# Special Triangles 1

## Task

Use the unit circle and indicated triangle below to find the exact value of the sine and cosine of the special angle $\pi/4.$

## 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 angles, in this case of $\pi/4$. 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, to derive the exact values of sine and cosine of $\pi/3$ and $\pi/6,$ are in progress. 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 vertex of the triangle that lies on the unit circle as point $P$. We know that $P$ has coordinates $(\cos(\pi/4), \sin(\pi/4)).$ Note that the given triangle is an isosceles right triangle with hypotenuse $1,$ so we begin by letting $l$ represent the length of each side of the triangle. Then we have $$ l^2 + l^2 = 1^2 \Longrightarrow 2l^2 = 1 \Longrightarrow l^2 = 1/2, $$ which means that $l = \sqrt{1/2},$ which can also be written as $l = \sqrt{2}/{2}.$ It follows that $P = (\sqrt{2}/2,\ \sqrt{2}/2),$ and we conclude that $$\cos (\pi/4) = \frac{\sqrt{2}}{2} \text{ and } \sin (\pi/4) = \frac{\sqrt{2}}{2}. $$ (Note that the last step of rationalizing the denominator is not really necessary, it is just an equivalent way of writing the solution.)

## Special Triangles 1

Use the unit circle and indicated triangle below to find the exact value of the sine and cosine of the special angle $\pi/4.$