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. 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 $y$ 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. 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? Oct 11 '18 at 16:47
• You can truncate the power series at some point. In this way you’ll obtain an approximation. 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.