Alternative expansion for the sine function

Question

Are there other polynomials expansions other than the Taylor/Maclaurin one? And other non-polynomial expansion? [for the $\sin(x)$ function]

Answer 1

Better than Taylor series, there are Pade approximants which, around a point $x=a$ approximate a function by $$f(x)=\frac{\sum_{i=0}^m a_i (x-a)^i}{1+\sum_{i=1}^n b_i (x-a)^i}$$ For example, for $\sin(x)$ around $x=0$, you would have

$$\sin(x)=\frac{x-\frac{7 }{60}x^3 } {1+\frac{1}{20}x^2 }\tag 1$$ $$\sin(x)=\frac{x-\frac{53 }{396}x^3+\frac{551 }{166320}x^5} {1+\frac{13 }{396}x^2+\frac{5 }{11088}x^4}\tag 2$$

If you plot them for $0 \leq x \leq \frac \pi 2$, you will not see almost any difference with $\sin(x)$. For example, using $(1)$, the maximum error is $0.0042271$ and using $(2)$, the maximum error is $0.0000030$.

Using them, you could have a good estimate of the first root of the equation $\sin(x)=k x$ with $(k <1)$ at the price of a linear or quadratic equation in $x^2$. For example, using $(1)$, the approximate solution would be $$x^2=\frac{60 (1-k)}{3 k+7}$$ Using $k=\frac 12$, this would give $x=2 \sqrt{\frac{15}{17}}\approx 1.87867$ while the exact solution would be $\approx 1.89549$. Using $(2)$, the approximate solution would be $\approx 1.89552$.

Edit

Even if the expression corresponds to polynomials, you can build quite accurate approximation (for $0\leq x \leq \pi$) such as $$\sin(x)=\sum_{i=1}^n a_i y^i\qquad \text{using} \qquad y=(\pi-x) x$$ (have a look here).

Using again PadeApproximants this would give for example $$\sin(x)=\frac{\frac{1}{\pi }y+\frac{\left(12-\pi ^2\right) }{\pi ^3 \left(\pi ^2-6\right)}y^2 } {1-\frac{2 \left(\pi ^2-9\right) }{\pi ^2 \left(\pi ^2-6\right)}y+\frac{\left(-36-6 \pi ^2+\pi ^4\right) }{6 \pi ^4 \left(\pi ^2-6\right)}y^2 }$$