CC The Common Trigonometric Functions

Subsection3.2The Common Trigonometric Functions

The Sine and Cosine Functions

Given an angle \(\theta \ \) (in either degrees or radians) and the \((x,y)\) coordinates of the corresponding point on the unit circle, we define cosine and sine as

\begin{equation*} \cos(\theta)=x \hspace{.25in} \text{and} \hspace{.25in} \sin(\theta)=y \end{equation*}

Note that sine and cosine are functions that take angles as inputs.

Using the \((x,y)\) coordinates we can find the cosine and sine values of common angles on the unit circle.

The applet below shows how the graphs of sine and cosine relate to the \(x\) and \(y\) coordinates of the point \(P\) on the unit circle. Move the slider to change the angle \(\theta\) and observe how the \(x\) and \(y\) values of point \(P\) relate to the outputs of \(\cos(\theta)\) and \(\sin(\theta)\text{.}\)

Figure3.33How the Graphs of Sine and Cosine Relate to the Unit Circle

The following identity is a consequence of the definitions of sine and cosine on a unit circle and the Pythagorean Theorem.

The Pythagorean Identity

For any angle \(\theta\text{,}\)

\begin{equation*} \cos^2(\theta)+\sin^2(\theta)=1 \end{equation*}

In addition to sine and cosine it is also useful to define several additional trigonometric functions.

The Tangent Function

Given an angle \(\theta \ \) (in either degrees or radians) and the \((x,y)\) coordinates of the corresponding point on the unit circle, we define tangent as

\begin{equation*} \tan(\theta)=\frac{y}{x} \end{equation*}

Just like sine and cosine, tangent is a function that takes angles as inputs.

Note that since \(\sin(\theta)=y\) and \(\cos(\theta)=x\text{,}\) we can also define tangent as

\begin{equation*} \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)} \end{equation*}

We can find the tangent values of common angles on the unit circle by using the sine and cosine values of the angles (or the corresponding \(y\) and \(x\) coordinates).

So far we have defined cosine, sine, and tangent with the \(x\) and \(y\) coordinates of points on the unit circle. We can also state equivalent, but more general definitions of sine, cosine, and tangent using a right triangle. On the right triangle, we label the hypotenuse as well as the side opposite the angle and the side adjacent (next to) the angle.

Given a right triangle with an angle of \(\theta\text{,}\) we define sine, cosine, and tangent as

\begin{align*} \sin(\theta) \amp =\frac{\text{opposite}}{\text{hypotenuse}} \\ \\ \cos(\theta) \amp =\frac{\text{adjacent}}{\text{hypotenuse}} \\ \\ \tan(\theta) \amp =\frac{\text{opposite}}{\text{adjacent}} \end{align*}

A common mnemonic for remembering these relationships is Soh-Cah-Toa formed by the first letters of "Sine is opposite over hypotenuse" and "Cosine is adjacent over hypotenuse." We will return to "Toa" in the next section.

Example3.34

Given the triangle shown below, find the value for \(\cos(\theta)\text{,}\) \(\sin(\theta)\text{,}\) and \(\tan(\theta)\text{.}\) right triangle example

Solution

The side adjacent to angle \(\theta\) is 15, and the hypotenuse of the triangle is 17. Using the definition of cosine given above, we get that

\begin{equation*} \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{15}{17}. \end{equation*}

The side opposite to angle \(\theta\) is 8, and the hypotenuse of the triangle is 17. Using the definition of sine given above, we get that

\begin{equation*} \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{8}{17}. \end{equation*}

The side opposite to angle \(\theta\) is 8, and side adjacent to angle \(\theta\) is 15. Using the definition of tangent given above, we get that

\begin{equation*} \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{8}{15}. \end{equation*}

Note that we could have also taken

\begin{equation*} \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{(\frac{8}{17})}{(\frac{15}{17})}=\frac{8}{15}. \end{equation*}

There are times when it can be useful to consider functions like \(\frac{1}{\sin(\theta)}\) or \(\frac{1}{\cos(\theta)}\text{.}\) This happens often enough that mathematicians have created special names for these types of functions. Generally, these functions are called reciprocal trigonometric functions, and they can be defined in terms of the sine, cosine, and tangent functions.

The Reciprocal Trigonometric Functions

Given an angle \(\theta\text{,}\) we define:

\begin{align*} \amp \text{the secant function as } \amp\amp \sec(\theta)=\frac{1}{\cos(\theta)}\\ \\ \amp \text{the cosecant function as } \amp\amp \csc(\theta)=\frac{1}{\sin(\theta)}\\ \\ \amp \text{the cotangent function as } \amp\amp \cot(\theta)=\frac{1}{\tan(\theta)} \end{align*}

Example3.35

Find \(\sec\left(\frac{5\pi}{6}\right)\text{,}\) \(\csc\left(\frac{5\pi}{6}\right)\text{,}\) and \(\cot\left(\frac{5\pi}{6}\right)\text{.}\)

Solution

Since \(\frac{5\pi}{6}\) is a common angle on the unit circle, we have that

\begin{align*} \sec\left(\frac{5\pi}{6}\right)\amp=\frac{1}{\cos\left(\frac{5\pi}{6}\right)} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}\\ \\ \csc\left(\frac{5\pi}{6}\right)\amp=\frac{1}{\sin\left(\frac{5\pi}{6}\right)} = \frac{1}{\frac{1}{2}} = 2\\ \\ \cot\left(\frac{5\pi}{6}\right)\amp=\frac{1}{\tan\left(\frac{5\pi}{6}\right)} = \frac{1}{-\frac{1}{\sqrt{3}}} = -\sqrt{3} \end{align*}

Example3.36

Using the triangle shown, evaluate \(\sec(\theta)\text{,}\) \(\csc(\theta)\text{,}\) and \(\cot(\theta)\text{.}\)

Solution

To find values for reciprocal trigonometric functions, we must first find values for \(\cos(\theta)\text{,}\) \(\sin(\theta)\text{,}\) and \(\tan(\theta)\text{.}\) We must also find the length of the hypotenuse. Letting the variable \(c\) represent the length of the hypotenuse and using the Pythagorean Theorem, we get that

\begin{align*} c^2 \amp= 8^2 + 15^2 \\ c^2 \amp= 64 + 225 \\ c^2 \amp= 289 \\ c \amp= \pm \sqrt{289} \\ c \amp= 17 \end{align*}

triangle for reciprocal trig example with hypotenuse labeled

Now solving for \(\cos(\theta)\text{,}\) \(\sin(\theta)\text{,}\) and \(\tan(\theta)\text{,}\) we get that

\begin{align*} \cos(\theta) \amp= \frac{15}{17} \\ \\ \sin(\theta) \amp= \frac{8}{17} \\ \\ \tan(\theta) \amp= \frac{8}{15} \end{align*}

Finally, solving for the reciprocal trigonometric functions, we get

\begin{align*} \sec(\theta) \amp= \frac{1}{\cos(\theta)} = \frac{17}{15} \\ \\ \csc(\theta) \amp= \frac{1}{\sin(\theta)} = \frac{17}{8} \\ \\ \cot(\theta) \amp= \frac{1}{\tan(\theta)} = \frac{15}{8} \end{align*}