# Secrets of the Mathematical Ninja: Sine and cosine (part II)

So why do some angles give you exact answers and some not? I’m not going to answer that today. Another time. Today, I’m just going to tell you how to remember the key values, the ones that live on your set square.

You can make an argument that the sine and cosine of any whole number of degrees, or any exact fraction of π radians, will give you an exact answer (although it may not be trivial to find). That’s sophistry, though: we’re really only interested in the ones we can easily figure out.

The Mathematical Ninja knows a handful of values for sine and can use them to figure out everything else. The ones you’re meant to know for C3 are: \(\\sin(0^\\circ) = \\sin(0) = 0 = \\frac{0}{2}\) \(\\sin(30^\\circ) = \\sin\\left(\\frac{\\pi}{6}\\right) = \\frac{1}{2} = \\frac{1}{2}\) \(\\sin(45^\\circ) = \\sin\\left(\\frac{\\pi}{4}\\right) = \\frac{1}{\\sqrt{2}} = \\frac{\\sqrt{2}}{2}\) \(\\sin(60^\\circ) = \\sin\\left(\\frac{\\pi}{3}\\right) = \\frac{\\sqrt{3}}{2}\) \(\\sin(90^\\circ) = \\sin\\left(\\frac{\\pi}{2}\\right) = 1 = \\frac{\\sqrt{4}}{2}\)

As you can see, the number under the square root goes up by one for each of the key angles.

For cosine, it’s a similar story, just the other way round: \(\\cos(0^\\circ) = \\sin\\left(\\frac{\\pi}{2}\\right) = 1 = \\frac{\\sqrt{4}}{2}\) \(\\cos(30^\\circ) = \\sin\\left(\\frac{\\pi}{3}\\right) = \\frac{\\sqrt{3}}{2}\) \(\\cos(45^\\circ) = \\sin\\left(\\frac{\\pi}{4}\\right) = \\frac{1}{\\sqrt{2}} = \\frac{\\sqrt{2}}{2}\) \(\\cos(60^\\circ) = \\sin\\left(\\frac{\\pi}{6}\\right) = \\frac{1}{2} = \\frac{1}{2}\) \(\\cos(90^\\circ) = \\sin(0) = 0 = \\frac{0}{2}\)

The answers go down instead of up.

There are a few other angles you can hang onto if you’re so inclined — as I’ve said before, for small angles in radians, $\sin(x)$ is approximately $x$ — even if you look at \frac{\pi}{6}, which works out to be 0.52 or so, the sine is only 4 or 5 percent off.

Also, it’s worth knowing the angles in a 3-4-5 triangle, which are about 37 and 53 degrees, or 0.644 and 0.927 radians. With those, you can say $\sin(37^\circ)$ is about 0.6 without even appearing to think.