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Any one seen this proof before?

$$\frac{d}{dx} \sin(x)^2=2\cos(x)\sin(x)$$ $$\frac{d}{dx} \cos(x)^2=-2\cos(x)\sin(x)$$

$$\frac{d}{dx} \sin(x)^2+\frac{d}{dx} \cos(x)^2=0$$ $$\sin(x)^2+\cos(x)^2=c$$ $$\sin(0)^2+\cos(0)^2=c$$ $$1=c,$$


Let C, A, and B be the hypotinuse, opposite, and ajacent sides of a right triangle, then $$((C\sin(x))^2+(C\cos(x))^2=C^2$$ $$A^2+B^2=C^2$$

Is this proof valid, i.e. is the Pythagorean theorem used in defining the above trig relations?

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Yes. The issue here is to ensure that the argument is not circular. – Andrés E. Caicedo Dec 18 '12 at 4:39
How is the argument circular, I guess the proofs of the later statements require some knowladge of trigonometry – Ethan Dec 18 '12 at 4:43
I've seen this before. (I'm sure no pun was intended with the word "circular". Or to be more precise: I'm sure a pun was intended by some people who've used it in this context.) – Michael Hardy Dec 18 '12 at 5:12
The argument seems circular to me, and in the "logically-flawed" sense. It works, but I would want to know how the $\sin$ and $\cos$ functions were defined, and how $d/dx\,\sin x=\cos x$ was proved. – Mario Carneiro Dec 18 '12 at 5:28
up vote 0 down vote accepted

I think this proof is OK.

You can get the derivatives from the trigonometric addition formulas, which can be in turn proved without the Pythagorean theorem, but only from (other) geometry.

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lol I came up with it about a month ago, I thought it was nice in the sense, it didn't use any geometry, but I understand that the proofs of the latter statements require more in depth knowladge of trigonometry and geometry. – Ethan Dec 18 '12 at 19:26
@Ethan - If you like an answer you're always welcome to upvote it as well. – nbubis Jan 2 '13 at 7:27

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