How to find inverse of sine without using calculator, example like sin^-1(4/7).......etc On this website i learnt how to find inverse cosine without using calculator with estimating formulated but the problem was it was for only cosine function but I want for sine and tangent as well 
Please can anyone introduce or share those estimating formulated 
I have been searching these formulated from long time
I appreciate that guy who explained everyone how to find inverse function for any given value.......
But Iit will be too great to me if someone also share this types of estimating formulaes.........
I think that guy was Grey Matters 
Thank you please try to share soon.....    : |
 A: If you have a compass, ruler and a protractor you can draw a unit circle then draw line y=x. The line y=x will intersect unit circle at one or two points. Connect those points with the origin and measure the angle formed by axis x and this line(s). 
A: In general the most accurate way to determine is to represent it by the cyclometric function value, eg.
$$\alpha = \arcsin \frac{4}{7}$$
For the angles that are combinations of finite number of angles $2^k30^{\circ}$ and $2^k90^{\circ}$ ($k \in \Bbb{Z}\backslash\{0\})$ there are are also methods using the trigonometric identities:
$$\sin\alpha\pm\beta = \sin\alpha\cos\beta\pm\sin\beta\cos\alpha\\
\sin\alpha\pm\beta =\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$$
and
$$\sin\frac{\alpha}{2}=\sqrt{\frac{1-\cos\alpha}{2}}\\
\cos\frac{\alpha}{2}=\sqrt{\frac{1+\cos\alpha}{2}}$$
So if we have sine value equal to $0,\frac{1}{2},\frac{\sqrt2}{2},\frac{\sqrt3}{2},1$ the argument value is trivial. 
In other cases you have to find out if the sine value is obtained with the above formulas. The difficulty increase with increasing number of elements in your combination and with the absolute value of $k$.
For example (only 2 elements, $\max|k|=1$):
$$\frac{\sqrt2+\sqrt6}{4}=\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}+\frac{1}{2}\frac{\sqrt2}{2}=\sin\frac{\pi}{4}\cos\frac{\pi}{6}+\sin\frac{\pi}{6}\cos\frac{\pi}{4}=\sin\frac{\pi}{4}+\frac{\pi}{6}=\sin\frac{5\pi}{12}$$
$$\arcsin \frac{\sqrt2+\sqrt6}{4} = \frac{5\pi}{12}$$
A: You can't render $\sin^{-1}(a)$ "nicely" for most rational values of $a$.  Except for the obvious cases $a\in \{0, \pm (1/2), \pm 1\}$ the inverse sine is not an algebraic number or an algebraic number times $\pi$.
For roots of certain special polynomial equations you can get the inverse sine values that are rational numbers times $\pi$.  Below are the cases involving quadratic roots:
$\sin^{-1}(\sqrt{2}/2)=\pi/4$, plus additive inverse
$\sin^{-1}(\sqrt{3}/2)=\pi/3$, plus additive inverse
$\sin^{-1}((\sqrt{5}-1)/4)=\pi/10$, plus additive inverse
$\sin^{-1}((\sqrt{5}+1)/4)=(3\pi)/10$, plus additive inverse
