Is this Pythagorean Theorem proof legitimate and correct?

[mohamadh95]

I found this proof of Pythagoras theorem, I need to find if it's "legitimate" and if it's correct. I mean by that: does the trigonometric relations I used need Pythagoras to be proved , or they were found by another method. Thank you in advance.

 

[PhotonCurve]

They are found in much simpler methods, but it looks right.

 

[symbolipoint]

A fairly simple proof using a trapezoid is that of James A. Garfield. Look at the article under the "Algebraic Proofs" section, http://en.wikipedia.org/wiki/Pythagorean_theorem

 

[Ray Vickson]

I found this proof of Pythagoras theorem, I need to find if it's "legitimate" and if it's correct. I mean by that: does the trigonometric relations I used need Pythagoras to be proved , or they were found by another method. Thank you in advance.

Your proof is invalid: it uses $\sin(\theta) = \sqrt{1-\cos^2(\theta)},$ but this property requires Pythagoras in its proof!

 

[vanhees71]

It depends on how you define the trigonometric functions. If you use the definition over their series
\cos x=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} x^{2k}, \quad \sin x=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{2k+1},
which defines these functions for all \mathbb{R}, you can easily prove that
\cos' x=-\sin x, \quad \sin' x=\cos x
and from this
\frac{\mathrm{d}}{\mathrm{d} x}[\cos^2 x+\sin^2 x]=0 \; \Rightarrow \cos^2 x+\sin^2 x=\text{const},
and from that, because of \cos 0=1, \quad \sin 0=0 you get \cos^2 x + \sin^2 x=1.
Then you can define the number \pi/2 as the smallest positive solution of \cos(\pi/2)=0.

In this way you can derive step by step all the properties of sin and cos without using any geometric definitions. Then you can prove the geometrical properties of these functions from the unit circle's standard parametrization
x=\cos \varphi, \quad y=\sin \varphi, \quad \varphi \in [0,2 \pi).
This is, however, obviously not precalculus anymore.

 

[mohamadh95]

Your proof is invalid: it uses $\sin(\theta) = \sqrt{1-\cos^2(\theta)},$ but this property requires Pythagoras in its proof!

Well about this equality I saw on Wikipedia that (sin(x))^2=(1-cos(2x))/2 is obtained from the cosine double-angle formula which can be proved without Pythagoras. And that's how I can find this equality without Pythagoras.

 

[mohamadh95]

It depends on how you define the trigonometric functions. If you use the definition over their series
\cos x=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} x^{2k}, \quad \sin x=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{2k+1},
which defines these functions for all \mathbb{R}, you can easily prove that
\cos' x=-\sin x, \quad \sin' x=\cos x
and from this
\frac{\mathrm{d}}{\mathrm{d} x}[\cos^2 x+\sin^2 x]=0 \; \Rightarrow \cos^2 x+\sin^2 x=\text{const},
and from that, because of \cos 0=1, \quad \sin 0=0 you get \cos^2 x + \sin^2 x=1.
Then you can define the number \pi/2 as the smallest positive solution of \cos(\pi/2)=0.

In this way you can derive step by step all the properties of sin and cos without using any geometric definitions. Then you can prove the geometrical properties of these functions from the unit circle's standard parametrization
x=\cos \varphi, \quad y=\sin \varphi, \quad \varphi \in [0,2 \pi).
This is, however, obviously not precalculus anymore.

Thank you very much, I should have thought of that earlier. Now I can confirm that the proof is right.

 

[Ray Vickson]

Thank you very much, I should have thought of that earlier. Now I can confirm that the proof is right.

Of course---Rudin does it this way. But that is hardly pre-calculus, as you say.