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I'm looking for a way to prove $\sin^2x+\cos^2x=1$ without using the Pythagorean Theorem or proving it.

The only thing I've tried was to play with the Taylor series, but I didn't get far with that as nothing seemed to 'mathify' nicely after a point.

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    $\begingroup$ Examine the derivative of $f(x)=\sin^2 x+\cos^2 x$. $\endgroup$ – David Mitra Aug 24 '14 at 10:43
  • $\begingroup$ I wonder if Euler's identity can be used here... $\endgroup$ – barak manos Aug 24 '14 at 10:50
  • $\begingroup$ All these proofs are very nice but I am not convinced that the demonstration of the auxillary results did not use the theorem itself. $\endgroup$ – Rene Schipperus Aug 24 '14 at 12:53
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Consider $g(x) = \sin^2 x + \cos^2 x $. Then,

$$g'(x) = 2 \sin{x} \cos{x} - 2 \cos{x} \sin{x} = 0.$$

This means that $g(x) $ is a constant. Since $g(0) = 1$, it follows that...

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Algebraic proof. $$1= \cos 0 = \cos(x-x) = \cos x \cos(-x) - \sin x \sin(-x) = \\ \cos x \cos x - (- \sin x \sin x ) = \cos^2 x + \sin^2 x.$$

Using Euler's formula. $$ \cos^2 x + \sin^2 x = (\cos x + i \sin x)(\cos x - i \sin x) = \\ (\cos x + i \sin x)(\cos(-x)+i \sin(-x)) = e^{ix}e^{-ix} = 1 $$

You can find other proofs and hints also for this two at ProofWiki.

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$$\sin^2 x+\cos^2 x= \sin x \sin x+\cos x \cos x= \\ \frac{1}{2} \left ( \cos{(x-x)}-\cos{(x+x)} \right ) +\frac{1}{2} \left ( \cos{(x-x)}+\cos{(x+x)} \right )=\frac{2\cos{(0)} }{2}=\cos{(0)}=1$$

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