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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?

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    Yes. The issue here is to ensure that the argument is not circular.2012-12-18
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    How is the argument circular, I guess the proofs of the later statements require some knowladge of trigonometry2012-12-18
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    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.)2012-12-18
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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.2012-12-18
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    @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.2018-04-18
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    @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.2018-04-18
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    @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.2018-04-18
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    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.2018-04-18
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    @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.2018-04-18
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    @Ethan sorry about that; just came across it; and it was an interesting proof of the theorem.2018-04-18

2 Answers 2

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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.

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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.2012-12-18
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You put a check mark beside nbubis' answer indicating that it solved your problem but I think that actually, you're missing something so it didn't solve your problem. I will give an answer that attempts to fill in what I think you're missing. If you treat distance as an undefined concept and assume that satisfies the following properties

  1. For any points $(x, y)$ and $(z, w)$ in $\mathbb{R}^2$, $d((0, 0), (z, w)) = d((x, y), (x + z, y + w))$
  2. For any point $(x, y)$ in $\mathbb{R}^2$, $d((0, 0), (x, y))$ is nonnegative
  3. For any nonnegative real number $x$, $d((0, 0), (x, 0)) = x$
  4. For any point $(x, y)$ in $\mathbb{R}^2$, $d((0, 0), (x, -y)) = d((0, 0), (x, y))$
  5. For any points $(x, y)$ and $(z, w)$ in $\mathbb{R}^2$, $d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$
  6. The area of any square in $\mathbb{R}^2$ is the square of the length of its edges
  7. $\forall x \in \mathbb{R}d((0, 0), (\cos(x) ,\sin(x))) = 1$

then yes you can prove it. The Pythagoren theorem is equivalent to the statement that distance satisfies properties 1 and 5. Property 1 just shows that the Pythagorean theorem holds for all right angle traingles in $\mathbb{R}^2$ whose legs are parallel to the axes. Using property 5, we can deduce from that that the Pythagorean theorem holds for all right angle triangles, not just the ones whose legs are parallel to the axes.

That still sweeps under the rug the issue of why we can treat it like distance is an undefined concept that we can assume satisfies those properties. As shown in this answer, $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$ is the unique function satisfying the first 5 properties and it also satisfies properties 6 and 7. We can decide that as a result of that, we can define the distance formula to be $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$.