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This document is intended to help you reproduce the unit circle on the fly without having to memorize loads and loads of trig values. Here is a summary of the steps we will perform:
- Step 1: Construct the unit circle for sin θ, where θ ranges between 0 and π 2 radians.
- Step 2: Complete the unit circle for sin θ and values of θ between π 2 and π.
- Step 3: Fill out the unit circle for sin θ by using a simple flip of steps 1 and 2.
- Step 4: Complete the unit circle for cos θ by rotating your sin θ circle by 90 degrees.
You can then use the unit circle for sin θ and cos θ to derive, for example, values of tan θ = (^) cossin^ θθ.
Before starting, recall that the x-coordinate on the unit circle corresponds to cos θ, and the y- coordinate represents sin θ. So, because we’re first filling out the sine values, we’re actually filling out all of the y-coordinates before we fill out the x-coordinates.
- Step 1: Start out by drawing a circle, and labelling the five angles in the first quadrant, in increasing order: these values (in order, measured in radians) are θ = 0, π 6 , π 4 , π 3 , and π 2. The equivalent values in degrees are θ = 0, 30 , 45 , 60 , 90.
π 6
π 4
π 3
π 2
The values of sin θ in the first quadrant have a nice symmetry about them if you write them in this form:
θ 0 π 6 π 4 π 3 π 2 sin θ
√ 0 2
√ 1 2
√ 2 2
√ 3 2
√ 4 2
When written in this fashion, all of them have a 2 in the denominator, a square root in the numerator, and the numbers inside the roots just count up monotonically from 0 to 4. Of course, some of these simplify:
4 = 2, so the sine values reduce to the following:
θ 0 π 6 π 4 π 3 π 2 sin θ (^0 )
√ 2 2
√ 3 2 1
Let’s include these in the circle as the second coordinate:
π 6
π 4
π 3
π 2 ( , (^12)
√ 2 2
√ 3 2
π 6
π 4
π 3
π 2 2 π 3 3 π 4 5 π 6
π
√ 2 2
√ 3 2
√ 2 2
√ 3 2
- Step 3: Let’s finish up the sine coordinates for θ between π and 2π. To get the angles, just add π radians to each angle starting from zero (i.e. 0 + π, π 6 + π = 76 π , π 4 + π = 54 π , etc.) while going around counterclockwise:
270 ◦^
225 ◦^315 ◦
π 6
π 4
π 3
π 2 2 π 3 3 π 4 5 π 6
π
7 π 6 5 π 4 4 π 3 3 π 2
5 π 3
7 π 4
11 π 6
√ 2 2
√ 3 2
√ 2 2
√ 3 2
In order to get the y-coordinates on the bottom half of the unit circle, you can just take the y-coordinate of the point directly above the point and negate it. For example, at θ = 53 π , the point
directly above it corresponds to the angle π 3 which has y-coordinate
√ 3
- Thus the^ y-coordinate at θ = 53 π is equal to −
√ 3
- The bottom line is that this step amounts to flipping the circle upside down to get the lower half and negating each of the y-coordinates. Filling these in gives the following:
your starting θ. For example, if you start at θ = 43 π , rotating counterclockwise by 90 degrees lands us at 116 π , for which the sine value is −^12. Therefore, the cosine of our starting angle, 43 π , is equal to −^12 , too.
Below is the final, finished unit circle constructed in this way. Notice that if you start from θ = 0 and read off the values of cos θ (the x-coordinates), you get the exact same list as if you had started from θ = π 2 and read off the values of sin θ (the y-coordinates).
270 ◦^
225 ◦^315 ◦
π 6
π 4
π 3
π 2 2 π 3 3 π 4 5 π 6
π
7 π 6 5 π 4 4 π 3 3 π 2
5 π 3
7 π 4
11 π 6
3 2 ,^
1 2
2 2 ,^
√ 2 2
1 2 ,^
√ 3 2
√ 3 2 ,^
1 2
√ 2 2 ,^
√ 2 2
−^12 ,
√ 3 2
√ 3 2 ,^ −
1 2
√ 2 2 ,^ −
√ 2 2
−^12 , −
√ 3 2
3 2 ,^ −
1 2
2 2 ,^ −
√ 2 2
1 2 ,^ −
√ 3 2
As a final suggestion, see if you can reproduce this on your own a couple of times (no peeking!). If you can, then you should be good to go for the exam!
