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## Section16.3Special Angles/Reference Angles

##### Trigonometric Values of $30^{\circ}\text{,}$ $45^{\circ}\text{,}$ and $60^{\circ}$.

As illustrated in FigureĀ 16.3.1, an angle of measurement $45^{\circ}$ drawn in standard position lies along the line with equation $y=x\text{.}$ This is because $45^{\circ}=\frac{1}{8}\left(360^{\circ}\right)\text{,}$ so $45^{\circ}$ is one-eight of a complete revolution, landing the terminal side of the angle midway between the positive $x$ and $y$ axes.

The point where the terminal side of the angle intersects the unit circle is labeled P. We can use the fact that P lies both on the unit circle and the line $y=x$ to determine the coordinates of P. We'll do this by substituting $x$ for $y$ in the equation of the unit circle and then solving for $x\text{.}$

\begin{align*} x^2+y^2\amp=1\\ x^2+x^2\amp=1\\ 2x^2\amp=1\\ x^2\amp=\frac{1}{2}\\ x\amp=\pm\sqrt{\frac{1}{2}}\\ x\amp=\pm\frac{1}{\sqrt{2}}\\ x\amp=\pm\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}\\ x\amp=\pm\frac{\sqrt{2}}{2} \end{align*}

Since PĀ lies in Quadrant I, its $x$-coordinate must be positive, so $x=\frac{\sqrt{2}}{2}\text{.}$ Since P lies on the line $y=x\text{,}$ its $y$-coordinate is also $\frac{\sqrt{2}}{2}\text{.}$ This is illustrated in FigureĀ 16.3.2.

###### Example16.3.3.

Determine the values of the six basic trigonometric at $\frac{\pi}{4}\text{.}$

Solution

Since $\frac{\pi}{4}$ is equivalent to $45^{\circ}\text{,}$ when drawn in standard position, an arc of measurement $\frac{\pi}{4}$ terminates at the point $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\text{.}$ The trigonometric values are derived below.

\begin{align*} \cos\left(\frac{\pi}{4}\right)\amp=x\\ \amp=\frac{\sqrt{2}}{2} \end{align*}
\begin{align*} \sin\left(\frac{\pi}{4}\right)\amp=y\\ \amp=\frac{\sqrt{2}}{2} \end{align*}
\begin{align*} \tan\left(\frac{\pi}{4}\right)\amp=\frac{y}{x}\\ \amp=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\\ \amp=1 \end{align*}
\begin{align*} \sec\left(\frac{\pi}{4}\right)\amp=\frac{1}{x}\\ \amp=\frac{1}{\frac{\sqrt{2}}{2}}\\ \amp=\frac{1}{1} \cdot \frac{2}{\sqrt{2}}\\ \amp=\frac{2}{\sqrt{2}}\\ \amp=\frac{2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}\\ \amp=\frac{2\sqrt{2}}{2}\\ \amp=\sqrt{2} \end{align*}
\begin{align*} \csc\left(\frac{\pi}{4}\right)\amp=\frac{1}{y}\\ \amp=\frac{1}{\frac{\sqrt{2}}{2}}\\ \amp=\frac{1}{1} \cdot \frac{2}{\sqrt{2}}\\ \amp=\frac{2}{\sqrt{2}}\\ \amp=\frac{2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}\\ \amp=\frac{2\sqrt{2}}{2}\\ \amp=\sqrt{2} \end{align*}
\begin{align*} \cot\left(\frac{7\pi}{4}\right)\amp=\frac{x}{y}\\ \amp=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\\ \amp=1 \end{align*}

Let's now turn our attention to an angle of measurement $30^{\circ}\text{.}$ In FigureĀ 16.3.4, $\angle \text{AOM}$ and $\angle \text{BOM}$ both have a measurement of $30^{\circ}\text{,}$ making the measurement of $\angle \text{AOB}$ $60^{\circ}\text{.}$ Because $\overline{\text{AB}}$ intersects the $x$-axis at a right angle, and the sums of the angles in $\bigtriangleup \text{AOM}$ and $\bigtriangleup \text{BOM}$ are both $180^{\circ}\text{,}$ it must be the case that the angles at the vertices label A and B are both $60^{\circ}\text{.}$

Because each angle in $\bigtriangleup \text{AOB}$ is $60^{\circ}\text{,}$ $\bigtriangleup \text{AOB}$ is an equilateral triangle and each of its sides consequently have equal length. Because $\overline{\text{OA}}$ and $\overline{\text{OB}}$ both start at the origin and terminate on the unit circle, we already knew that they both have a length of one unit. Because $\bigtriangleup \text{AOB}$ is equilateral, it must be the case that $\overline{\text{AB}}$ also has a length of one unit. Because the point labeled M lies halfway between the points A and B, it must be the case that $\overline{\text{AM}}$ has a length of $\frac{1}{2}$ unit. Because $\bigtriangleup \text{AOM}$ is a right triangle, we can use the Pythagorean Theorem to determine the length of $\overline{\text{OM}}\text{.}$

\begin{align*} \overline{\text{OM}}^{\,2}+\left(\frac{1}{2}\right)^2\amp=1^2\\ \overline{\text{OM}}^{\,2}+\frac{1}{4}\amp=1\\ \overline{\text{OM}}^{\,2}\amp=\frac{3}{4}\\ \overline{\text{OM}}\amp=\pm\sqrt{\frac{3}{4}}\\ \overline{\text{OM}}\amp=\pm\frac{\sqrt{3}}{2} \end{align*}

So the length of $\overline{\text{OM}}$ is $\frac{\sqrt{3}}{2}$ units. Let's note that the length of $\overline{\text{Om}}$ is also the $x$-coordinate of the point A and the length of $\overline{\text{AM}}$ is also the $y$-coordinate of the point B. As shown in in FigureĀ 16.3.5, an angle of measurement $30^{\circ}$ (or, equivalently, $\frac{\pi}{6}$) intersects the unit circle at the point $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)\text{.}$

###### Example16.3.6.

Determine the values of the six basic trigonometric functions at $\frac{\pi}{6}$

Solution

An arc of measurement $\frac{\pi}{6}\text{,}$ when drawn in standard position, terminates at the point $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)\text{.}$ We can use the coordinates of this point to determine the six trigonometric values.

\begin{align*} \cos\left(\frac{\pi}{6}\right)\amp=x\\ \amp=\frac{\sqrt{3}}{2} \end{align*}
\begin{align*} \sin\left(\frac{\pi}{6}\right)\amp=y\\ \amp=\frac{1}{2} \end{align*}
\begin{align*} \tan\left(\frac{\pi}{6}\right)\amp=\frac{y}{x}\\ \amp=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\\ \amp=\frac{1}{2} \cdot \frac{2}{\sqrt{3}}\\ \amp=\frac{1}{\sqrt{3}}\\ \amp=\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}\\ \amp=\frac{\sqrt{3}}{3} \end{align*}
\begin{align*} \sec\left(\frac{\pi}{6}\right)\amp=\frac{1}{x}\\ \amp=\frac{1}{\frac{\sqrt{3}}{2}}\\ \amp=\frac{1}{1} \cdot \frac{2}{\sqrt{3}}\\ \amp=\frac{2}{\sqrt{3}}\\ \amp=\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}\\ \amp=\frac{2\sqrt{3}}{3} \end{align*}
\begin{align*} \csc\left(\frac{\pi}{6}\right)\amp=\frac{1}{y}\\ \amp=\frac{1}{\frac{1}{2}}\\ \amp=\frac{1}{1} \cdot \frac{2}{1}\\ \amp=2 \end{align*}
\begin{align*} \cot\left(\frac{\pi}{6}\right)\amp=\frac{x}{y}\\ \amp=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\\ \amp=\frac{\sqrt{3}}{2} \cdot \frac{2}{1}\\ \amp=\sqrt{3} \end{align*}

We can determine the coordinates of the point where the terminal side of an angle of measurement $60^{\circ}\text{,}$ drawn in standard position, intersects the unit circle in a manner very similar to that used for $30^{\circ}\text{.}$

The triangle $\bigtriangleup \text{ABC}$ in FigureĀ 16.3.7 is equilateral, and, consequently, $\overline{\text{AB}}$ has a length of one unit. Since the point labeled M lies midway between A and B, $\overline{\text{MB}}$ has a length of one-half unit. Because $\bigtriangleup \text{AMO}$ is a right triangle and we know the lengths of two of the sides, we can use the Pythagorean Theorem to determine the length of the third side, $\overline{\text{OM}}\text{.}$ This calculation results in a value of $\frac{\sqrt{3}}{2}\text{.}$

As shown in in FigureĀ 16.3.8, an angle of measurement $60^{\circ}$ (or, equivalently, $\frac{\pi}{3}$) intersects the unit circle at the point $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\text{.}$ We could use the coordinates of this point to determine the six basic trigonometric values at $\frac{\pi}{3}$ (or, equivalently, $60^{\circ}$).

Altogether there are three points on the unit circle in Quadrant I that those well versed in trigonometry memorize. In FigureĀ 16.3.9, each of the points is shown along with the degree measure of the angle that intersects the unit circle at the point. In FigureĀ 16.3.10, each of the points is shown along with the radian measure of the angle that intersects the unit circle at the point.

##### Reference Angles.

Suppose that an angle, $\theta\text{,}$ is drawn in standard position. Regardless of the rotation that produced the terminal side of the angle $\theta\text{,}$ the reference angle for $\theta$ is the smallest angle formed between the terminal side of $\theta$ and the $x$-axis. We shall use the symbol $\theta^{\,\prime}$ to represent the reference angle of $\theta\text{.}$

###### Example16.3.11.

Determine the reference angle for the angle $\theta=217^{\circ}\text{.}$

Solution

The measurement of $\theta$ is just a little more than $180^{\circ}\text{,}$ so $\theta$ terminates in Quadrant III. The reference angle for $\theta$ is shown in FigureĀ 16.3.12.

Since one side of $\theta^{\,\prime}$ corresponds to $180^{\circ}$ and the other side corresponds to $217^\circ\text{,}$ we can calculate the reference angle as follows.

\begin{align*} \theta^{\,\prime}\amp=217^{\circ}-180^{\circ}\\ \amp=37^{\circ} \end{align*}
###### Example16.3.13.

Determine the reference angle for the angle $\theta=\frac{8\pi}{5}\text{.}$

Solution

The first thing we need to determine is the quadrant in which $\theta$ terminates. $\frac{8\pi}{5}$ is greater than $\pi$ and less than $2\pi\text{,}$ so $\theta$ definitely terminates in either Quadrant III or Quadrant IV. One measurement for an angle in standard position that terminates along the negative $y$-axis is $\frac{3\pi}{2}$ which is equivalent to $1.5\pi\text{.}$ Our angle, $\theta\text{,}$ is equivalent to $1.6\pi\text{,}$ so it rotates just past the negative $y$-axis, terminating in Quadrant IV. The reference angle for $\theta$ is shown in FigureĀ 16.3.14.

Since one side of $\theta^{\,\prime}$ corresponds to $\frac{8\pi}{5}$ and the other side corresponds to $2\pi\text{,}$ we can calculate the reference angle as follows.

\begin{align*} \theta^{\,\prime}\amp=2\pi-\frac{8\pi}{5}\\ \amp=\frac{10\pi}{5}-\frac{8\pi}{5}\\ \amp=\frac{2\pi}{5} \end{align*}
###### Example16.3.15.

Determine the reference angle for the angle $\theta=-575^{\circ}\text{.}$

Solution

The first thing we need to determine is the quadrant in which $\theta$ terminates. $\frac{575}{360} \approx 1.6\text{,}$ so $\theta$ completes a little more than 1.5 complete revolutions. Because the measurement is negative, the rotation is in the clockwise direction, so $\theta$ terminates in the second quadrant. An angle of measurement $720^{\circ}$represents two complete revolutions, so $\theta$ is coterminal with $-575^{\circ}+720^{\circ}$ which simplifies to $145^{\circ}\text{.}$ The reference angle for $\theta$ is shown in FigureĀ 16.3.16.

Since one side of $\theta^{\,\prime}$ corresponds to $145^{\circ}$ and the other side corresponds to $180^\circ\text{,}$ we can calculate the reference angle as follows.

\begin{align*} \theta^{\,\prime}\amp=180^{\circ}-145^{\circ}\\ \amp=35^{\circ} \end{align*}
###### Example16.3.17.

Determine the reference angle for the angle $\theta=-\frac{21\pi}{11}\text{.}$

Solution

Because $\frac{22\pi}{11}=2\pi\text{,}$ $-\frac{21\pi}{11}$ must be just short of one complete revolution. Because the measurement is negative, the rotation is clockwise and the angle terminates in Quadrant I. The angle is coterminal with $-\frac{21\pi}{11}+2\pi$ which simplifies to $\frac{\pi}{11}\text{.}$ Any angle whose measurement is between $0$ and $\frac{\pi}{2}$ is its own reference angle, so $\frac{\pi}{12}$ is its own reference angle and also the reference angle for $-\frac{21\pi}{11}\text{.}$

###### Example16.3.18.

Determine the reference angle for the angle $\theta=\frac{141\pi}{7}\text{.}$

Solution

One way we can make this determination is to recognize the following.

\begin{align*} \frac{141\pi}{7}\amp=\frac{140\pi}{7}+\frac{\pi}{7}\\ \amp=20\pi+\frac{\pi}{7} \end{align*}

Because $20\pi$ results in ten complete revolutions, $\frac{141\pi}{7}$ is coterminal with $\frac{\pi}{7}\text{.}$ Because $\frac{\pi}{7}$ is between $0$ and $\frac{\pi}{2}\text{,}$ it is its own reference angle and, consequently, also the reference angle for $\frac{141\pi}{7}\text{.}$

##### A Fundamental Property Related to Reference Angles.

If $\theta_1$ and $\theta_2$ have the same reference angle, then the absolute values of both their $x$-coordinates and their $y$-coordinates are equal. As a consequence, if $\theta_1$ and $\theta_2$ have the same reference angle, then:

\begin{equation*} \abs{\cos(\theta_1)}=\abs{\cos(\theta_2)} \end{equation*}
\begin{equation*} \abs{\sin(\theta_1)}=\abs{\sin(\theta_2)} \end{equation*}
\begin{equation*} \abs{\tan(\theta_1)}=\abs{\tan(\theta_2)} \end{equation*}
\begin{equation*} \abs{\sec(\theta_1)}=\abs{\sec(\theta_2)} \end{equation*}
\begin{equation*} \abs{\csc(\theta_1)}=\abs{\csc(\theta_2)} \end{equation*}
\begin{equation*} \abs{\cot(\theta_1)}=\abs{\cot(\theta_2)} \end{equation*}
###### Example16.3.19.

Determine the values of $\cos\left(150^{\circ}\right)$ and $\sin\left(150^{\circ}\right)\text{.}$

Solution

The reference angle for $150^{\circ}$ is $180^{\circ}-150^{\circ}$ which simplifies to $30^{\circ}\text{.}$ Because $30^{\circ}$ intersects the unit circle at the point $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)\text{,}$ the absolute values of the coordinates where $150^{\circ}$ intersects the unit circle are $\abs{x}=\frac{\sqrt{3}}{2}$ and $\abs{y}=\frac{1}{2}\text{.}$ Because $150^{\circ}$terminates in the second quadrant, it's $x$-coordinate is negative and its $y$-coordinate is positive. Taken all together, we conclude that $150^{\circ}$ intersects the unit circle at the point $\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)\text{.}$ Consequently, $\cos\left(150^{\circ}\right)=-\frac{\sqrt{3}}{2}$ and $\sin\left(150^{\circ}\right)=\frac{1}{2}$

###### Example16.3.21.

Determine the value of $\tan\left(-\frac{\pi}{4}\right)\text{.}$

Solution

An angle of measurement $-\frac{\pi}{4}$ terminates in Quadrant IV and has a reference angle of $\frac{\pi}{4}\text{.}$ Because the angle terminates in Quadrant IV, its $x$-coordinate is positive and its $y$-coordinate is negative. Because $\frac{\pi}{4}$ intersects the unit circle at the point $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\text{,}$ we can conclude that $-\frac{\pi}{4}$ intersects the unit circle at the point $\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)\text{.}$ The tangent value is calculated below.

\begin{align*} \amp=\frac{y}{x}\\ \amp=\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\\ \amp=-1 \end{align*}

In FigureĀ 16.3.23, the four angles whose measurements fall between $0^{\circ}$ and $360^{\circ}$ that have a reference angle of $30^{\circ}$ are shown along with the coordinates of the points where the terminal sides of the angles intersect the unit circle. The picture is repeated in FigureĀ 16.3.24, this time with the angles labeled in terms of radians.

In FigureĀ 16.3.25, the four angles whose measurements fall between $0^{\circ}$ and $360^{\circ}$ that have a reference angle of $45^{\circ}$ are shown along with the coordinates of the points where the terminal sides of the angles intersect the unit circle. The picture is repeated in FigureĀ 16.3.26, this time with the angles labeled in terms of radians.

In FigureĀ 16.3.27, the four angles whose measurements fall between $0^{\circ}$ and $360^{\circ}$ that have a reference angle of $60^{\circ}$ are shown along with the coordinates of the points where the terminal sides of the angles intersect the unit circle. The picture is repeated in FigureĀ 16.3.28, this time with the angles labeled in terms of radians.

### ExercisesExercises

For each stated angle, sketch the angle in standard position, mark the reference angle, and determine the measurement of the reference angle.

###### 1.

$\theta=312^{\circ}$

Solution
\begin{align*} \theta^{\,\prime}\amp=360^{\circ}-312^{\circ}\\ \amp=48^{\circ} \end{align*}
###### 2.

$\theta=-248^{\circ}$

Solution

Let's begin by noting that $\theta$ is coterminal with $-248^{\circ}+360^{\circ}$ which simplifies to $112^{\circ}\text{.}$

\begin{align*} \theta^{\,\prime}\amp=180^{\circ}-112^{\circ}\\ \amp=68^{\circ} \end{align*}
###### 3.

$\theta=-310^{\circ}$

Solution

Let's begin by noting that $\theta$ is coterminal with $-310^{\circ}+360^{\circ}$ which simplifies to $50^{\circ}\text{.}$ Since $50^{\circ}$ is its own reference angle, it is also the reference angle for $-310^{\circ}\text{,}$

###### 4.

$\theta=\frac{4\pi}{9}$

Solution

Because $\frac{4\pi}{9}$ is between $0$ and $\frac{\pi}{2}\text{,}$ it is its own reference angle.

###### 5.

$\theta=-\frac{17\pi}{13}$

Solution

Let's begin by noting that $\theta$ is coterminal with $-\frac{17\pi}{13}+2\pi$ which simplifies to $\frac{9\pi}{13}\text{.}$

\begin{align*} \theta^{\,\prime}\amp=\pi-\frac{9\pi}{13}\\ \amp=\frac{13\pi}{13}-\frac{9\pi}{13}\\ \amp=\frac{4\pi}{13} \end{align*}
###### 6.

$\theta=\frac{29\pi}{20}$

Solution
\begin{align*} \theta^{\,\prime}\amp=\frac{29\pi}{20}-\pi\\ \amp=\frac{29\pi}{20}-\frac{20\pi}{20}\\ \amp=\frac{9\pi}{20} \end{align*}

For each stated angle, draw the angle in standard position and indicate the coordinates at the point at which the terminal side of the angle intersects the unit circle. Then determine the indicated trigonometric values.

###### 7.

$\theta=\frac{7\pi}{6}\text{.}$ Determine both $\sin{(\theta)}$ and $\cos{(\theta)}\text{.}$

Solution

The reference angle, $\theta^{\,\prime}\text{,}$ is $\frac{\pi}{6}\text{,}$ so the absolute values of the coordinates where the terminal side of the angle intersects the unit circle are $\abs{x}=\frac{\sqrt{3}}{2}$ and $\abs{y}=\frac{1}{2}\text{.}$ Because the terminal side of the angle intersects the unit circle in Quadrant III, both coordinates at the point of intersection of are negative. Putting it all together we derive the following.

\begin{align*} \cos{(\theta)}\amp=x\\ \amp=-\frac{\sqrt{3}}{2} \end{align*}
\begin{align*} \sin{(\theta)}\amp=y\\ \amp=-\frac{1}{2} \end{align*}
###### 8.

$\theta=\frac{3\pi}{4}\text{.}$ Determine both $\sec{(\theta)}$ and $\csc{(\theta)}\text{.}$

Solution

The reference angle, $\theta^{\,\prime}\text{,}$ is $\frac{\pi}{4}\text{,}$ so the absolute values of the coordinates where the terminal side of the angle intersects the unit circle are $\abs{x}=\frac{\sqrt{2}}{2}$ and $\abs{y}=\frac{\sqrt{2}}{2}\text{.}$ Because the terminal side of the angle intersects the unit circle in Quadrant II, the $x$-coordinate at the point of intersection is negative whereas the $y$-coordinate is positive. Putting it all together we derive the following.

\begin{align*} \sec{(\theta)}\amp=\frac{1}{x}\\ \amp=\frac{1}{-\frac{\sqrt{2}}{2}}\\ \amp=-\frac{2}{\sqrt{2}}\\ \amp=-\frac{2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}\\ \amp=-\frac{2\sqrt{2}}{2}\\ \amp=-\sqrt{2} \end{align*}
\begin{align*} \csc{(\theta)}\amp=\frac{1}{y}\\ \amp=\frac{1}{\frac{\sqrt{2}}{2}}\\ \amp=\frac{2}{\sqrt{2}}\\ \amp=\frac{2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}\\ \amp=\frac{2\sqrt{2}}{2}\\ \amp=\sqrt{2} \end{align*}
###### 9.

$\theta=-330^{\circ}\text{.}$ Determine both $\sin{(\theta)}$ and $\cos{(\theta)}\text{.}$

Solution

The reference angle, $\theta^{\,\prime}\text{,}$ is $30^{\circ}\text{,}$ so the absolute values of the coordinates where the terminal side of the angle intersects the unit circle are $\abs{x}=\frac{\sqrt{3}}{2}$ and $\abs{y}=\frac{1}{2}\text{.}$ Because the terminal side of the angle intersects the unit circle in Quadrant I, both coordinates at the point of intersection are positive. Putting it all together we derive the following.

\begin{align*} \tan{(\theta)}\amp=\frac{y}{x}\\ \amp=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\\ \amp=\frac{1}{2} \cdot \frac{2}{\sqrt{3}}\\ \amp=\frac{1}{\sqrt{3}}\\ \amp=\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}\\ \amp=\frac{\sqrt{3}}{3} \end{align*}
\begin{align*} \cot{(\theta)}\amp=\frac{x}{y}\\ \amp=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\\ \amp=\frac{\sqrt{3}}{2} \cdot \frac{2}{1}\\ \amp=\sqrt{3} \end{align*}
###### 10.

$\theta=270^{\circ}\text{.}$ Determine both $\tan{(\theta)}$ and $\cot{(\theta)}\text{.}$

Solution

By inspection, the terminal side of $\theta$ intersects the unit circle at the point $(0,-1)\text{.}$ This gives the following.

\begin{align*} \cot{(\theta)}\amp=\frac{x}{y}\\ \amp=\frac{0}{-1}\\ \amp=0 \end{align*}

The tangent value doe not exist at $\theta$ because the ration of the $y$ and $x$ coordinates is $\frac{y}{x}=\frac{-1}{0}$ and $\frac{-1}{0}$ is an undefined quantity.

###### 11.

$\theta=\frac{5\pi}{3}\text{.}$ Determine both $\tan{(\theta)}$ and $\cot{(\theta)}\text{.}$

Solution

The reference angle, $\theta^{\,\prime}\text{,}$ is $\frac{\pi}{3}\text{,}$ so the absolute values of the coordinates where the terminal side of the angle intersects the unit circle are $\abs{x}=\frac{1}{2}$ and $\abs{y}=\frac{\sqrt{3}}{2}\text{.}$ Because the terminal side of the angle intersects the unit circle in Quadrant IV, the $x$-coordinate at the point of intersection is positive whereas the $y$-coordinate is negative. Putting it all together we derive the following.

\begin{align*} \tan{(\theta)}\amp=\frac{y}{x}\\ \amp=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}\\ \amp=-\frac{\sqrt{3}}{2} \cdot \frac{2}{1}\\ \amp=-\sqrt{3} \end{align*}
\begin{align*} \cot{(\theta)}\amp=\frac{x}{y}\\ \amp=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}\\ \amp=\frac{1}{2} \cdot -\frac{2}{\sqrt{3}}\\ \amp=-\frac{1}{\sqrt{3}}\\ \amp=-\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}\\ \amp=-\frac{\sqrt{3}}{3} \end{align*}
###### 12.

$\theta=-\frac{7\pi}{4}\text{.}$ Determine both $\tan{(\theta)}$ and $\cot{(\theta)}\text{.}$

Solution

The reference angle, $\theta^{\,\prime}\text{,}$ is $\frac{\pi}{4}\text{,}$ so the absolute values of the coordinates where the terminal side of the angle intersects the unit circle are $\abs{x}=\frac{\sqrt{2}}{2}$ and $\abs{y}=\frac{\sqrt{2}}{2}\text{.}$ Because the terminal side of the angle intersects the unit circle in Quadrant I, both coordinates at the point of intersection are positive. Putting it all together we derive the following.

\begin{align*} \tan{(\theta)}\amp=\frac{y}{x}\\ \amp=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\\ \amp=1 \end{align*}
\begin{align*} \cot{(\theta)}\amp=\frac{x}{y}\\ \amp=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\\ \amp=1 \end{align*}

Determine the value of each indicated trigonometric expression.

###### 13.

$\cos{\left(\frac{4\pi}{3}\right)}$

Solution

$\frac{4\pi}{3}$ intersects the unit circle at the point $\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)\text{.}$ This gives us the following.

\begin{align*} \cos{\left(\frac{4\pi}{3}\right)}\amp=x\\ \amp=-\frac{1}{2} \end{align*}
###### 14.

$\sin{\left(-\frac{\pi}{6}\right)}$

Solution

$-\frac{\pi}{6}$ intersects the unit circle at the point $\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\text{.}$ This gives us the following.

\begin{align*} \sin{\left(-\frac{\pi}{6}\right)}\amp=y\\ \amp=-\frac{1}{2} \end{align*}
###### 15.

$\tan{\left(180^{\circ}\right)}$

Solution

$180^{\circ}$ intersects the unit circle at the point $(-1,0)\text{.}$ This gives us the following.

\begin{align*} \tan{\left(180^{\circ}\right)}\amp=\frac{y}{x}\\ \amp=\frac{0}{-1}\\ \amp=0 \end{align*}
###### 16.

$\cot{\left(135^{\circ}\right)}$

Solution

$135^{\circ}$ intersects the unit circle at the point $\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\text{.}$ This gives us the following.

\begin{align*} \cot{\left(135^{\circ}\right)}\amp=\frac{x}{y}\\ \amp=\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\\ \amp=-1 \end{align*}
###### 17.

$\sec{\left(-\frac{3\pi}{2}\right)}$

Solution

$-\frac{3\pi}{2}$ intersects the unit circle at the point $(0,1)\text{.}$ The secant value is not defined at $-\frac{3\pi}{2}$ because the ration $\frac{1}{x}$ becomes $\frac{1}{0}$ which is an undefined quantity.

###### 18.

$\cot{\left(\frac{2\pi}{3}\right)}$

Solution

$\frac{2\pi}{3}$ intersects the unit circle at the point $\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\text{.}$ This gives us the following.

\begin{align*} \cot{\left(\frac{2\pi}{3}\right)}\amp=\frac{x}{y}\\ \amp=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\\ \amp=-\frac{1}{2} \cdot \frac{2}{\sqrt{3}}\\ \amp=-\frac{1}{\sqrt{3}}\\ \amp=-\frac{1}{\sqrt{3}} \cdot \frac{sqrt{3}}{\sqrt{3}}\\ \amp=-\frac{\sqrt{3}}{3} \end{align*}
###### 19.

$\cos{\left(-\frac{7\pi}{6}\right)}$

Solution

$-\frac{7\pi}{6}$ intersects the unit circle at the point $\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)\text{.}$ This gives us the following.

\begin{align*} \cos{\left(-\frac{7\pi}{6}\right)}\amp=x\\ \amp=-\frac{\sqrt{3}}{2} \end{align*}
###### 20.

$\cot{\left(225^{\circ}\right)}$

Solution

$225^{\circ}$ intersects the unit circle at the point $\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)\text{.}$ This gives us the following.

\begin{align*} \cot{\left(225^{\circ}\right)}\amp=\frac{x}{y}\\ \amp=\frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}}\\ \amp=1 \end{align*}
###### 21.

$\csc{\left(\frac{\pi}{4}\right)}$

Solution

$\frac{\pi}{4}$ intersects the unit circle at the point $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\text{.}$ This gives us the following.

\begin{align*} \csc{\left(\frac{\pi}{4}\right)}\amp=\frac{1}{y}\\ \amp=\frac{1}{\frac{\sqrt{2}}{2}}\\ \amp=\frac{2}{\sqrt{2}}\\ \amp=\frac{2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}\\ \amp=\frac{2\sqrt{2}}{2}\\ \amp=\sqrt{2} \end{align*}
###### 22.

$\sin{\left(450^{\circ}\right)}$

Solution

$450^{\circ}$ intersects the unit circle at the point $(0,1)\text{.}$ This gives us the following.

\begin{align*} \sin{\left(450^{\circ}\right)}\amp=y\\ \amp=1 \end{align*}
###### 23.

$\tan{\left(-330^{\circ}\right)}$

Solution

$-330^{\circ}$ intersects the unit circle at the point $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)\text{.}$ This gives us the following.

\begin{align*} \tan{\left(-330^{\circ}\right)}\amp=\frac{y}{x}\\ \amp=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\\ \amp=\frac{1}{3} \cdot \frac{2}{\sqrt{3}}\\ \amp=\frac{1}{\sqrt{3}}\\ \amp=\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}\\ \amp=\frac{\sqrt{3}}{3} \end{align*}
###### 24.

$\sec{\left(-\frac{5\pi}{6}\right)}$

Solution

$-\frac{5\pi}{6}$ intersects the unit circle at the point $\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\text{.}$ This gives us the following.

\begin{align*} \sec{\left(-\frac{5\pi}{6}\right)}\amp=\frac{1}{x}\\ \amp=\frac{1}{-\frac{\sqrt{3}}{2}}\\ \amp=-\frac{2}{\sqrt{3}}\\ \amp=-\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}\\ \amp=-\frac{2\sqrt{3}}{3} \end{align*}