# Approximate Trig Functions without the use of Taylor Series

I am familiar with how a trig function, i.e. $$\sin(x)$$, can be approximated by a MacLauren series;

\begin{align} \sin(x_0) &\approx \sin(0) + \cos(0) x_0 - \frac{1}{2}\sin(0) x_0^2 - \frac{1}{3!}\cos(0) x_0^3 + \dots \\&= x_0 - \frac{1}{6} x_0^3 +\dots. \end{align} However, this makes me wonder if there is any other way to approximate trig functions. I would be surprised if a mathematical approximation did not exist until after the advent of calculus. Is there any way to approximate trig functions without the use of a Taylor Series?

• My friend, that is the definition of sinx. – John Mitchell Oct 11 '18 at 16:36
• @JohnMitchell isn't that more of something derived from the definition of $\sin x$? I would think the more fundamental definition would be: given $\theta$ radians, $\sin(\theta) = y/\sqrt{x^2+y^2}$ for a right triangle with base length $x$ and height $y$ – Andrew Gazelka Oct 11 '18 at 16:40
• That’s the geometric meaning of sinx, not the definition. The modern definition of sinx is precisely the power series you’ve given. – John Mitchell Oct 11 '18 at 16:43
• @JohnMitchell Trigonometry has been used for thousands of years without calculus having been discovered and mathematical work being primarily geometric. Surely there is another way to approximate trig functions? – Andrew Gazelka Oct 11 '18 at 16:47
• You can truncate the power series at some point. In this way you’ll obtain an approximation. – John Mitchell Oct 11 '18 at 16:49

A special case of the angle-sum formulas $$\cos (x+y)=\cos x \sin y -\sin x \cos y$$ and $$\sin (x+y)=\sin x \cos y + \cos x \sin y$$ (which is for all $$x,y$$), when $$x=y\in (0,\pi /2)$$ by elementary geometry:

In $$\triangle ABC$$ with $$BA=CA=1 ,$$ with $$D$$ being the mid-point of $$BC$$, and $$\angle BAD=x,$$ we have $$BD =\sin x.$$ And the Cosine Formula gives $$BD^2=\frac {1}{4}BC^2=$$ $$=\frac {1}{4}(BA^2+CA^2-2BA\cdot CA \cos \angle BAC)=$$ $$=\frac {1}{2}(1-\cos 2x).$$ Therefore $$\sin^2 x =\frac {1-\cos 2x}{2}.$$

So when $$z=2x\in (0,\pi)$$ we have the half-angle formulas $$\sin z/2=\sqrt {(1-\cos z)/2}$$ and $$\cos z/2=\sqrt {1-\sin^2 z/2}=\sqrt {(1+\cos z)/2}.$$

The Cosine Formula is a direct consequence of "Pythagoras". And $$\sin^2 z/2+\cos^2 z/2=1$$ IS "Pythagoras".

Between 21 and 22 centuries ago Archimedes used this to obtain the sin , cos, and tan of $$\pi/(3\cdot 2^n)$$ for $$n=1,2,3,4$$ starting with $$\cos \pi /6=\sqrt 3\;/2,$$ obtaining $$3+\frac {10}{71}<48\sin \pi/48<\pi <48 \tan \pi/48<3+\frac {1}{7}.$$

The general angle-sum formulas can be proved by elementary geometry. There is a very simple proof in the old classic Trigonometry, by Hobson (likely still available from Dover Publications, still a great source for cheap re-prints).

With the general angle-sum formulas, and knowing $$\sin \pi /3$$ and $$\sin \pi/4 ,$$ we can use the half-angle formulas to compute $$\sin (\,A\pi/(3\cdot 2^m)+B\pi/(4\cdot 2^n)\,)$$ for any $$A,B\in \Bbb Z$$ and any $$m,n \in \Bbb Z^+\cup \{0\},$$ which can be as close to any $$\sin x$$ as we like.

For if $$x=x'+d \;$$ then $$|\sin x-\sin x'|=$$ $$=|(\sin x'\cos d+\cos x'\sin d)-\sin x'|=$$ $$=|(\sin x')(-1+\cos d)+\cos x' \sin d|=$$ $$=|(\sin x')(-2\sin^2 d/2)+\cos x' \sin d| \leq$$ $$\leq |\sin x'|\cdot d^2/2+|\cos x'|\cdot |\sin d|\;\leq\; d^2/2+|d|.$$

Also, one of the Comments recommends the wiki article History Of trigonometry.