# 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. An acute angle is one that is between 0 and π/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 in the first quadrant, and an angle in the second quadrant which is symmetric with π:

You can see that to measure this symmetric angle from the postive x-axis, you just subtract it from π. 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. So the following identities are evident from this picture:

cos(π – π) = -cosπ
sin(π – π) = sinπ
tan(π – π) = -cosπ/sinπ = -tanπ

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

As an example, let π = π/3, (60Β°). The following is true for π/3:

cos(π/3) = 1/2
sin(π/3) = β3Μ/2
tan(π/3) = β3Μ

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

cos(2π/3) = -1/2
sin(2π/3) = β3Μ/2
tan(2π/3) = -β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. The corresponding coordinates of the intersected point on the unit circle are both the negative of the coordinates for π. So the following identities are shown in this picture:

So these identities are

cos(π + π) = -cosπ
sin(π + π) = -sinπ
tan(π + π) = -cosπ/-sinπ = tanπ

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

cos(4π/3) = -1/2
sin(4π/3) = -β3Μ/2
tan(4π/3) = β3Μ

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) = cosπ
sin(-π) = -sinπ
tan(-π) = cosπ/-sinπ = -tanπ

So,

cos(-π/3) = 1/2
sin(-π/3) = -β3Μ/2
tan(-π/3) = -β3Μ

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