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Whenever I have to calculate the value of a given trigonometric function for an angle, I always refer to a table similar to this:

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But what if I want to find the value for sin$\theta$, where $\theta$ = 32$^{\circ}$ or $\theta$ = 49$^{\circ}$ or for any function, for that matter.

hjpotter92
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Always Learning Forever
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    Here is a website on the matter. She discusses two methods: the first is using radians and the infinite series representation of $\sin{x}$, and the second is using degrees and the various trig identites: http://www.marypat.org/stuff/nylife/010206.html – layman Aug 18 '14 at 03:57
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    In the case of rational multiples of $\pi$, there is a rather tedious procedure for getting exact results using trig identities and the special angles. For example, one can find $\sin(\pi/12)$ by using the half-angle formula with $\theta=\pi/6$. It is a lot harder with many angles, however. – Ian Aug 18 '14 at 04:45
  • @Ian It should probably be pointed out that there is only such a procedure for some rational multiplies of $\pi$, specifically the rational multiplies corresponding to increments of $3^\circ$ (and repeated bisection of these angles). But it can't be used to calculate, say, sine of $1^\circ=\frac{\pi}{180}$ and $2^\circ=\frac{\pi}{90}$ – David H Aug 18 '14 at 05:01
  • You're right. However, you can get a series expansion with the same technique. For example, $\pi/180 = \pi/60 \sum_{k=1}^\infty 4^{-k}$ generates a procedure for getting a series representation of $\sin(\pi/180)$. – Ian Aug 18 '14 at 05:19

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  • I think it could be a good idea you switch from degrees to radians.

  • You can find more detailed tables than the one you put in your post

  • Now, suppose than you need the value of $\sin(\frac{32 \pi}{180})$ (which is your $32$ degrees). One of the ways is to use series centered at a point where you know the values. For this specific case, let use write $$\sin(\frac{32 \pi}{180})=\sin(\frac{30 \pi}{180}+\frac{2 \pi}{180})=\sin(\frac{ \pi}{6}+\frac{\pi}{90})$$ Now, we shall consider the development of $\sin(a+x)$ built at $x=0$ $$\sin(a+x)=\sin (a)+x \cos (a)-\frac{1}{2} x^2 \sin (a)-\frac{1}{6} x^3 \cos (a)+\frac{1}{24} x^4 \sin (a)+O\left(x^5\right)$$ in which you know the values of $\sin(a)$ and $\cos(a)$. For $a=\frac{\pi}{6}$, this just write $$\sin(\frac{ \pi}{6}+x)=\frac{1}{2}+\frac{\sqrt{3} x}{2}-\frac{x^2}{4}-\frac{x^3}{4 \sqrt{3}}+\frac{x^4}{48}+O\left(x^5\right)$$ Now, replace $x=\frac{ \pi}{90}$ and compute; you would obtain a value of $0.529919263860$ while the exact number is $0.529919264233$. For sure, if you need less accuracy, use less terms.

Claude Leibovici
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