TITLE: Proof of the Pythagorean Theorem via $\frac{d}{dx}\sin^2 x + \frac{d}{dx}\cos^2 x = 0$
QUESTION [11 upvotes]: Any one seen this proof before?
$$\frac{d}{dx} \sin(x)^2=2\cos(x)\sin(x)$$
$$\frac{d}{dx} \cos(x)^2=-2\cos(x)\sin(x)$$
$$\frac{d}{dx} \sin(x)^2+\frac{d}{dx} \cos(x)^2=0$$
$$\sin(x)^2+\cos(x)^2=c$$
$$\sin(0)^2+\cos(0)^2=c$$
$$1=c,$$
$$\sin(x)^2+\cos(x)^2=1$$
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?

REPLY [2 votes]: 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.