Derive $\sin x$ expansion without using calculus

Question

We know that $$\sin x = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \cdots = \sum_{n\ge 0} \frac{(-1)^n}{(2n+1)!}x^{2n+1}$$

But how could we derive this without calculus?

There were some approach using $e^{ix} = \cos x + i \sin x$, pls notice I also would like to avoid such definition as to prove $e^{ix} = \cos x + i \sin x$ again we need the expansion of $\sin x$ and $\cos x$.

One approach I tried is to start with $\sin^2 x$: let $$\sin^2 x := \sum_{n\ge 1} a_n x^{2n}$$ Note: I guess such an ansatz as $\sin x$ is an odd function -- so that $\sin x$ has only $x$'s power of odd numbers, so $\sin^2 x$ only has $x$'s power of even numbers.

Now if I can arrive $$\sin^2 x = \sum_{n\ge 1} \frac{(-1)^{n+1} 2^{2n-1} }{(2n)!} x^{2n},$$ then via $\cos^2x = 1-2\sin^2 x$ I can get the expansion of $\cos x$ then $\sin x$.

To derive $a_n$, first I use $\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$ from geometric interpretation (the arc is almost the opponent edge for small angle $x$), to get $$a_1=1$$

Then from $$\sin^2 2x = 4 \sin^2 x ( 1 - \sin^2 x) \Rightarrow \sin^2 x - \frac14 \sin^2 2x = \sin^4 x, $$ I get $$\sum_{n\ge1} (1-2^{2n-2}) a_n x^{2n} \equiv \left(\sum_n a_n x^{2n}\right)^2$$ This leads to the recursive formula $$ (1-2^{2n}) a_{n+1} = \sum_{k=1}^{n} a_k a_{n-k}, $$ with $a_1=1$, I do get $$a_2=-\frac13, a_3=\frac{2}{45}, a_4 = -\frac{1}{315}, a_5 = \frac{2}{14175}$$ etc, however, I can only get such results via manual calculation -- there is convolution involved, I could not derive a formula for $a_n$.

Is there a way out pls?

Answer 1

Let us first correct your recursive formula, which should rather be $$ \left(1-2^{2n-2}\right) a_n= \sum_{k=1}^{n-1} a_k a_{n-k}.$$

To derive from it (and from $a_1=1$) that $$\forall n\in\Bbb N\quad a_n=\frac{(-1)^{n+1} 2^{2n-1} }{(2n)!},$$ just check that this sequence satifies our recursive formula, i.e. check that $$ \left(1-2^{2n-2}\right)\frac{(-1)^{n+1} 2^{2n-1} }{(2n)!}= \sum_{k=1}^{n-1}\frac{(-1)^{(k+1)+(n-k+1)} 2^{(2k-1)+(2n-2k-1)} }{(2k)!(2n-2k)!}.$$ After some easy simplifications, the task is to prove that $$\sum_{k=0}^n\binom{2n}{2k}=2^{2n-1}.$$ This is done for instance here.

Answer 2

"Without calculus" is a pretty big ask, given that the modern definition of sine is the power series representation you gave, which is the result of analysis. It arises from solving the differential equation $f(x)=-f''(x), f(0)=0, f'(0)=1$. I think the best anyone will be able to do is to show you how to recover the power series of sine from these conditions. So, without any further ado:

Let $f(x)=\sum_{n=0}^{\infty}a_nx^n=a_0+a_1x+a_2x^2+a_3x^3+...$

be the solution to our differential equation. Then for $f''(x)$ we have the following:

$f''(x)=\sum_{n=0}^{\infty}(n+1)(n+2)a_{n+2}x^n=2a_2+6a_3x+12a_4x^2+20a_5x^3+...$

Now it's time to find the $a_n$'s. Given that $f(0)=0$, we may conclude that $a_0=0$. Furthermore, since $f'(0)=1$, we must have $a_1=1$. From the equation $f(x)=-f''(x)$, we know that the sum of the coefficients of each power series for each power of $x$ must be $0$. This gives us the following infinite set of equations:

$$a_0=0$$ $$a_1=1$$ $$a_0+2a_2=0\Longrightarrow a_2=0$$ $$a_1+6a_3=0\Longrightarrow a_3=-\frac{a_1}{6}=-\frac{1}{6}$$ $$a_2+12a_4=0\Longrightarrow a_4=0$$ $$a_3+20a_5=0\Longrightarrow a_5=-\frac{a_3}{20}=\frac{1}{120}$$

This pattern continues, with $a_n=0$ when $n$ is even. It therefore makes sense to re-index $a_n$ so that we get only odd values of $n$. We can accomplish this by setting $a_{2n+1}=\frac{(-1)^n}{(2n+1)!}$. Then for $n=0$ we get $a_1=1$; for $n=1$, $a_3=-\frac{1}{6}$; for $n=2$, $a_5=\frac{1}{120}$, and so on. Our solution is therefore $\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$. One may verify this solution by differentating it twice and checking that the differential equation is satisfied.