Using the multitude of formulas surrounding the trigonometric functions, mathematicians have been able to find exact values for many arguments of the trigonometric functions. For example, $$\sin \frac{\pi}{12}=\frac{\sqrt 3-1}{2 \sqrt 2}$$ In fact, it is even the possible to find the value of the sine of one degree. However, my question is about how to calculate exact values for stranger values of the sine function involving "irrational amounts of $\pi$", such as $$\sin(\pi \sqrt 2)$$ or $$\sin(\pi e)$$ does anybody know how to calculate either of these, or how I should go about doing it? Many of the currently-used trigonometric identities will become useless if one attempts to use them for irrational numbers, since they involve sums and products, and there are no "special values" like this from which to start.

Any ideas?

Edit: What I mean by "exact value" is a value expressed not in terms of the trigonometric functions, or not as a Riemann Sum or Product; something closed-form that isn't circular.

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    $\begingroup$ its exact value is $\sin (\pi \sqrt 2)$ $\endgroup$ – windircurse Jul 8 '17 at 14:40
  • $\begingroup$ In a word ... No. $\endgroup$ – Oscar Lanzi Jul 8 '17 at 14:43
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    $\begingroup$ The sine of rational multiples of $\pi$ can be evaluated by resolution of rational equations of a degree related to the value of the denominator. This doesn't hold for irrational multiples. $\endgroup$ – Yves Daoust Jul 8 '17 at 14:43
  • $\begingroup$ $\sin(\pi\sqrt 2)$ is a symbol just like $\log 2$ or $\sqrt{e}$ when you need to USE the number then you need a rational approximation, like in computer programming, where "real" variables are obviously rational approximations. $-0.9639$ or $-0.96390253284987733029$ according to the need of the user $\endgroup$ – Raffaele Jul 8 '17 at 15:32
  • $\begingroup$ @Raffaele The OP probably meant that they want $\sin(\pi\sqrt2)$ in terms of algebraic operations, like $\sqrt{~}$ and $\sqrt[3]{~}$. $\endgroup$ – Simply Beautiful Art Jul 8 '17 at 15:35

One cannot derive values of trig functions for non-rational multiples of $\pi$. The approach for any rational multiple of $\pi$ is the same though:

$$x=\cos\left(\frac ab\pi\right),~\gcd(a,b)=1$$

$$\cos(a\pi)=\frac b2\sum_{k=0}^{\lfloor b/2\rfloor}\frac{(-1)^k}{b-k}\binom{b-k}k(2x)^{b-2k}$$

Thus, the cosine and sine of any rational multiple of $\pi$ may be algebraically evaluated from a polynomial, since we know the values of $\cos(a\pi)$ for natural $a$.

On the other hand, if $\theta$ is not a rational multiple of $\pi$, then we cannot evaluate it in a closed form, unless you allow things such as


Furthermore, it can be shown that $\cos(\theta)$ and $\sin(\theta)$ must either both be transcendental (so nothing pretty like $\sqrt{3/5}$), or $\theta$ must be a transcendental non-rational multiple of $\pi$ (of the form $\theta=\arcsin(x)$ for an algebraic $x$). This follows directly from Baker's theorem and the Lindemann–Weierstrass theorem, that $\cos(\sqrt2)$ and values of $\sin(a)$ for algebraic $a$, must be algebraically independent of $\sin(b)$ for another algebraic $b$, following from the complex forms of trig functions.

Particularly, this means that:

$$x=\frac{\cos(\sqrt2)}{\cos(\sqrt3)}\implies x\text{ is transcendental}$$

So even if we knew a few values, we'd be unable to derive much from them. Likewise, $e^{i\theta}$ is transcendental for non-zero algebraic $\theta$, so $\cos(\sqrt2)$ and $\cos(\sqrt3)$ are both transcendental.

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    $\begingroup$ Nice! Thanks for the answer! :) $\endgroup$ – Frpzzd Jul 8 '17 at 15:08
  • $\begingroup$ @MarkViola $x$ is defined by the first line, and $b$ is the denominator in $\frac ab$. $\endgroup$ – Simply Beautiful Art Jul 8 '17 at 15:30
  • $\begingroup$ Ah. And so $x^2\le 1\implies 1-x^{-2}\le 0$ So, the square root is purely imaginary then. $\endgroup$ – Mark Viola Jul 8 '17 at 15:33
  • $\begingroup$ @MarkViola :P That's besides the point. But yeah, that occurs because for large enough $b$, the algebraic forms of $\cos(a\pi/b)$ must necessarily involve complex numbers (most of the time) (casus irreducibilis) $\endgroup$ – Simply Beautiful Art Jul 8 '17 at 15:34
  • $\begingroup$ But the left-hand side is purely real, while the right-hand side is complex. $\endgroup$ – Mark Viola Jul 8 '17 at 15:36

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