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How do I go about deriving say, $\cfrac{d}{dx}[\sin x]=\cos x$ without using the limit definition of a derivative? I believe with that approach I'd have to know how to do Taylor Series expansions, and so I haven't gone that far yet.

So far the best argument I can come up with is that the rate of change at the minimum and maximums, k, of $\sin x$ is $0$, yet it is also the case that $\cos k=0$. This isn't really a derivation, though.

Any simple ways of doing this without needing calc 2 and up?

Lex_i
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  • I'm not sure there's a way to prove these without the limit definition - after all, that's the definition. I will say, though, that you can use the angle addition formulas instead of a taylor expansion. – HallaSurvivor Mar 21 '20 at 05:14
  • It depends on your definition of $\sin{x}$. A common way is to start defining $\arctan{x} = \int_0^x \frac{1}{1+t^2}dt$ and then you can define $\sin{x}$ in terms of the tan function. property $\sin'{x} = \cos{x}$ follows immediately. there are other definitions. In the book Calculus by Spivak, its chapter 15 is dedicated to the trigonometric functions and he presents several definitions. – Donlans Donlans Mar 21 '20 at 05:19
  • https://math.stackexchange.com/questions/3336/how-many-ways-are-there-to-define-sine-and-cosine you can find the spivaks defintion in the answer of Pedro Tamaroff – Donlans Donlans Mar 21 '20 at 05:24

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If you can buy yourself just a teeny bit of multivariate calculus, you can get this with no limits. All we need are three facts about the function $t\mapsto (\cos(t),\sin(t))$. For brevity and to remind ourselves we can't use any other special properties of the trig functions we'll just write these as $(c(t),s(t))$:

  1. It moves around the unit circle - the Pythagorean identity $c^2+s^2=1$
  2. It moves at unit speed - so $\dot{c}^2+\dot{s}^2=1$
  3. It moves counterclockwise.

The tiny bit of multivariate calc is in (2). We know the length of the circular arc subtended by an angle, measured in radians, is the circle's radius times the angle; that's what defines radians in the first place. So with unit radius we just have $L(t)=t$ and $\frac{dL}{dt}=1$. And $dL$ is the hypotenuse of the right triangle with sides $dx$ and $dy$ so $1=\frac{dL}{dt}^2=\frac{dx}{dt}^2+\frac{dy}{dt}^2=\dot{c}^2+\dot{s}^2$.

By (1) and the chain rule (and dividing by two), $c\dot{c}+s\dot{s}=0$.

We solve for $\dot{c}$ and substitute into (2), yielding $(\frac{-s\dot{s}}{c})^2+\dot{s}^2=1$ and $(s^2+c^2)\dot{s}^2=c^2$. By (2) we simplify that to $\dot{s}^2=c^2$.

Clockwise or counterclockwise just tells us whether it's the negative or positive square root. CCW means the signs of $\dot{s}$ and $c$ agree: when we're on the right-half plane we're moving up, on the left we're moving down. So we take the positive square root, $\dot{s}=c$. Using $c\dot{c}+s\dot{s}=0$ we also know $\dot{c}=-s$.

Actually, it's enough to know it's CCW at just one point. We know $\cos(0)>0$, and for small $t$, $\sin(t)\geq \sin(0)$ so $\sin'(0)\geq 0$.

In terms of a little more multivariable calc, a constant radius means our position and velocity vectors are perpendicular; their dot product is just the chain rule result $c\dot{c}+s\dot{s}=0$. Remembering how to come up with a perpendicular vector, you can skip the algebra step. I have to give credit to Michael O’Connor for the above way of putting it with less multivariate calc.

Prodicus
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