# Special Triangles 2

Alignments to Content Standards: F-TF.A.3

Use the unit circle and indicated triangle below to find the exact value of the sine and cosine of the special angle $\pi/6.$ ## IM Commentary

Using known facts about the unit circle and isosceles triangles together with the Pythagorean Theorem, we can derive the sine and cosine of special angels, in this case of $\pi/6$. This task can be done as a mini lecture soliciting responses from the students, or as a challenge problem for students to ponder and discuss. It is a very nice connection between geometry and algebra that uses quite simply the symmetry of the triangle. A natural mathematical practice to focus on in this task is SMP 3 - Make a viable argument and critique the reasoning of others. Similar tasks derive the exact values of sine and cosine of $\pi/4$ and $\pi/3$. A variant of this task would be to write down the steps in the proof and to ask students to supply the justification.

## Solution

First we label the vertices of the triangle that lie on the unit circle as points $P$ and $Q$. We know that $P$ has coordinates $(\cos(\pi/6), \sin(\pi/6)).$ Note  that the blue triangle is isosceles since two of its sides are radii of the circle and therefore both have length $1.$  Therefore, the angles opposite these sides have the same measure, and therefore must both measure $\pi/3$ in order for the angles in the blue triangle to add up to $\pi.$  This means that the blue triangle is equilateral, and so $\overline{PQ} = 1.$  But we can also observe that the two smaller triangles that make up the blue triangle are congruent by SAS (or since $Q$ is the reflection of $P$ across the $x$-axis); therefore, $y=1/2,$ and by using the Pythagorean Theorem, we have $x^2 + (1/2)^2 = 1^2,$ which implies $x^2 = 3/4$ and $x = \sqrt{3}/2.$  Thus, $P = (\sqrt{3}/2,\ 1/2),$ and we conclude that

$$\cos \pi/6 = \frac{\sqrt{3}}{2} \text{ and } \sin \pi/6 = \frac{1}{2}.$$