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I had an idea for a proof by contradiction of the Pythagorean theorem. Where the angle between the base, x and the hypotenuse z is $\theta$, of a right angled triangle.

And the side opposite to the angle is y. Assume that $$x^2 + y^2 \neq z^2$$ Then you have to accept that $$\sin^2 \theta + \cos^2 \theta \neq 1$$ But if you use the Maclaurin expansions of $\sin^2 \theta$ and $\cos^2 \theta$ and add then you get $1$ This is a clear contradiction, so $$x^2 + y^2 = z^2$$

Is this valid?

benjamincantor
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    At some stage you would need to mention right-angled triangles, and that the functions representing the ratios of sides are the same functions as the power series – Henry Apr 12 '19 at 09:25
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    To avoid circularity you have to make sure you can identify sin/cos in a right-angled triangle with the Maclaurin expansion, without using Pythagoras as part of the proof. – Mark Bennet Apr 12 '19 at 09:26
  • Sorry I didn't explain what each thing meant, so I added that. – benjamincantor Apr 12 '19 at 09:29

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Yes. Actually, in Modern Maths the sine and cosine are defined by their series expansions. Of course, you need to know something about the multiplication of two series.

The terms in the series of $\sin^2(x)$ are all of even power since $\sin^2(x)$ is an even function. We can write its general term as $$ \sin^2(x)=\sum_{n=1}^\infty (-1)^{n+1}\left(\sum_{k=1}^n \frac{x^{2n}}{(2k-1)!(2n-2k+1)!}\right) $$ By playing some tricks on conbinactories$n\choose k$, we get $$ \sum_{k=1}^n {2n\choose 2k-1}=\sum_{k=0}^n {2n\choose 2k}=2^{2n-1}. $$ Plugging this into the series of $\sin^2(x)$ above yields $$ \sin^2(x)=\sum_{n=1}^\infty (-1)^{n+1} \frac{2^{2n-1}}{(2n)!} x^{2n}. $$ NB Now you can check whether it is correct by using $\sin^2x=1/2-(1/2)\cos2x.$ However, do not use it to obtain the series expansion at the first place. This trig identity depends on $\sin^2x+\cos^2x=1$, so you cannot assume $\sin^2x=1/2-(1/2)\cos2x$ before you have proved $\sin^2x+\cos^2x=1$. Just in case you do it in the "wrong" way.

Repeat the process for $\cos(x)$, we get $$ \cos^2(x)=1+\sum_{n=1}^\infty (-1)^{n} \frac{2^{2n-1}}{(2n)!} x^{2n}. $$

So we have proven that $\sin^2x+\cos^2x=1$. You may also verify that those series converges.

Now we need to use this result to prove the Pythagorean theorem. This is not easy at all, because you got to link the trig identity to the ratio of sides of a triangle. You have to do this by using Hilbert's axioms of Euclidean spaces. However, relating the geometry to such a power series is clearly much harder than proving the Pythagorean theorem directly. If you want to treat $\sin\theta=b/c$ or something as an axiom, then the proof is completed.

Ma Joad
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  • If I say x^2+y^2 is not equal to z^2, I get z^2(cos^2x+sin^2x) is not equal to z^2, and if I divide each side by z^2, that leaves cos^2x+sin^2x is not equal to 1. But using the expansions I've shown it's not true, so x^2+y^2=z^2 is that valid? – benjamincantor Apr 12 '19 at 11:16
  • No. because you have to show that $x/z=\cos\theta$ first. You have not shown that. – Ma Joad Apr 12 '19 at 12:06
  • Ah okay. How would I go about showing that? – benjamincantor Apr 12 '19 at 23:11
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    Well, you can assume it to be true as an axiom of geometry. However, if you really want to prove that by geometrical arguments, then it is unlikely to work. It would be much harder than proving $x^2+y^2=z^2$ – Ma Joad Apr 13 '19 at 04:16