Chapter 14.1, Problem 62PPSE

Solution

a.

To explain: The reason for the y-coordinate of the point P is $\sin \left(\theta +\pi \right)$ .

Given information:

The diagram of a unit circle with two points $\left(\cos \theta ,\sin \theta \right)$ and $P$ .

Calculation:

The point $\left(\cos \theta ,\sin \theta \right)$ has $y$ -coordinate $\sin \theta$ . The point $\left(\cos \theta ,\sin \theta \right)$ is rotated by angle of $\pi$ to get the point $P$ . So, the angle made by the point $P$ with the positive $x$ -axis is $\theta +\pi$ . Therefore, the $y$ -coordinate of the point $P$ is $\sin \left(\theta +\pi \right)$ .

b.

To prove: The two triangles shown in the diagram are congruent.

Given information:

The diagram of a unit circle with two points $\left(\cos \theta ,\sin \theta \right)$ and $P$ .

Calculation:

Name the points of the triangles as shown below.

  

Observe that the triangles OAQ and OBP are right triangles with hypotenuses OQ and OP respectively.

The angles AOQ and BOP are opposite angles. So, they are equal. Now, since the two angles of the two triangles are equal, the third will also be equal. That means, the triangles are similar.

Since the triangles are similar right triangles with same equal hypotenuses, the triangles are congruent.

c.

To show: The two blue segments are congruent.

Given information:

The diagram of a unit circle with two points $\left(\cos \theta ,\sin \theta \right)$ and $P$ .

Calculation:

Name the points of the triangles as shown below.

  

Since triangles are congruent and the angles AOQ and BOP are equal, the sides AQ and BP are equal. That means, the two blue segments are congruent.

d.

To show: The y-coordinate of the point $P$ is $-\sin \theta$ .

Given information:

The diagram of a unit circle with two points $\left(\cos \theta ,\sin \theta \right)$ and $P$ .

Calculation:

The point $Q\left(\cos \theta ,\sin \theta \right)$ has the y-coordinate $\sin \theta$ . That means, the length the segment AQ is $\sin \theta$ .

The segments AQ and PB are congruent. So, the length PB is also $\sin \theta$ . But the point $P$ lies below the x-axis. So, the y-coordinate of the point $P$ is $-\sin \theta$ .

e.

To conclude: $\sin \left(\theta +\pi \right)=-\sin \theta$

Given information:

The diagram of a unit circle with two points $\left(\cos \theta ,\sin \theta \right)$ and $P$ .

Calculation:

From part (a), the y-coordinate of the point $P$ is $\sin \left(\theta +\pi \right)$ , and in part (d), the y-coordinate is obtained as $-\sin \theta$ . The expression $\sin \left(\theta +\pi \right)$ and $-\sin \theta$ different representations of the same point. So, it is concluded that $\sin \left(\theta +\pi \right)=-\sin \theta$ .