Find exact values of the trigonometric functions secant, cosecant, tangent, and cotangent

To define the remaining functions, we will once again draw a unit circle with a point [latex]\left(x,y\right)[/latex] corresponding to an angle of [latex]t[/latex], as shown in Figure 1. As with the sine and cosine, we can use the [latex]\left(x,y\right)[/latex] coordinates to find the other functions.

Figure 1

The first function we will define is the tangent. The tangent of an angle is the ratio of the y-value to the x-value of the corresponding point on the unit circle. In Figure 1, the tangent of angle [latex]t[/latex] is equal to [latex]\frac{y}{x},x\ne 0[/latex]. Because the y-value is equal to the sine of [latex]t[/latex], and the x-value is equal to the cosine of [latex]t[/latex], the tangent of angle [latex]t[/latex] can also be defined as [latex]\frac{\sin t}{\cos t},\cos t\ne 0[/latex]. The tangent function is abbreviated as [latex]\tan[/latex]. The remaining three functions can all be expressed as reciprocals of functions we have already defined.

• The secant function is the reciprocal of the cosine function. In Figure 1, the secant of angle [latex]t[/latex] is equal to [latex]\frac{1}{\cos t}=\frac{1}{x},x\ne 0[/latex]. The secant function is abbreviated as [latex]\sec[/latex].
• The cotangent function is the reciprocal of the tangent function. In Figure 1, the cotangent of angle [latex]t[/latex] is equal to [latex]\frac{\cos t}{\sin t}=\frac{x}{y},y\ne 0[/latex]. The cotangent function is abbreviated as [latex]\cot[/latex].
• The cosecant function is the reciprocal of the sine function. In Figure 1, the cosecant of angle [latex]t[/latex] is equal to [latex]\frac{1}{\sin t}=\frac{1}{y},y\ne 0[/latex]. The cosecant function is abbreviated as [latex]\csc[/latex].

Example 1: Finding Trigonometric Functions from a Point on the Unit Circle

The point [latex]\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)[/latex] is on the unit circle, as shown in Figure 2. Find [latex]\sin t,\cos t,\tan t,\sec t,\csc t[/latex], and [latex]\cot t[/latex].

Figure 2

Show Solution

Because we know the [latex]\left(x,y\right)[/latex] coordinates of the point on the unit circle indicated by angle [latex]t[/latex], we can use those coordinates to find the six functions:

[latex]\begin{gathered}\sin t=y=\frac{1}{2}\\ \cos t=x=-\frac{\sqrt{3}}{2}\\ \tan t=\frac{y}{x}=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{1}{2}\left(-\frac{2}{\sqrt{3}}\right)=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}\\ \sec t=\frac{1}{x}=\frac{1}{\frac{-\frac{\sqrt{3}}{2}}{}}=-\frac{2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3}\\ \csc t=\frac{1}{y}=\frac{1}{\frac{1}{2}}=2\\ \cot t=\frac{x}{y}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\frac{\sqrt{3}}{2}\left(\frac{2}{1}\right)=-\sqrt{3}\end{gathered}[/latex]

Try It

The point [latex]\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)[/latex] is on the unit circle, as shown in Figure 3. Find [latex]\sin t,\cos t,\tan t,\sec t,\csc t[/latex], and [latex]\cot t[/latex].

Figure 3

Show Solution

[latex]\sin t=-\frac{\sqrt{2}}{2},\cos t=\frac{\sqrt{2}}{2},\tan t=-1,\sec t=\sqrt{2},\csc t=-\sqrt{2},\cot t=-1[/latex]

Because we know the sine and cosine values for the common first-quadrant angles, we can find the other function values for those angles as well by setting [latex]x[/latex] equal to the cosine and [latex]y[/latex] equal to the sine and then using the definitions of tangent, secant, cosecant, and cotangent. The results are shown in the table below.

Using Reference Angles to Evaluate Tangent, Secant, Cosecant, and Cotangent

We can evaluate trigonometric functions of angles outside the first quadrant using reference angles as we have already done with the sine and cosine functions. The procedure is the same: Find the reference angle formed by the terminal side of the given angle with the horizontal axis. The trigonometric function values for the original angle will be the same as those for the reference angle, except for the positive or negative sign, which is determined by x– and y-values in the original quadrant. Figure 4 shows which functions are positive in which quadrant.

To help us remember which of the six trigonometric functions are positive in each quadrant, we can use the mnemonic phrase “A Smart Trig Class.” Each of the four words in the phrase corresponds to one of the four quadrants, starting with quadrant I and rotating counterclockwise. In quadrant I, which is “A,” all of the six trigonometric functions are positive. In quadrant II, “Smart,” only sine and its reciprocal function, cosecant, are positive. In quadrant III, “Trig,” only tangent and its reciprocal function, cotangent, are positive. Finally, in quadrant IV, “Class,” only cosine and its reciprocal function, secant, are positive.

Figure 4

Using Even and Odd Trigonometric Functions

To be able to use our six trigonometric functions freely with both positive and negative angle inputs, we should examine how each function treats a negative input. As it turns out, there is an important difference among the functions in this regard.

Consider the function [latex]f\left(x\right)={x}^{2}[/latex], shown in Figure 5. The graph of the function is symmetrical about the y-axis. All along the curve, any two points with opposite x-values have the same function value. This matches the result of calculation: [latex]{\left(4\right)}^{2}={\left(-4\right)}^{2}[/latex], [latex]{\left(-5\right)}^{2}={\left(5\right)}^{2}[/latex], and so on. So [latex]f\left(x\right)={x}^{2}[/latex] is an even function, a function such that two inputs that are opposites have the same output. That means [latex]f\left(-x\right)=f\left(x\right)[/latex].

Figure 5. The function [latex]f\left(x\right)={x}^{2}[/latex] is an even function.

Now consider the function [latex]f\left(x\right)={x}^{3}[/latex], shown in Figure 6. The graph is not symmetrical about the y-axis. All along the graph, any two points with opposite x-values also have opposite y-values. So [latex]f\left(x\right)={x}^{3}[/latex] is an odd function, one such that two inputs that are opposites have outputs that are also opposites. That means [latex]f\left(-x\right)=-f\left(x\right)[/latex].

Figure 6. The function [latex]f\left(x\right)={x}^{3}[/latex] is an odd function.

We can test whether a trigonometric function is even or odd by drawing a unit circle with a positive and a negative angle, as in Figure 7. The sine of the positive angle is [latex]y[/latex]. The sine of the negative angle is −y. The sine function, then, is an odd function. We can test each of the six trigonometric functions in this fashion. The results are shown in in the table below.

Figure 7

Recognize and Use Fundamental Identities

We have explored a number of properties of trigonometric functions. Now, we can take the relationships a step further, and derive some fundamental identities. Identities are statements that are true for all values of the input on which they are defined. Usually, identities can be derived from definitions and relationships we already know. For example, the Pythagorean Identity we learned earlier was derived from the Pythagorean Theorem and the definitions of sine and cosine.

Alternate Forms of the Pythagorean Identity

We can use these fundamental identities to derive alternative forms of the Pythagorean Identity, [latex]{\cos }^{2}t+{\sin }^{2}t=1[/latex]. One form is obtained by dividing both sides by [latex]{\cos }^{2}t:[/latex]

The other form is obtained by dividing both sides by [latex]{\sin }^{2}t:[/latex]

Example 7: Using Identities to Relate Trigonometric Functions

If [latex]\text{cos}\left(t\right)=\frac{12}{13}[/latex] and [latex]t[/latex] is in quadrant IV, as shown in Figure 8, find the values of the other five trigonometric functions.

Figure 8

Show Solution

We can find the sine using the Pythagorean Identity, [latex]{\cos }^{2}t+{\sin }^{2}t=1[/latex], and the remaining functions by relating them to sine and cosine.

[latex]\begin{gathered}{\left(\frac{12}{13}\right)}^{2}+{\sin }^{2}t=1 \\ {\sin }^{2}t=1-{\left(\frac{12}{13}\right)}^{2} \\ {\sin }^{2}t=1-\frac{144}{169} \\ {\sin }^{2}t=\frac{25}{169} \\ \sin t=\pm \sqrt{\frac{25}{169}} \\ \sin t=\pm \frac{\sqrt{25}}{\sqrt{169}} \\ \sin t=\pm \frac{5}{13} \end{gathered}[/latex]

The sign of the sine depends on the y-values in the quadrant where the angle is located. Since the angle is in quadrant IV, where the y-values are negative, its sine is negative, [latex]-\frac{5}{13}[/latex].

The remaining functions can be calculated using identities relating them to sine and cosine.

[latex]\begin{gathered}\tan t=\frac{\sin t}{\cos t}=\frac{-\frac{5}{13}}{\frac{12}{13}}=-\frac{5}{12} \\ \sec t=\frac{1}{\cos t}=\frac{1}{\frac{12}{13}}=\frac{13}{12} \\ \csc t=\frac{1}{\sin t}=\frac{1}{-\frac{5}{13}}=-\frac{13}{5}\\ \cot t=\frac{1}{\tan t}=\frac{1}{-\frac{5}{12}}=-\frac{12}{5}\end{gathered}[/latex]