Sine and Cosine of Complementary Angles

Task

1. Suppose $0^\circ \lt a \lt 90^\circ$ is the measure of an acute angle. Draw a picture and explain why $\sin{a} = \cos{(90 -a)}$
2. Are there any angle measures $0^\circ \lt a \lt 90^\circ$ for which $\sin{a} = \cos{a}$. Explain.

IM Commentary

The goal of this task is to provide a geometric explanation for the relationship between the sine and cosine of acute angles. In addition, students will need to think about the sine and cosine as functions of an angle in order to reason about when $\sin{a} = \cos{a}$.

The picture is vital for both parts of this task. If done shortly after students have seen definitions of the sine and cosine students will hopefully quickly think about drawing and labeling an appropriate right triangle. If students are struggling to get started the teacher may need to push them in this direction.

Solution

1. To find $\sin{a}$ we draw a right triangle $ABC$ and we assume that $m(\angle A) = a$:

   

   The sine of $\angle A$ is the length of the side opposite $\angle A$ divided by the length of the hypotenuse: $$ \sin{a} = \frac{|BC|}{|AC|}. $$ Since $\angle B$ is a right angle we have $m(\angle A) + m(\angle C) + 90^\circ = 180^\circ$ and so $m(\angle C) = 90^\circ - a$. Since the cosine of $\angle C$ is the length of the side adjacent to $C$ divided by the length of the hypotenuse we have $$ \cos{(90 -a)} = \frac{|BC|}{|AC|}. $$ Thus $\sin{a} = \cos{(90-a)}$. The only property of $a$ that we have used in this argument is that it is one of the (non right) angles in a right triangle. So $\sin{a} = \cos{(90-a)}$ for any $0 \lt a \lt 90$.

2. Reasoning from the picture above, we have

   $$ \begin{align} \sin{a} &= \frac{|BC|}{|AC|},\\ \cos{a} &= \frac{|AB|}{|AC|}. \end{align} $$

   So if $\sin{a} = \cos{a}$ then $|AB| = |BC|$. This means that $\triangle ABC$ is isosceles and so $a = 45$.

   Alternatively, $\sin{0} = 0$ and $\sin{90} = 1$ and the function $\sin{x}$ increases as $x$ increases from 0 to 90. On the other hand $\cos{0} = 1$ and $\cos{90} = 0$ and the function $\cos{x}$ decreases as $x$ increases from 0 to 90. This means that $\sin{x}$ and $\cos{x}$ can take the same value for at most one number $x$ between 0 and 90. They do take the same value when $x = 45$ since $\sin{45} = \cos{45} = \frac{\sqrt{2}}{2}$.