How to calc arc sine without a calculator?

Solution 1:

To compute $\arcsin(x)$ we might need to use a bit of calculus.

Note that $$ \arcsin(x)=\int_0^x\frac{\mathrm{d}t}{\sqrt{1-t^2}} $$ Using the binomial theorem, we get that $$ \frac1{\sqrt{1-x^2}}=\sum_{k=0}^\infty\binom{2k}{k}\left(\frac{x}{2}\right)^{2k} $$ Integrating term by term, we get $$ \arcsin(x)=\sum_{k=0}^\infty\frac2{2k+1}\binom{2k}{k}\left(\frac{x}{2}\right)^{2k+1} $$

Iterative Method Requiring Square Roots

We can use the identity $$ \begin{align} \sin^2(x/2) &=\frac{1-\sqrt{1-\sin^2(x)}}{2}\\[6pt] &=\frac{\sin^2(x)}{2+2\sqrt{1-\sin^2(x)}} \end{align} $$ and the limit $$ \lim_{n\to\infty}2^n\sin(x/2^n)=x $$ to compute $x$ from $\sin(x)$.

Solution 2:

$\arcsin $ is defined to be the inverse of $\sin$ but restricted to a certain range. Hence $\arcsin(\sin(x))=x$ if $x$ is within this range (generally either $0$ to $2\pi$ or $-\pi$ to $\pi$) or a value $y$ such that $\sin(y)=\sin(x)$ i.e. $y=x+2\pi n$ or $y=\pi -x +2\pi m$ for some $n\in \mathbb{Z}$ or $m\in \mathbb{Z}$ and $y$ is in this range.