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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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    @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 w$i$th 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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Without any assumptions about what properties distance follows, you can't prove anything about what the distance formula is and all you can prove is that $\forall$x$\in \mathbb{R}\cos^2(x) + \sin^2(x) = 1$. We could make the assumptions that distance is a binary function from $\mathbb{R}^2$ to $\mathbb{R}$, in otherwords, a function from $(\mathbb{R}^2)^2$ to $\mathbb{R}$ satisfying 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. $\forall x \in \mathbb{R}\forall z \in \mathbb{R}^+d((0, 0), (z\cos(x) ,z\sin(x))) = z$

Then we could show that the distance formula is $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$. This just shows that for any right angle triangle whose legs are parallel to the axes, the Pythagoren theorem holds. To prove the Pythagorean theorem holds for all right angle triangles, we have to show that distance also satisfies the following property

  • For any points $(x, y)$ and $(z, w)$ in $\mathbb{R}^2$, $d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$

That can be done as follows. $d((0, 0), (xz - yw, xw + yz)) = \sqrt{(xz - yw)^2 + (xw + yz)^2} = \sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = \sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = \sqrt{(x^2 + y^2)(z^2 + w^2)} = \sqrt{x^2 + y^2}\sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$

Now how do we show that there actually exists a way of defining distance that satisfies the assumptions I made? Because it's trivial to show that the function $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$ in fact does satisfy those properties.

Some people make other assumptions about what properties distance follows. Here are some assumptions about distance in $\mathbb{R}$ each of which some people make.

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

How do we know there exists a way of defining distance that satisfies all 7 properties? Because it has been proven in this answer that $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$ is the unique function satisfying the first 5 properties from this list and it also satisfies properties 6 and 7 from this list.

The second assumption I made earlier does not appear as one of them. That's because using properties 3, 5, and 7, we can deduce that that definition of distance satisfies the second assumption I made earlier.