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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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    $\begingroup$ Yes. The issue here is to ensure that the argument is not circular. $\endgroup$ Dec 18, 2012 at 4:39
  • $\begingroup$ How is the argument circular, I guess the proofs of the later statements require some knowladge of trigonometry $\endgroup$
    – Ethan
    Dec 18, 2012 at 4:43
  • $\begingroup$ 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.) $\endgroup$ Dec 18, 2012 at 5:12
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    $\begingroup$ 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. $\endgroup$ Dec 18, 2012 at 5:28
  • $\begingroup$ @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. $\endgroup$ Apr 18, 2018 at 2:32

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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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