more informal exploration of the topic. There is also a mention of the poster's self-identified weakness in algebra, which may influence the discussion dynamics.

FrogPad
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I'm working with this expression and I do not understand how to simplify it by hand:

[tex]\frac{1}{\sqrt{\left( \frac{y}{x}\right)^2 +1}}[/tex]

My TI-89 reduces it to:
[tex]\frac{|x|}{\sqrt{x^2+y^2}}[/tex]

How is it doing this? This is not homework. I'm sure it would be acceptiable to just put the simplification down on paper... but if you would rather give hints, that's fine. Thanks :)

The original expression was taken from:
[tex]\cos \tan^{-1} \frac{y}{x}[/tex]
 
Last edited:
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Under the squareroot you have the expression [tex]\frac{x^2}{y^2}+1[/tex]

Remember that
[tex]\frac{a}{b}\pm\frac{c}{d}=\frac{ad\pm bc}{bd}[/tex]

And also, remember that
[tex]\frac{1}{\frac{a}{b}}=\frac{b}{a}[/tex]

Do you see know how your calculator did it?
 
:)

hehe

god my algebra is WEAK.
Thanks.
 
Or: multiply both numerator and denominator by |x|:
[tex]\frac{|x|}{|x|\sqrt{\frac{y^2}{x^2}+ 1}}= \frac{|x|}{\sqrt{x^2(\frac{y^2}{x^2+ 1)}}= \frac{|x|}{\sqrt{y^2+ x^2}}[/tex]
 

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