Proving the Pythagorean Identity: A Brief Analysis

[Apost8]

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].

 

[VietDao29]

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? :)

 

[Apost8]

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]?

 

[VietDao29]

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? :)

 

[Apost8]

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]

 

[VietDao29]

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?
:)

 

[Apost8]

Makes sense to me. Thanks for your help!