Elementary proof that the MacLaurin series of $\sin x$ converges to $\sin x$ for all $x$
By John Thompson
In my book it is given:
$\sin x = x- \dfrac {x^3}{3!}+\dfrac{x^5}{5!}- \dfrac{x^7}{7!}...$
I googled around for a proof but couldn't understand any of them. I would like to know if there's any elementary high school level proof the series
$\endgroup$ 55 Answers
$\begingroup$A proper proof would require ideas from analysis/calculus and you should get familiar with these (if not already). The typical proof is via Taylor's theorem. A non-rigorous version uses the formula $$\sin nx=n\sin x-\frac{n(n^2-1^2)}{3!}\sin^3x+\frac{n(n^2-1^2)(n^2-3^2)}{5!}\sin^5x-\cdots$$ (convince yourself that the above formula is true by trying out odd values of $n$ like $3,5$) and then puts $nx=t$ where $t$ is constant, $n\to\infty, x\to 0$ to get $$\sin t=t-\frac{t^3}{3!}+\dots$$
Newton used another procedure to obtain infinite series for $\sin x$. Using the geometric definition of $\sin x$ (as mentioned in your comment) one obtains $$x=\int_{0}^{\sin x}\frac{dt} {\sqrt{1-t^2}}$$ for $x\in[-\pi/2,\pi/2]$ and expanding the integrand via binomial theorem and integrating term by term we get $$x=\sin x+\frac{1}{2}\frac{\sin^3x}{3}+\frac{1\cdot 3}{2\cdot 4}\frac{\sin^5x}{5}+\dots$$ and then assuming $$\sin x=ax+bx^3+cx^5+\dots$$ and comparing coefficients we can easily find $a, b, c, d$.
$\endgroup$ 2 $\begingroup$Well the expansion above is merely applying Taylor's Theorem to the function $\sin(x)$.
If you want to rigorously understand this expansion, you will most likely just have to look at a proof of Taylor's theorem, which involves creating a better approximation for any function $f(x)$, by continually adding terms which mitigate the error of the polynomial approximation.
$\endgroup$ $\begingroup$$$f(x)=e^x$$
$$f'(x)=e^x \tag 1$$ (I assumed you know this property of $e^x$. ) $$f''(x)=f'(x)=e^x$$ $$f^{(n)}(x)=e^x$$
$$f^{(n)}(0)=1$$
If we find the Taylor series for a function $f(x)$ is:
$$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)x^n}{n!}=\sum_{n=0}^{\infty} \frac{x^n}{n!}$$
$$f(ix)=e^{ix}=\sum_{n=0}^{\infty} \frac{i^nx^n}{n!}$$
$$i=\sqrt{-1}$$ $$i^2=-1$$ $$i^3=-i$$ $$i^4=1$$ $$i^5=i$$ . .
$$e^{ix}=1+\frac{ix}{1!}+\frac{i^2x^2}{2!}+\frac{i^3x^3}{3!}+\frac{i^4x^4}{4!}+\frac{i^5x^5}{5!}+\frac{i^6x^6}{6!}+\frac{i^7x^7}{7!}+\frac{i^8x^8}{8!}+......$$
$$e^{ix}=1+\frac{ix}{1!}+\frac{-x^2}{2!}+\frac{-ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}+\frac{-x^6}{6!}+\frac{-ix^7}{7!}+\frac{x^8}{8!}+......$$
$$e^{ix}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+..... +i(\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+......) \tag2$$
This is Euler Formula: link$$e^{ix}=\cos x +i \sin x \tag 3$$
Take both side derivative Equation $2$
$$(e^{ix})'=(-\frac{x}{1!}+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}+......) +i(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+.....) $$
$$(e^{ix})'=-\sin x +i\cos x \tag 4 $$
$$(e^{ix})'=\cos 'x +i \sin' x \tag 5 $$ ///If we want equal (4) and (5)
$$ \cos 'x +i \sin' x =-\sin x +i\cos x$$
we equal imaginary and real parts separately, we will get
$$\cos 'x= - \sin x$$
$$\sin 'x= \cos x$$
They are trigonometric function properties. link
$$\cos x=\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}$$ $$\sin x=\sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$$
$\endgroup$ $\begingroup$Initially, I was looking for a simple proof that doesn't use the Taylor theorem but according to the answers I have received so far, I don't think its possible (except the answer by @Paramanand).
Finally, I found this video very helpful. It's exactly what I was looking for. Now I have understood how to derive it using the Taylor Theorem.
$\endgroup$ $\begingroup$The Taylor series for a function $f(x)$ is: $$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}.$$ We need to calculate $\sin^{(n)}(0)$ for all $n.$ When $n \equiv 0\mod{2},$ $$\sin^{(n)}(0)=0.$$ When $n \equiv 1\mod{4},$ $$\sin^{(n)}(0)=1.$$ When $n \equiv 3\mod{4},$ $$\sin^{(n)}(0)=-1.$$ Using these values, all the even terms disappear from the series, obtaining the formula: $$\sin x=\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}.$$
$\endgroup$ 2![How can I play the content from another account on my new Xbox 360 console? [duplicate]](https://edu.smbbmcl.edu.pk/uploads/how-can-i-play-the-content-from-another-account-on-my-new-xbox-360-console-duplicate_720.jpg)
