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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?

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    $\begingroup$ My friend, that is the definition of sinx. $\endgroup$ – John Mitchell Oct 11 '18 at 16:36
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    $\begingroup$ @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 $y$ $\endgroup$ – Andrew Gazelka Oct 11 '18 at 16:40
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    $\begingroup$ That’s the geometric meaning of sinx, not the definition. The modern definition of sinx is precisely the power series you’ve given. $\endgroup$ – John Mitchell Oct 11 '18 at 16:43
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    $\begingroup$ @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? $\endgroup$ – Andrew Gazelka Oct 11 '18 at 16:47
  • $\begingroup$ You can truncate the power series at some point. In this way you’ll obtain an approximation. $\endgroup$ – John Mitchell Oct 11 '18 at 16:49
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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.

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