Can the fact $(\sin x)'=\cos x$ be proved not using the fact $\lim_{x\to 0} \frac{\sin x}{x}=1$?

Question

Is it possible to prove $$\left(\sin(x)\right)^\prime = \cos(x)$$ without using the fact that $\displaystyle\lim_{x\to 0} \frac{\sin x}{x}=1$?

If so, Please give me the proof of that case.

Answer 1

Note that the definition of the derivative of sine at $0$ is $\lim_{h \to 0} \frac{\sin h}{h}$, so you are asking "can you differentiate sine at an arbitrary point $x$ without knowing how to differentiate it at $0$?". The answer is pretty clearly no. The best you can hope for is to change the definition of sine and cosine so that the difficulty of the calculation is suppressed. Here are two possibilities:

1. Define sine and cosine by: $$\sin x = x - \frac{x^3}{3} + \frac{x^5}{5} - \ldots$$ $$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{4} - \ldots$$ It is not hard to show that the power series on the right-hand side converge on the entire real line, and thus they can be differentiated term-by-term to prove what you want. The downside is that it takes a great deal of effort to show that the functions defined here have anything to do with the sine and cosine functions that you're used to; for example, try proving that $\sin^2 x + \cos^2 x = 1$ or that sine and cosine have period $2 \pi$ this way.

2. By general theory from ODE's, the differential equation $$y'' + y = 0$$ has a two dimensional solution space so define sine to be the unique solution $y_1$ satisfying the initial condition $y_1(0) = 0$ and $y_1'(0) = 1$, and define cosine to be the unique solution $y_2$ satisfying $y_2(0) = 1$ and $y_2'(0) = 0$. Now set $y_3 = y_1'$. Observe that $y_3(0) = y_1'(0) = 1$ and $y_3'(0) = y_1''(0) = -y_1(0) = 0$. Moreover $y_3'' + y_3 = y_1''' + y_1' = (y_1'' + y_1)' = 0$, so $y_3$ solves the same differential equation as $y_2$ with the same initial data. By uniqueness, $y_1' = y_3 = y_2$.

The downsides of this approach are (at least) twofold. First, it is once again not easy to prove that the functions defined this way have anything in common with the usual sine and cosine functions (of course, if you know how to differentiate them already then you can use uniqueness). Second, the proof of the existence and uniqueness theorem for ordinary differential equations is much harder than calculating the limit of $\frac{\sin x}{x}$.

Answer 2

You probably won't like this, but here goes: $$\sin x={e^{ix}-e^{-ix}\over2i}$$ Differentiating, $$(\sin x)'={ie^{ix}+ie^{-ix}\over2i}={e^{ix}+e^{-ix}\over2}=\cos x$$

Answer 3

Some speculations.

Let $f(x)=(\sin x)', g(x)=(\cos x)'$.
Then $(\sin 2x)'=2f(2x)=2f(x)\cos x+2g(x)\sin x$.

$\cos^2 x+\sin^2 x=1$ so $g(x)\cos x + f(x)\sin x =0$ and $g(x)=-f(x)\tan x$.

And finally we have $f(2x)\cos x=f(x)\cos 2x$. $f(x)=\cos x$ is obviously a solution.
Assuming $f(x)$ is continuous and periodic, may be it is possible to obtain that it is unique, I don't know.