Trigonometry, Part 2

Now let’s use the unit circle to see some of the common trig identities. These identities (rules) will be used in future posts.

Let’s assume we have an acute angle πœƒ. An acute angle is one that is between 0 and πœ‹/2 (or 0 to 90Β°). The following identities are valid for any angle, not just acute ones – it is just easier to see the logic in the diagram if we assume this.

The following picture shows the relationship between an angle πœƒ in the first quadrant, and an angle in the second quadrant which is symmetric with πœƒ:

You can see that to measure this symmetric angle from the postive x-axis, you just subtract it from πœ‹. The coordinates of the intersected point on the unit circle are negative for the x coordinate but the same y coordinate as the original angle πœƒ. So the following identities are evident from this picture:

cos(πœ‹ – πœƒ) = -cosπœƒ
sin(πœ‹ – πœƒ) = sinπœƒ
tan(πœ‹ – πœƒ) = -cosπœƒ/sinπœƒ = -tanπœƒ

Again, these are true for any angle, not just acute ones.

As an example, let πœƒ = πœ‹/3, (60Β°). The following is true for πœ‹/3:

cos(πœ‹/3) = 1/2
sin(πœ‹/3) = √3Μ…/2
tan(πœ‹/3) = √3Μ…

Now πœ‹ – πœ‹/3 = 2πœ‹/3. So using these identities, we know that

cos(2πœ‹/3) = -1/2
sin(2πœ‹/3) = √3Μ…/2
tan(2πœ‹/3) = -√3Μ…

Now let’s look at a symmetric angle in the third quadrant. To measure this angle from the positive x-axis, you add it to πœ‹. The corresponding coordinates of the intersected point on the unit circle are both the negative of the coordinates for πœƒ. So the following identities are shown in this picture:

So these identities are

cos(πœ‹ + πœƒ) = -cosπœƒ
sin(πœ‹ + πœƒ) = -sinπœƒ
tan(πœ‹ + πœƒ) = -cosπœƒ/-sinπœƒ = tanπœƒ

Using our same example, πœ‹ + πœ‹/3 = 4πœ‹/3. Using these identities:

cos(4πœ‹/3) = -1/2
sin(4πœ‹/3) = -√3Μ…/2
tan(4πœ‹/3) = √3Μ…

One more quadrant to go:

As was mentioned before, angles measured clockwise from the positive x-axis are negative. So the following trig identities are shown in the figure above:

cos(-πœƒ) = cosπœƒ
sin(-πœƒ) = -sinπœƒ
tan(-πœƒ) = cosπœƒ/-sinπœƒ = -tanπœƒ

So,

cos(-πœ‹/3) = 1/2
sin(-πœ‹/3) = -√3Μ…/2
tan(-πœ‹/3) = -√3Μ…

There are a couple more identities I would like to show but I’ll save that for next time.