Lesson 7.4 : The Other Trigonometric Functions

New Concept

Beyond sine and cosine lie four more trigonometric functions: tangent, cotangent, secant, and cosecant. This lesson defines them using unit circle coordinates $(x, y)$ and explores their relationships, values for key angles, and fundamental identities.

What’s next

Next, you’ll master these functions through interactive examples and practice cards, learning to find their exact values and apply fundamental trigonometric identities.

Property

If $t$ is a real number and $(x, y)$ is a point where the terminal side of an angle of $t$ radians intercepts the unit circle, then

Examples

• The point $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ is on the unit circle. For the angle $t$ corresponding to this point, $\tan t = \frac{y}{x} = \frac{\sqrt{2}/2}{\sqrt{2}/2} = 1$ and $\sec t = \frac{1}{x} = \frac{1}{\sqrt{2}/2} = \sqrt{2}$.

• For the angle $t = \frac{\pi}{3}$, we know the point on the unit circle is $(\frac{1}{2}, \frac{\sqrt{3}}{2})$. We can find $\cot t = \frac{x}{y} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$.

Property

Given an angle not in the first quadrant, use reference angles to find all six trigonometric functions.

1. Measure the angle formed by the terminal side of the given angle and the horizontal axis. This is the reference angle.
2. Evaluate the function at the reference angle.
3. Observe the quadrant where the terminal side of the original angle is located. Based on the quadrant, determine whether the output is positive or negative. The mnemonic “A Smart Trig Class” helps remember which functions are positive in each quadrant.

Examples

• To find $\tan(\frac{3\pi}{4})$, the reference angle is $\frac{\pi}{4}$. Since $\frac{3\pi}{4}$ is in quadrant II where tangent is negative, $\tan(\frac{3\pi}{4}) = -\tan(\frac{\pi}{4}) = -1$.

• To find $\sec(\frac{7\pi}{6})$, the reference angle is $\frac{\pi}{6}$. The angle $\frac{7\pi}{6}$ is in quadrant III, where secant (reciprocal of cosine) is negative. So, $\sec(\frac{7\pi}{6}) = -\sec(\frac{\pi}{6}) = -\frac{2\sqrt{3}}{3}$.

Property

An even function is one in which $f(-x) = f(x)$.
An odd function is one in which $f(-x) = -f(x)$.
Cosine and secant are even:

Sine, tangent, cosecant, and cotangent are odd:

Examples

• Secant is an even function. If $\sec(t) = 5$, then the secant of the opposite angle is the same, so $\sec(-t) = 5$.

• Tangent is an odd function. If $\tan(t) = -2.5$, then the tangent of its opposite is the negative of that value, so $\tan(-t) = -(-2.5) = 2.5$.

Property

We can derive some useful identities from the six trigonometric functions. The other four trigonometric functions can be related back to the sine and cosine functions using these basic relationships:

Examples

• Given $\sin(30^\circ) = \frac{1}{2}$ and $\cos(30^\circ) = \frac{\sqrt{3}}{2}$, we can find $\tan(30^\circ) = \frac{\sin(30^\circ)}{\cos(30^\circ)} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$.

• To simplify the expression $(\cot t)(\sin t)$, we can rewrite cotangent in terms of sine and cosine: $(\frac{\cos t}{\sin t})(\sin t) = \cos t$.

Property

Examples

• If $\tan t = 2$ and $t$ is in quadrant I, we can find $\sec t$. Using $1 + \tan^2 t = \sec^2 t$, we get $1 + (2)^2 = \sec^2 t$, so $\sec^2 t = 5$. Since $t$ is in quadrant I, $\sec t = \sqrt{5}$.

• If $\cot t = -3$ and $t$ is in quadrant IV, we can find $\csc t$. Using $\cot^2 t + 1 = \csc^2 t$, we get $(-3)^2 + 1 = \csc^2 t$, so $\csc^2 t = 10$. Since $t$ is in quadrant IV, $\csc t = -\sqrt{10}$.

Property

The period $P$ of a repeating function $f$ is the number representing the interval such that $f(x + P) = f(x)$ for any value of $x$.
The period of the cosine, sine, secant, and cosecant functions is $2\pi$.
The period of the tangent and cotangent functions is $\pi$.

Examples

• The tangent function has a period of $\pi$. Therefore, $\tan(\frac{9\pi}{8}) = \tan(\frac{\pi}{8} + \pi) = \tan(\frac{\pi}{8})$.

• The sine function has a period of $2\pi$. So, the value of $\sin(\frac{13\pi}{6})$ is the same as $\sin(\frac{\pi}{6} + 2\pi)$, which equals $\sin(\frac{\pi}{6}) = \frac{1}{2}$.