Proving the Pythagorean Identity: A Brief Analysis

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Homework Help Overview

The discussion revolves around proving the Pythagorean identity, specifically that \(\cos^2 \theta + \sin^2 \theta = 1\) holds for all angles \(\theta\). Participants are examining different proofs and questioning their validity, particularly regarding potential circular reasoning in the first proof and the applicability of the second proof in various contexts.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster presents two proofs and questions the validity of the first proof due to its reliance on the identity being proven. Participants discuss the circular nature of the first proof and suggest using a unit circle for the second proof to accommodate negative angles.

Discussion Status

Participants are actively engaging with the proofs, with some suggesting modifications to enhance clarity and applicability. There is recognition of the need to demonstrate the identity for negative angles, and guidance is being offered on how to approach this using the unit circle.

Contextual Notes

Participants are considering the implications of using different representations (right triangle vs. unit circle) and the assumptions related to the angles involved, particularly negative angles.

Apost8
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For homework, we were asked to prove that [tex]\cos^2 \theta + \sin^2 \theta = 1[/tex] is true for all angles [tex]\theta[/tex]. Can someone please take a look at these and let me know if they are acceptable. I'm pretty sure the second one works, but I'm not sure of the first one, mainly because the premise of the proof is derived from the identity I'm trying to prove. Is that allowable or is that a circular argument? Thanks in advance.

P.S. I'm still trying to get used to LATEX, so please forgive me if I've screwed anything up.

PROOF #1:

Given that: [tex]\tan^2 \theta + 1= \sec^2 \theta[/tex]

[tex]\frac{ sin^2 \theta}{cos^ 2 \theta} + 1= \frac {1}{cos^ 2\theta}[/tex]

multiplying by: [tex]\cos^ 2 \theta[/tex]

I get:

[tex]\sin^2 \theta + \cos^2 \theta =1[/tex]


PROOF #2:

(First I drew a right triangle, labeling x, y, r, and [tex]\theta[/tex] ).

Given that: [tex]x^2 + y^2 = r^2[/tex]

dividing by: [tex]r^2[/tex]:

[tex]\frac {x^2} {r^2} + \frac {y^2} {r^2} = \frac {r^2}{r^2}[/tex]

I get:

[tex]\cos^2 \theta + \sin^2 \theta = 1[/tex].
 
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The first proof is a circular argument. How can one prove that:
tan2x + 1 = sec2x without using the Pythagorean Identity (sin2x + cos2x = 1)?
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For proof 2, what if your [tex]\theta[/tex] is negative? I think it will be better if you do it in a unit circle, then to draw a right triangle.
Do you get it? :)
 
So, for proof two, I need to draw the triangle in a unit circle to show that the statement holds true for a negative value of [tex]\theta[/tex]?
 
Last edited:
Apost8 said:
So, for proof two, I need to draw the triangle in a unit circle to show that the statement holds true for [tex]\-theta[/tex]?
Yes, it's correct.
Can you go from there? Note that (-a)2 = a2. And the radius of a unit circle is 1.
Can you go from here? :)
 
I'm not sure how I would show that [tex]\cos^2\theta + \sin^2 \theta = 1[/tex] is true for [tex]-\theta[/tex].

[tex]\cos^2 (-{\theta}) + \sin^2 (-{\theta}) = 1[/tex]
 
Let C be the center of a unit circle, and D be a point on the circle, whose coordinate is (cos t, sin t). Now that you'll always have cos2 t + sin2t = x2 + y2 = r2 = 1. (Since the radius of a unit circle is indeed 1). Can you get it? Or is there anything still unclear?
:)
 
Makes sense to me. Thanks for your help!
 

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