# Ask Uncle Colin: Remembering the Unit Circle

Dear Uncle Colin,

How do you remember the values for the trig functions in the unit circle?

- Can’t Abide Stupid Trigonometry

Hi, CAST, and thanks for your message!

The short answer is, by knowing what sine and cosine *mean* on a unit circle, knowing a couple of special triangles, and knowing a bit about symmetry.

### What the functions mean

Mentally draw a radius to any point on a unit circle, and mentally measure an angle from the x-axis, anticlockwise, until you hit the radius. Call that angle $\theta$.

$\sin(\theta)$ is, *by definition*, the $y$-coordinate of that point.

$\cos(\theta)$ is, *by definition*, the $x$-coordinate.

Try a few (mentally). If you pick the point at the top of the circle, $\theta$ is a right angle; $\sin(90º) = 1$ and $\cos(90º) = 0$. ((I’ve picked degrees here because I’m in that sort of a mood. Lay off, OK, sensei?))

Pick the rightmost point, at $(-1,0)$. The angle there is 180º, so $\cos(180º)=-1$ and $\sin(180º) = 0$. The corners of the circle ((you heard me)) are all straightforward to work out.

**The sine and cosine functions pick out the co-ordinates of a point on a unit circle, as a function of its angle from the $x$-axis.** This is important.

### Special triangle number 1

There are two especially useful triangles, the first of which is the isosceles right triangle, with two angles of 45º. Knowing this allows you to work out the values for any angle that’s a multiple of 45º.

The isosceles right-angled triangle has side lengths in the ratio of $1:1:\sqrt{2}$, which you can verify with Pythagoras. That means $\sin(45º) = \cos(45º) = \frac{1}{\sqrt{2}}$, using basic trigonometry.

So, if you put a point on the circle at 45º anticlockwise from the x-axis, its coordinates are $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ ((No, I’m not going to rationalise. Rationalising is irrational.))

If you think about reflecting the circle across the $y$-axis, this point will become the point at 135º, with coordinates of $\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$, which gives you $\cos(135º) = -\frac{1}{\sqrt{2}}$ and $\sin(135º)= \frac{1}{\sqrt{2}}$.

You can continue around the circle to find the values for 225º and 315º.

### Special triangle number 2

The second special triangle is an equilateral triangle cut in half (I usually imagine the base horizontal, and slice straight up the middle vertically). Then the angles are 60º (at the bottom), 30º (at the top) and 90º (in the corner). The side from the 60º corner to the right angle is half the distance from the 60º to the top (because we’ve cut an equilateral triangle in half), and with a bit of Pythagoras, it becomes clear that the side lengths are in the ratio $1 : \sqrt{3} : 2$, with the longer leg opposite 60º.

That means that $\cos(60º) = \frac{1}{2}$ and $\sin(60º) = \frac{\sqrt{3}}{2}$ - I find that remembering $\sqrt{3} > 1$ is helpful for not muddling these up. Whenever I have to figure out (say) $\cos(300º)$, I imagine where I’d put the triangle to put my point in the right place. For 300º, the short base goes on the positive $x$-axis and the ‘top’ is directly below it. That means $\cos(300º) = \frac{1}{2}$ and $\sin(300º) = -\frac{\sqrt{3}}{2}$.

If I wanted 330º instead, I’d need to put the base vertically on the negative $y$-axis and point the top to the right, so $\cos(330º) = \frac{\sqrt{3}}{2}$ and $\sin(330º) = -\frac{1}{2}$.

I hope that helps you start to get the ideas, CAST!

- Uncle Colin