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Any one seen this proof before?

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

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

$$\sin(x)^2+\cos(x)^2=1$$

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?

Blue
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Ethan Splaver
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    Yes. The issue here is to ensure that the argument is not circular. – Andrés E. Caicedo Dec 18 '12 at 04:39
  • How is the argument circular, I guess the proofs of the later statements require some knowladge of trigonometry – Ethan Splaver Dec 18 '12 at 04:43
  • 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 05:12
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    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 05:28
  • @MarioCarneiro I think you have to go back to basic algebra with linear equations on this one. Instead of thinking of the $\sin(x)$ and $\cos(x)$ as circular functions with a period in terms of $\pi$, but in the terms of being related to the slope of a line $$\frac{ rise } {run} = \frac{y_2 - y_1}{x_2 - x_1} \text{ as } \frac{\sin x}{\cos x} = \tan x$$ where $$x \text{ is the angle above the horizontal }$$ In relation to the length of the sides of the triangle its interior angles and its area. – Francis Cugler Apr 18 '18 at 02:32
  • @FrancisCugler That equation doesn't bring any angle anything into the picture, though. The slope of the line does not uniquely determine the point $(\sin x,\cos x)$; you need the angle and the distance to do that. But obviously locating it on the circle $x^2+y^2=1$ would be... circular. – Mario Carneiro Apr 18 '18 at 05:11
  • @MarioCarneiro Kind of, but not when you are referring to the $\sin x$ and $\cos x$ in regards the the properties of a triangle where the slope or ratio of the two is mutually inclusive to the hypotenuse. Here's another example... aside from this: consider the points (-1,0), (0,0), and (1,0) on the x-axis of a unit circle. Not that we are concerned with the circle but just the two points they form a straight line between 3 points. If you take the dot product between the outer points it is equivalent to the $cos$ of the angle between them 180 degrees. This relationship isn't circular. – Francis Cugler Apr 18 '18 at 10:59
  • I keep getting pinged from your guy(s) comments to this question I asked over five years ago. Going back and reading my old questions/comments from early high school makes me cringe, please don't lol. – Ethan Splaver Apr 18 '18 at 11:01
  • @MarioCarneiro Basically the OP is using the properties of the trig functions to express the relation ship based on the properties of triangles and not circles although the two are related. Here it is expressed as Sides A,B, & C. This implies the definition by right triangle and not it's circular definition. – Francis Cugler Apr 18 '18 at 11:01
  • @Ethan sorry about that; just came across it; and it was an interesting proof of the theorem. – Francis Cugler Apr 18 '18 at 11:03

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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.

Nathaniel Bubis
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