The Extended Unit Circle

A solid understanding of the unit circle makes trigonometry much easier to understand. You should absolutely understand that the East, North, West, and South points (shown in green below) have angle measures that are multiples of $\pi/2$ and their coordinates are $$(1,0), \; (0,1) \; (-1,0) \; \text{ and } \; (0,-1).$$ This makes it very easy to read off the values of the sine and cosine. The trig values of $\pi/6$ and $\pi/3$ can be derived from an equilateral triangle and the trig values of $\pi/4$ can be derived from an isosceles right triangle. Again, an understanding of the unit circle helps you quickly see the trig values of related angles, like $-7\pi/6$.

You can use the interactive version of the unit circle to check values of the standard angles between zero and $2\pi$. You can also hit the "Show more angles" button, if you're curious about some crazier angles.

Show more angles:

Finding the values at the crazier angles

The values shown in the demo above were all computed using Mathematica's FunctionExpand command. You can do this with more angles as well. For example, you can compute $\cos(11\pi/24)$ like so:

FunctionExpand[Cos[11 Pi/24]]
$\displaystyle \frac{1}{4} \sqrt{2-\sqrt{2}} \left(1+\sqrt{2}\right)-\frac{1}{4} \left(\sqrt{2}-1\right) \sqrt{3 \left(2+\sqrt{2}\right)} $

You can even pass the result to TeXForm to generate the TeX commands needed to typeset the expression. That's exactly how the output above was created.

Even crazier angles!

The sine and cosine of any rational multiple of $\pi$ is always an algebraic number, i.e. the root of some polynomial with integer coefficients. For example, $\sin(3\pi/4) = 1/\sqrt{2}$ is a root of $2x^2-1$. But some angles are harder than others. In principle, we can express the sine or cosine of $\pi/n$ as an algebraic number as follows. By de Moivre's formula,

$$(\cos(\pi/n)+i\sin(\pi/n))^n = \cos(\pi)+i\sin(\pi)=-1.$$

Thus, if we expand out the term on the left, set the real part equal to $-1$ and the imaginary part equal to zero, we obtain a pair of equations that $\cos(\pi/n)$ and $\sin(\pi/n)$ must satisfy. If we can solve those, we've got our expressions for the sine and cosine. Even if we can't, we at least have them both expressed as roots of a polynomial system.

For example, here's Mathematica code to find $\sin(\pi/7)$.

n = 7;
expr = ComplexExpand[(Cos[Pi/n] + I*Sin[Pi/n])^n];
subbed = expr /. {Cos[Pi/n] -> x, Sin[Pi/n] -> y};
eqs = {ComplexExpand[Re[subbed]] == -1,
   ComplexExpand[Im[subbed]] == 0};
roots = {x, y} /. Solve[eqs, {x, y}, Reals];
Last[Last[SortBy[roots, N]]]
$\displaystyle \frac{1}{6} \sqrt{\frac{1}{2} \left(42-\frac{14 \sqrt[3]{7}}{\left(\frac{1}{2} \left(-1+3 i \sqrt{3}\right)\right)^{2/3}}-\frac{4\ 7^{2/3}}{\sqrt[3]{\frac{1}{2} \left(-1+3 i \sqrt{3}\right)}}-2\ 2^{2/3} \sqrt[3]{7 \left(-1+3 i \sqrt{3}\right)}-\sqrt[3]{2} \left(7 \left(-1+3 i \sqrt{3}\right)\right)^{2/3}\right)} $