How to evaluate trigonometric functions by pen and paper? How did people determined the values of trigonometric functions before calculators, like e.g. $\sin 37^\circ$ up to five decimal places? Was that possible to find before series were invented?
 A: Before computers, trigonometric functions were something you looked up in tables. The computation of these tables was a lot of work, and often tables were copied and reused for centuries after they were computed, until eventually more precision was needed and someone had to start it all over from scratch.
The basic tools for constructing trigonometric tables from Hipparchus up to the computer age were formulas that allow you to find the values from smaller angles if you already know them for larger ones. In particular, the half-angle formula
$$ \cos \frac{\theta}{2} = \sqrt{\frac{1+\cos\theta}{2}} $$
Then, starting from $\cos 90^\circ=0$ you can successively find the cosines of $45^\circ$, $22^\circ30'$, $11^\circ15'$ and so forth, down to as small an angle as you'd like your precision to be. Then you get the corresponding sines by $\sin\theta = \sqrt{1-\cos^2\theta}$.
From there, you can build up the value for each degree value by using the angle addition formulas -- for example, the values for $67^\circ 30'$ can be gotten by combining the values for $45^\circ$ and $22^\circ30'$, and in this way you can fill in the gaps in your table, to any desired precision, given enough labor.
(The ancients thought in terms of chords rather than sines and cosines, and their algebra was set up slightly differently, but the basic idea is the same).

Doing pure halvings leaves you with your base values being sines of rather clumsy angles. Various tricks to get around this were developed. For example in the 1400s Islamic mathematicians developed a method where you first compute the sine of $3^\circ$ exactly (my source doesn't say how, but possibly by using the angle-difference formulas on the sines and cosines of $72^\circ$ and $60^\circ$, which can be derived by considering pentagrams and equilateral triangles, and then halving twice). From there the sine of $1^\circ$ is found by solving the cubic equation
$$ \sin 3^\circ = 3\sin 1^\circ - 4\sin^3 1^\circ$$
numerically, and then you can build up your trig table in whole-degree steps as before.
A: You could also compute it from the series.  Not that this is better than the accepted answer, but just for completeness.
http://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions
