<s>[INST] How does WolframAlpha simplify sine and cosine? When I feed WolframAlpha an expression like $\sin({\pi\frac{2}{3}})$, it correctly prints that this is equal to $\frac{\sqrt3}{2}$, instead of the decimal expansion $0.866025403\ldots$.
Perhaps it has a lookup table for common fractions of $\pi$. Or is it likely to be more sophisticated? Does it solve the convergence of the series expansion for $\sin(x)$?
 [/INST] It must be that $\sin\left(\dfrac{p}{q}\pi\right)$ is algebraic. To see why check out this question.
I am almost certain that W|A doesn't use power series unless the value is very small simply because it would be too slow to calculate the value of arbitrary trig functions using power series. It is more likely that there is a certain class of rational numbers such as $\frac{1}{3}$ where the forms of $\sin(\frac{\pi}{3})$ and $\cos(\frac{\pi}{3})$ are known and then formulae such as the double angle formula gives results for other rational numbers such as $\sin({\frac{2\pi}{3}})$. The result is then simplified and sent to the user.
This is only a conjecture as I do not have access to any Mathematica source code.
Here is an example. Suppose you know that $$\sin\left(\dfrac{\pi}{2}\right) = 1 \quad\text{and}\quad \cos\left(\dfrac{\pi}{2}\right) = 0.$$
We know that $$\sin^2(\theta) = \dfrac{1-\cos(2\theta)}{2}.$$
It must be that:
$$\sin^2\left(\dfrac{\pi}{4}\right) = \dfrac{1}{2} \quad\text{and}\quad \cos^2\left(\dfrac{\pi}{4}\right) = 1-\dfrac{1}{2} = \dfrac{1}{2}$$
$$\sin\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2} \quad\text{and}\quad \cos\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}.$$
Continuing...
$$\sin\left(\dfrac{\pi}{8}\right) = \dfrac{\sqrt{{2-\sqrt{2}}}}{2} \quad\text{and}\quad \cos\left(\dfrac{\pi}{8}\right) = \dfrac{\sqrt{{2+\sqrt{2}}}}{2}$$
$$\sin\left(\dfrac{\pi}{16}\right) = \dfrac{\sqrt{2-\sqrt{{2+\sqrt{2}}}}}{2} \quad\text{and}\quad \cos\left(\dfrac{\pi}{16}\right) = \dfrac{\sqrt{2+\sqrt{{2+\sqrt{2}}}}}{2} $$
Which agrees with W|A.
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