Evaluating Sine and Cosine at Special Acute Angles - Problem 2

Another special angle is pi over 6. I want to find the cosine and sine of pi over 6. I’ve drawn the angle pi over 6 on my unit circle and I’ve drawn the reflection of point p across the x axis. Remember pi over 6 is 30 degrees so angle POQ will be 60 degrees and by symmetry this is going to be an equilateral triangle. Not hat means all three sides are going to be the same length.

Now this side has length 2y, so all three sides have length 2y and that 2y equals 1. The radius of the circle is 1, but the radius also has to have length 2y. So 2y equals 1 and that means y equals ½. This is precisely the sine of pi over 6. The second coordinate gives you the sine of the angle, so sine pi over 6 is ½.

To get the cosine I have to remember that this point is on the unit circle so x² plus y² equals 1. Y is ½ so I substitute and ½² is ¼ so I subtract 1/4 from both sides, I get ¾ and then I take square roots. It’s plus or minus root 3 over 2. But of course I am dealing with a point in the first quadrant so the x coordinate should be positive and I get x equals root 3 over 2.

X is the cosine of pi over 6 so cosine of pi over 6 is root 3 over 2. Sine of pi over 6 is ½, cosine of pi over 6 is root 3 over 2.