Understanding why the Maclaurin series for $\sin \theta$ converges to $\sin \theta$ and not something else

Question

Question: Why does

$$ \sin\theta = \sum_{n=0}^{\infty} (-1)^n \frac{\theta^{2n+1}}{(2n+1)!} $$

converge to $\sin \theta$?

What I understand:

I understand that this formula comes from viewing $\sin \theta$ as a Maclaurin series.

I understand that this is the case because $\sin \theta$ can be viewed as a Taylor expansion:

$$ \sin x = \sum_{k=0}^{n} \frac{\sin^{(k)}(x)(x-0)^k}{k!} + \frac{\sin^{(n+1)}(c)(x-0)^n}{(n+1)!} $$

Moreover, I understand that since the derivative of $\sin \theta$ is at most $1$, that the error terms converge to $0$ (so that this series converges).

What I don't understand is why this series converges to the desired target, $\sin x$. Why is this the case?

NOTE: Other explanations on this site seem to only explain the steps I outlined above, but not why the Maclaurin series actually converges to $\sin \theta$ (unless I'm missing something).

Answer 1

The thing that you do not tell us is: how do you define the function $\sin$? Most mathematicians define it precisely as the function $x \mapsto \sum \limits _{n=0} ^\infty (-1)^n \frac {x ^{2n+1}} {(2n+1)!}$, in which case your question has a trivial answer.

Answer 2

Part 1. For $n\geq 1$ we have $$ (\bullet ) \quad\sin y-\sin x=(n+1)!^{-1}(y-x)^{n+1}\sin^{(n+1)}c_n+\sum_{j=1}^nj!^{-1}(y-x)^j\sin^{(j)}x$$ where $c_n \in [x,y]\cup [y,x].$ (See footnote.) For $y=0$ this gives $$-\sin x=(n+1)!^{-1}(-x)^{n+1}\sin^{(n+1)}c_n+\sum_{j=1}^nj!^{-1}(-x)^{j}\sin^{(j)}x$$ $$ \text {where }\quad c_n\in [0,x]\cup [x,0].$$ $$\text {Therefore }\quad\sin x=(n+1)!^{-1}(-x)^{n+1}\sin^{(n+1)}c_n+\sum_{j=1}^nj!^{-1}(-x)^{j-1}\sin^{(j)}x=$$ $$= R_n(x)+S_n(x)$$ $$ \text {where }\quad R_n(x)=(n+1)!^{-1}(-x)^{n+1}\sin^{(n+1)}c_n.$$ Since $R_n(x)\to 0$ as $n\to \infty$, we have $S_n(x)\to \sin x$ as $n\to \infty.$ That is, $$\sin x=\lim_{n\to \infty}S_n(x)=\sum_{j=1}^{\infty}j!^{-1}(-x)^{j-1}\sin^{(j)}x.$$

Part 2. When $n\geq 1,$ and $x=0$ in $(\bullet)$ we have $$\sin y=\sin y-\sin 0=(n+1)!^{-1}y^{n+1}\sin^{(n+1)}c_n+\sum_{j=1}^nj!^{-1}y^j\sin^{(j)}(0)=T_n(y)+U_n(y)$$ $$\text { where } \quad T_n(x)=(n+1)!^{-1}y^{n+1}\sin^{(n+1)}c_n.$$ Since $T_n(y)\to 0 $ as $n\to \infty,$ and since $\sin^{(j)}(0)=0$ when $j$ is even , while $\sin^{(j)}(0)=(-1)^{(j-1)/2}$ when $j$ is odd, we have $$\sin y=\lim_{j\to\infty}U_n(y)=\sum_{k=0}^{\infty}(2k+1)!^{-1}(-1)^k y^{2k+1}.$$

Footnote. $c_n\in [x,y]\cup [y,x]$ is just a convenient easy of saying that $c_n$ belongs to the closed interval from x to y.

Remarks. (1).It does not make sense to write $\sin x=R_n(x)+\sum_{n=1}^{\infty}B_n(x).$..... (2). There is an error in the summation for $\sin x$ in your Q.