Deriving power series for $\sin x$ without using Taylor's Theorem or $\exp z$

Another method is using the inequality

$$\sin(x)\leq x$$

Integrate this between $0$ and $t$

$$-\cos(t)+\cos(0)\leq t^2/2$$ $$1-t^2/2\leq \cos(t)$$

Integrate this between $0$ and $x$:

$$x-x^3/3!\leq \sin(x)$$

Repeat this and you will get a lower and upper bound, which are just partial sums of the Taylor series. Note that you directly get the taylor series for $\cos(x)$.

I think this works:

Starting with defining $(cos(t), sin(t))$ from the unit circle, it can be shown by elementary means that $sin(t)$ satisfies $f''=-f$. Also by observation $f(0)=0$.

Consider a power series $f(x) = a_0 + a_1 x + a_2 x^2 + ... $ with the same two constraints, we have $a_0=0$ and, by differentiating twice and comparing coefficients, $(k+2)(k+1)a_{k+2} = -a_k$.

It can be immediately seen that for even $k$, $a_k=0$.

Also that:

$a_3 = -\frac{1}{(3)(2)}a_1$ = -$\frac{a_1}{3!}$

$a_5 = -\frac{1}{(5)(4)}a_3 = +\frac{a_1}{5!}$

... etc.

So $f(x) = a_1 (x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...)$ which (by the Ratio Test) converges.

$a_1$ must have value 1 if $f$ is to behave similarly to $sin$ in the neighbourhood of 0 (as $sin(x) \to x$ as $x \to 0$).

So we have $sin(x)$ and $f(x)$ both satisfying $f''=-f$ and having value 0 at $x=0$.

EDIT:

However, as pointed out in the comments, there might be more than one function satisfying this property.

If the function can be written as a power series, we have found it! It must be $f$ with $a_1=1$.

But maybe there is some weird function that can't be written as a power series... so this proof is not quite complete. Can anyone put the last nail in?

I thought that you might want to derive the series without calculus. From angle addition formulas we have $$\sin(n-1)x=\sin nx\cos x-\cos nx\sin x$$ $$\sin(n+1)x=\sin nx\cos x+\cos nx\sin x$$ Adding, we get $$\sin(n+1)x+\sin(n-1)x=2\sin nx\cos x$$ And the key identity $$\sin(n+1)x=2\sin nx\cos x-\sin(n-1)x$$ So we can show that $$\sin2x=2\sin x\cos x-0=2\sin x\cos x$$ $$\sin3x=2(2\sin x\cos x)\cos x-\sin x=4\sin x(1-\sin^2x)-\sin x=3\sin x-4\sin^3x$$ This implies $$\sin x=3\sin\frac x3-4\sin^3\frac x3\tag{1}$$ And for a for a first-quadrant angle $x\le\frac{\pi}2$, $$\sin x\ge\sin\frac x3(4-4\sin^2\frac{\pi}6)=2\sin\frac x3\tag{2}$$ At this point we will try to find the series for $\sin^{-1}x=x+bx^3+cx^5+\dots$. We know the leading term is $x$ and since $x=3\left(\frac x3\right)$ we start from eq $(1)$ with $$\sin x=3\left(\sin\frac x3-\frac43\sin^3\frac x3\right)$$ So far so good. Now for the next term: $$\begin{align}\sin x+b\sin^3x&=3\sin\frac x3-4\sin^3\frac x3+b\left(3\sin\frac x3-4\sin^3\frac x3\right)^3\\ &=3\sin\frac x3+(27b-4)\sin^3\frac x3-108b\sin^5\frac x3+144b\sin^7\frac x3-64b\sin^9\frac x3\\ &=3\left(\sin\frac x3+b\sin^3\frac x3\right)+O\left(\sin^5\frac x3\right)\end{align}$$ Comparing coefficients of $\sin^3\frac x3$ on the last two lines we get $27b-4=3b$ or $b=\frac16$. Then $$\sin x+\frac16\sin^3x=3\sin\frac x3+\frac12\sin^3\frac x3-18\sin^5\frac x3+24\sin^7\frac x3-\frac{32}3\sin^9\frac x3$$ Let's get one more term: $$\begin{align}\sin x+\frac16\sin^3x+c\sin^5x&=3\sin\frac x3+\frac12\sin^3\frac x3-18\sin^5\frac x3+24\sin^7\frac x3\\ &-\frac{32}3\sin^9\frac x3+c\left(3\sin\frac x3-4\sin^3\frac x3\right)^3\\ &=3\sin\frac x3+\frac12\sin^3\frac x3+\left(243c-18\right)\sin^5\frac x3+\left(24-1620c\right)\sin^7\frac x3\\ &+\left(4320c-\frac{32}3\right)\sin^9\frac x3-5760c\sin^{11}\frac x3+3840c\sin^{13}\frac x3-1024c\sin^{15}\frac x3\\ &=3\left(\sin\frac x3+\frac16\sin^3\frac x3+c\sin^5\sin^5\frac x3\right)+O\left(\sin^7\frac x3\right)\end{align}$$ Again comparing coefficients of the $\sin^5\frac x3$ terms in the last two lines we find that $243c-18=3c$, so $c=\frac3{40}$ . Then $$\begin{align}\sin x+\frac16\sin^3x+\frac3{40}\sin^5x&=3\left(\sin\frac x3+\frac16\sin^3\frac x3+\frac3{40}\sin^5\frac x3\right)+\frac{\sin^7\frac x3}{30}\left(-149-184\cos^2\frac x3\right.\\ &\left.-864\cos^4\frac x3+576\cos^6\frac x3-2304\cos^8\frac x3\right)\end{align}$$ As can be seen, given enough patience and accurate algebra we could derive as many terms as we wished of the series for $\sin^{-1}(x)$ in this fashion. Applying inequality $(2)$ $n$ times, we find that $$\sin x\ge2^n\sin\frac x{3^n}$$ For first quadrant angles $x$, also $$0<149+184\cos^2\frac x3+864\cos^4\frac x3-576\cos^6\frac x3+2304\cos^8\frac x3<3501$$ so for such angles, $$\begin{align}-\frac{389}{10}\frac{3^n}{2^{7n}}\sin^7x&\le3^{n-1}\left(\sin\frac x{3^{n-1}}+\frac16\sin^3\frac x{3^{n-1}}+\frac3{40}\sin^5\frac x{3^{n-1}}-\frac x{3^{n-1}}\right)\\ &-3^n\left(\sin\frac x{3^n}+\frac16\sin^3\frac x{3^n}+\frac3{40}\sin^5\frac x{3^n}-\frac x{3^n}\right)\le0\tag{3}\end{align}$$ Since $$\frac{3^n}{2^{7n}}=\frac{\frac{3^n}{2^{7n}}}{1-\frac3{2^7}}-\frac{\frac{3^{n+1}}{2^{7(n+1)}}}{1-\frac3{2^7}}$$ We can sum inequality $(3)$ from $n=1$ to $n=N$ to get $$\begin{align}-\frac{1167}{1250}\sin^7x&\le-\frac{24896}{625}\cdot\frac3{128}\sin^7x+\frac{24896}{625}\cdot\frac{3^{N+1}}{2^{7N+7}}\sin^7x\\ &\le\sin x+\frac16\sin^3x+\frac3{40}\sin^5x-x\\ &-3^N\left(\sin\frac x{3^N}+\frac16\sin^3\frac x{3^N}+\frac3{40}\sin^5\frac x{3^N}-\frac x{3^N}\right)\le0\end{align}$$ Since the $N$-dependent part is $$f(x,N)=x\cdot\frac{\sin\frac x{3^N}+\frac16\sin^3\frac x{3^N}+\frac3{40}\sin^5\frac x{3^N}-\frac x{3^N}}{\frac x{3^N}}$$ And $$\lim_{N\rightarrow\infty}f(x,N)=0$$ For any first quadrant angle $x$, it follows that for any first quadrant angle $x$, $$-\frac{1167}{1250}\sin^7x\le\sin x+\frac16\sin^3x+\frac3{40}\sin^5x-x\le0$$ The knowledge we obtained from that limit above we could have gotten more fundamentally without actually taking a limit, so we have the first $3$ terms of the series for $\sin^{-1}(x)$ from algebra.

Now we want to work backwards to find the series for $\sin x$. $$\begin{align}\sin x&=x+px^3+qx^5+O(x^7)\\ &=\sin^{-1}(\sin x)+p\left(\sin^{-1}(\sin x)\right)^3+q\left(\sin^{-1}(\sin x)\right)^5+O\left(\left(\sin^{-1}(\sin x)\right)^7\right)\\ &=\sin x+\left(p+\frac16\right)\sin^3x+\left(q+\frac p2+\frac3{40}\right)\sin^5x+O(\sin^7x)\end{align}$$ Equating coefficients of like powers of $\sin x$, we find that $$p=-\frac16$$ $$q=\frac p2-\frac3{40}=\frac1{120}$$ Hopefully you can see how to work out the series for $\sin x$ with just algebra and trigonometry. How many terms would you have to derive like this before it would be easier to learn enough calculus to write down the Taylor series directly?