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Guide allo studio > Precalculus II

The Other Trigonometric Functions

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 (x,y)\left(x,y\right) corresponding to an angle of tt, as shown in Figure 1. As with the sine and cosine, we can use the (x,y)\left(x,y\right) coordinates to find the other functions.
Graph of circle with angle of t inscribed. Point of (x, y) is at intersection of terminal side of angle and edge of circle. 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 tt is equal to yx,x≠0\frac{y}{x},x\ne 0. Because the y-value is equal to the sine of tt, and the x-value is equal to the cosine of tt, the tangent of angle tt can also be defined as sin⁡tcos⁡t,cos⁡t≠0\frac{\sin t}{\cos t},\cos t\ne 0\\. The tangent function is abbreviated as tan.\text{tan}\text{.}\\ 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 tt is equal to 1cos⁡t=1x,x≠0\frac{1}{\cos t}=\frac{1}{x},x\ne 0\\. The secant function is abbreviated as sec.\text{sec}\text{.}\\
  • The cotangent function is the reciprocal of the tangent function. In Figure 1, the cotangent of angle tt is equal to cos⁡tsin⁡t=xy,y≠0\frac{\cos t}{\sin t}=\frac{x}{y},y\ne 0\\. The cotangent function is abbreviated as cot.\text{cot}\text{.}\\
  • The cosecant function is the reciprocal of the sine function. In Figure 1, the cosecant of angle tt is equal to 1sin⁡t=1y,y≠0\frac{1}{\sin t}=\frac{1}{y},y\ne 0\\. The cosecant function is abbreviated as csc.\text{csc}\text{.}\\

A General Note: Tangent, Secant, Cosecant, and Cotangent Functions

If tt is a real number and (x,y)\left(x,y\right) is a point where the terminal side of an angle of tt radians intercepts the unit circle, then
tan⁡t=yx,x≠0sec⁡t=1x,x≠0csc⁡t=1y,y≠0cot⁡t=xy,y≠0\begin{array}{l}\tan t=\frac{y}{x},x\ne 0\\ \sec t=\frac{1}{x},x\ne 0\\ \csc t=\frac{1}{y},y\ne 0\\ \cot t=\frac{x}{y},y\ne 0\end{array}\\

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

The point (−32,12)\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)\\ is on the unit circle, as shown in Figure 2. Find sin⁡t,cos⁡t,tan⁡t,sec⁡t,csc⁡t\sin t,\cos t,\tan t,\sec t,\csc t, and cot⁡t\cot t\\.
Graph of circle with angle of t inscribed. Point of (negative square root of 3 over 2, 1/2) is at intersection of terminal side of angle and edge of circle. Figure 2

Solution

Because we know the (x,y)\left(x,y\right) coordinates of the point on the unit circle indicated by angle tt, we can use those coordinates to find the six functions:
sin⁡t=y=12cos⁡t=x=−32tan⁡t=yx=12−32=12(−23)=−13=−33sec⁡t=1x=1−32=−23=−233csc⁡t=1y=112=2cot⁡t=xy=−3212=−32(21)=−3\begin{array}{l}\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{array}\\

Try It 1

The point (22,−22)\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)\\ is on the unit circle, as shown in Figure 3. Find sin⁡t,cos⁡t,tan⁡t,sec⁡t,csc⁡t\sin t,\cos t,\tan t,\sec t,\csc t\\, and cot⁡t\cot t\\.
Graph of circle with angle of t inscribed. Point of (square root of 2 over 2, negative square root of 2 over 2) is at intersection of terminal side of angle and edge of circle. Figure 3
Solution

Example 2: Finding the Trigonometric Functions of an Angle

Find sin⁡t,cos⁡t,tan⁡t,sec⁡t,csc⁡t\sin t,\cos t,\tan t,\sec t,\csc t\\, and cot⁡t\cot t\\ when t=π6t=\frac{\pi }{6}\\.

Solution

We have previously used the properties of equilateral triangles to demonstrate that sin⁡π6=12\sin \frac{\pi }{6}=\frac{1}{2}\\ and cos⁡π6=32\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}\\. We can use these values and the definitions of tangent, secant, cosecant, and cotangent as functions of sine and cosine to find the remaining function values.
tan⁡ π6=sin⁡ π6cos⁡ π6=1232=13=33\begin{array}{l}\begin{array}{l}\\ \tan \text{ }\frac{\pi }{6}=\frac{\sin \text{ }\frac{\pi }{6}}{\cos \text{ }\frac{\pi }{6}}\end{array}\hfill \\ =\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\hfill \end{array}\\
sec⁡π6=1cos⁡π6=132=23=233\begin{array}{l}\sec \frac{\pi }{6}=\frac{1}{\cos \frac{\pi }{6}}\hfill \\ =\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\hfill \end{array}\\
csc⁡π6=1sin⁡π6=112=2\csc \frac{\pi }{6}=\frac{1}{\sin \frac{\pi }{6}}=\frac{1}{\frac{1}{2}}=2\\
cot⁡π6=cos⁡π6sin⁡π6=3212=3\begin{array}{l}\cot \frac{\pi }{6}=\frac{\cos \frac{\pi }{6}}{\sin \frac{\pi }{6}}\hfill \\ =\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\hfill \end{array}\\

Try It 2

Find sin⁡t,cos⁡t,tan⁡t,sec⁡t,csc⁡t\sin t,\cos t,\tan t,\sec t,\csc t, and cot⁡t\cot t when t=π3t=\frac{\pi }{3}. Solution
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 xx equal to the cosine and yy equal to the sine and then using the definitions of tangent, secant, cosecant, and cotangent. The results are shown in the table below.
Angle 00 π6, or 30∘\frac{\pi }{6},\text{ or }{30}^{\circ}\\ π4, or 45∘\frac{\pi }{4},\text{ or } {45}^{\circ }\\ π3, or 60∘\frac{\pi }{3},\text{ or }{60}^{\circ }\\ π2, or 90∘\frac{\pi }{2},\text{ or }{90}^{\circ }\\
Cosine 1 32\frac{\sqrt{3}}{2}\\ 22\frac{\sqrt{2}}{2}\\ 12\frac{1}{2}\\ 0
Sine 0 12\frac{1}{2}\\ 22\frac{\sqrt{2}}{2}\\ 32\frac{\sqrt{3}}{2}\\ 1
Tangent 0 33\frac{\sqrt{3}}{3}\\ 1 3\sqrt{3}\\ Undefined
Secant 1 233\frac{2\sqrt{3}}{3}\\ 2\sqrt{2}\\ 2 Undefined
Cosecant Undefined 2 2\sqrt{2}\\ 233\frac{2\sqrt{3}}{3}\\ 1
Cotangent Undefined 3\sqrt{3}\\ 1 33\frac{\sqrt{3}}{3}\\ 0

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.
Graph of circle with each quadrant labeled. Under quadrant 1, labels fro sin t, cos t, tan t, sec t, csc t, and cot t. Under quadrant 2, labels for sin t and csc t. Under quadrant 3, labels for tan t and cot t. Under quadrant 4, labels for cos t, sec t. Figure 4

How To: 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.

Example 3: Using Reference Angles to Find Trigonometric Functions

Use reference angles to find all six trigonometric functions of −5π6-\frac{5\pi }{6}.

Solution

The angle between this angle’s terminal side and the x-axis is π6\frac{\pi }{6}, so that is the reference angle. Since −5π6-\frac{5\pi }{6} is in the third quadrant, where both xx and yy are negative, cosine, sine, secant, and cosecant will be negative, while tangent and cotangent will be positive.
cos⁡(−5π6)=−32,sin⁡(−5π6)=−12,tan(−5π6)=33sec(−5π6)=−233,csc(−5π6)=−2,cot⁡(−5π6)=3\begin{array}{l}\cos \left(-\frac{5\pi }{6}\right)=-\frac{\sqrt{3}}{2},\sin \left(-\frac{5\pi }{6}\right)=-\frac{1}{2},\text{tan}\left(-\frac{5\pi }{6}\right)=\frac{\sqrt{3}}{3}\hfill \\ \text{sec}\left(-\frac{5\pi }{6}\right)=-\frac{2\sqrt{3}}{3},\text{csc}\left(-\frac{5\pi }{6}\right)=-2,\cot \left(-\frac{5\pi }{6}\right)=\sqrt{3}\hfill \end{array}

Try It 3

Use reference angles to find all six trigonometric functions of −7π4-\frac{7\pi }{4}. Solution

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 f(x)=x2f\left(x\right)={x}^{2}, 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: (4)2=(−4)2{\left(4\right)}^{2}={\left(-4\right)}^{2}, (−5)2=(5)2{\left(-5\right)}^{2}={\left(5\right)}^{2}, and so on. So f(x)=x2f\left(x\right)={x}^{2} is an even function, a function such that two inputs that are opposites have the same output. That means f(−x)=f(x)f\left(-x\right)=f\left(x\right). Graph of parabola with points (-2, 4) and (2, 4) labeled.

Figure 5. The function f(x)=x2f\left(x\right)={x}^{2} is an even function.

Now consider the function f(x)=x3f\left(x\right)={x}^{3}, 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 f(x)=x3f\left(x\right)={x}^{3} is an odd function, one such that two inputs that are opposites have outputs that are also opposites. That means f(−x)=−f(x)f\left(-x\right)=-f\left(x\right). Graph of function with labels for points (-1, -1) and (1, 1).

Figure 6. The function f(x)=x3f\left(x\right)={x}^{3} 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 yy. 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.
Graph of circle with angle of t and -t inscribed. Point of (x, y) is at intersection of terminal side of angle t and edge of circle. Point of (x, -y) is at intersection of terminal side of angle -t and edge of circle. Figure 7
sin⁡t=ysin⁡(−t)=−ysin⁡t≠sin⁡(−t)\begin{array}{l}\sin t=y\hfill \\ \sin \left(-t\right)=-y\hfill \\ \sin t\ne \sin \left(-t\right)\hfill \end{array} cost=xcos⁡(−t)=xcos⁡t=cos⁡(−t)\begin{array}{l}\text{cos}t=x\hfill \\ \cos \left(-t\right)=x\hfill \\ \cos t=\cos \left(-t\right)\hfill \end{array} tan(t)=yxtan⁡(−t)=−yxtan⁡t≠tan⁡(−t)\begin{array}{l}\text{tan}\left(t\right)=\frac{y}{x}\hfill \\ \tan \left(-t\right)=-\frac{y}{x}\hfill \\ \tan t\ne \tan \left(-t\right)\hfill \end{array}
sec⁡t=1xsec⁡(−t)=1xsec⁡t=sec⁡(−t)\begin{array}{l}\sec t=\frac{1}{x}\hfill \\ \sec \left(-t\right)=\frac{1}{x}\hfill \\ \sec t=\sec \left(-t\right)\hfill \end{array} csc⁡t=1ycsc⁡(−t)=1−ycsc⁡t≠csc⁡(−t)\begin{array}{l}\csc t=\frac{1}{y}\hfill \\ \csc \left(-t\right)=\frac{1}{-y}\hfill \\ \csc t\ne \csc \left(-t\right)\hfill \end{array} cot⁡t=xycot⁡(−t)=x−ycot⁡t≠cot(−t)\begin{array}{l}\cot t=\frac{x}{y}\hfill \\ \cot \left(-t\right)=\frac{x}{-y}\hfill \\ \cot t\ne cot\left(-t\right)\hfill \end{array}

A General Note: Even and Odd Trigonometric Functions

An even function is one in which f(−x)=f(x)f\left(-x\right)=f\left(x\right). An odd function is one in which f(−x)=−f(x)f\left(-x\right)=-f\left(x\right). Cosine and secant are even:
cos⁡(−t)=costsec⁡(−t)=sec⁡t\begin{array}{l}\cos \left(-t\right)=\text{cos}t\hfill \\ \sec \left(-t\right)=\sec t\hfill \end{array}
Sine, tangent, cosecant, and cotangent are odd:
sin⁡(−t)=−sin⁡ttan⁡(−t)=−tan⁡tcsc⁡(−t)=−csc⁡tcot⁡(−t)=−cot⁡t\begin{array}{l}\sin \left(-t\right)=-\sin t\hfill \\ \tan \left(-t\right)=-\tan t\hfill \\ \csc \left(-t\right)=-\csc t\hfill \\ \cot \left(-t\right)=-\cot t\hfill \end{array}

Example 4: Using Even and Odd Properties of Trigonometric Functions

If the secant of angle tt is 2, what is the secant of −t?-t?

Solution

Secant is an even function. The secant of an angle is the same as the secant of its opposite. So if the secant of angle t is 2, the secant of −t-t is also 2.

Try It 4

If the cotangent of angle tt is 3\sqrt{3}, what is the cotangent of −t?-t? Solution
https://youtu.be/YbU8Sq0quWE

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.

A General Note: Fundamental Identities

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:
tant=sin⁡tcos⁡t\text{tan}t=\frac{\sin t}{\cos t}\\
sec⁡t=1cos⁡t\sec t=\frac{1}{\cos t}\\
csc⁡t=1sin⁡t\csc t=\frac{1}{\sin t}\\
cott=1tant=costsint\text{cot}t=\frac{1}{\text{tan}t}=\frac{\text{cos}t}{\text{sin}t}\\

Example 5: Using Identities to Evaluate Trigonometric Functions

  1. Given sin⁡(45∘)=22,cos⁡(45∘)=22\sin \left(45^\circ \right)=\frac{\sqrt{2}}{2},\cos \left(45^\circ \right)=\frac{\sqrt{2}}{2}\\, evaluate tan⁡(45∘)\tan \left(45^\circ \right)\\.
  2. Given sin⁡(5π6)=12,cos(5π6)=−32,evaluatesec⁡(5π6)\sin \left(\frac{5\pi }{6}\right)=\frac{1}{2},\text{cos}\left(\frac{5\pi }{6}\right)=-\frac{\sqrt{3}}{2},\text{evaluate}\sec \left(\frac{5\pi }{6}\right)\\.

Solution

Because we know the sine and cosine values for these angles, we can use identities to evaluate the other functions.
  1. tan⁡(45∘)=sin⁡(45∘)cos⁡(45∘)=2222=1\begin{array}{l}\tan \left(45^\circ \right)=\frac{\sin \left(45^\circ \right)}{\cos \left(45^\circ \right)}\hfill \\ =\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\hfill \\ =1\hfill \end{array}\\
  2. sec⁡(5π6)=1cos⁡(5π6)=1−32=−231=−23=−233\begin{array}{l}\sec \left(\frac{5\pi }{6}\right)=\frac{1}{\cos \left(\frac{5\pi }{6}\right)}\hfill \\ =\frac{1}{-\frac{\sqrt{3}}{2}}\hfill \\ \begin{array}{l}=\frac{-2\sqrt{3}}{1}\hfill \\ =\frac{-2}{\sqrt{3}}\hfill \\ =-\frac{2\sqrt{3}}{3}\hfill \end{array}\hfill \end{array}\\

Try It 5

Evaluate csc(7π6)\text{csc}\left(\frac{7\pi }{6}\right)\\. Solution

Example 6: Using Identities to Simplify Trigonometric Expressions

Simplify sec⁡ttan⁡t\frac{\sec t}{\tan t}\\.

Solution

We can simplify this by rewriting both functions in terms of sine and cosine.

sec⁡ttan⁡t=1cos⁡tsin⁡tcos⁡tTo divide the functions, we multiply by the reciprocal.=1cos⁡tcos⁡tsin⁡tDivide out the cosines.=1sin⁡tSimplify and use the identity.=csc⁡t\begin{array}\hfill \frac{\sec t}{\tan t}=\frac{\frac{1}{\cos t}}{\frac{\sin t}{\cos t}}\hfill& \text{To divide the functions, we multiply by the reciprocal.} \\=\frac{1}{\cos t}\frac{\cos t}{\sin t}\hfill&\text{Divide out the cosines.} \\ =\frac{1}{\sin t}\hfill&\text{Simplify and use the identity.}\\ =\csc t \end{array}\\

By showing that sec⁡ttan⁡t\frac{\sec t}{\tan t}\\ can be simplified to csc⁡t\csc t, we have, in fact, established a new identity.
sec⁡ttan⁡t=csc⁡t\frac{\sec t}{\tan t}=\csc t\\

Try It 6

Simplify tan⁡t(cos⁡t)\tan t\left(\cos t\right)\\. Solution

Alternate Forms of the Pythagorean Identity

We can use these fundamental identities to derive alternative forms of the Pythagorean Identity, cos⁡2t+sin⁡2t=1{\cos }^{2}t+{\sin }^{2}t=1\\. One form is obtained by dividing both sides by cos⁡2t:{\cos }^{2}t:\\
cos⁡2tcos⁡2t+sin⁡2tcos⁡2t=1cos⁡2t1+tan⁡2t=sec⁡2t\begin{array}{c}\frac{{\cos }^{2}t}{{\cos }^{2}t}+\frac{{\sin }^{2}t}{{\cos }^{2}t}=\frac{1}{{\cos }^{2}t}\\ 1+{\tan }^{2}t={\sec }^{2}t\end{array}\\
The other form is obtained by dividing both sides by sin⁡2t:{\sin }^{2}t:\\
cos⁡2tsin⁡2t+sin⁡2tsin⁡2t=1sin⁡2tcot⁡2t+1=csc⁡2t\begin{array}{c}\frac{{\cos }^{2}t}{{\sin }^{2}t}+\frac{{\sin }^{2}t}{{\sin }^{2}t}=\frac{1}{{\sin }^{2}t}\\ {\cot }^{2}t+1={\csc }^{2}t\end{array}\\

A General Note: Alternate Forms of the Pythagorean Identity

1+tan⁡2t=sec⁡2t1+{\tan }^{2}t={\sec }^{2}t\\
cot⁡2t+1=csc⁡2t{\cot }^{2}t+1={\csc }^{2}t\\

Example 7: Using Identities to Relate Trigonometric Functions

If cos(t)=1213\text{cos}\left(t\right)=\frac{12}{13}\\ and tt is in quadrant IV, as shown in Figure 8, find the values of the other five trigonometric functions.
Graph of circle with angle of t inscribed. Point of (12/13, y) is at intersection of terminal side of angle and edge of circle. Figure 8
 

Solution

We can find the sine using the Pythagorean Identity, cos⁡2t+sin⁡2t=1{\cos }^{2}t+{\sin }^{2}t=1, and the remaining functions by relating them to sine and cosine.
(1213)2+sin⁡2t=1 sin⁡2t=1−(1213)2 sin⁡2t=1−144169 sin⁡2t=25169 sin⁡t=±25169 sin⁡t=±25169 sin⁡t=±513\begin{array}{l}{\left(\frac{12}{13}\right)}^{2}+{\sin }^{2}t=1\hfill \\ \text{ }{\sin }^{2}t=1-{\left(\frac{12}{13}\right)}^{2}\hfill \\ \text{ }{\sin }^{2}t=1-\frac{144}{169}\hfill \\ \text{ }{\sin }^{2}t=\frac{25}{169}\hfill \\ \text{ }\sin t=\pm \sqrt{\frac{25}{169}}\hfill \\ \text{ }\sin t=\pm \frac{\sqrt{25}}{\sqrt{169}}\hfill \\ \text{ }\sin t=\pm \frac{5}{13}\hfill \end{array}\\
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, −513-\frac{5}{13}\\. The remaining functions can be calculated using identities relating them to sine and cosine.
tan⁡t=sin⁡tcos⁡t=−5131213=−512sec⁡t=1cos⁡t=11213=1312csc⁡t=1sin⁡t=1−513=−135cot⁡t=1tan⁡t=1−512=−125\begin{array}{l}\begin{array}{l}\begin{array}{l}\tan t=\frac{\sin t}{\cos t}=\frac{-\frac{5}{13}}{\frac{12}{13}}=-\frac{5}{12}\hfill \\ \begin{array}{l}\\ \sec t=\frac{1}{\cos t}=\frac{1}{\frac{12}{13}}=\frac{13}{12}\end{array}\hfill \end{array}\hfill \\ \csc t=\frac{1}{\sin t}=\frac{1}{-\frac{5}{13}}=-\frac{13}{5}\hfill \end{array}\\ \cot t=\frac{1}{\tan t}=\frac{1}{-\frac{5}{12}}=-\frac{12}{5}\end{array}\\

Try It 7

If sec⁡(t)=−178\sec \left(t\right)=-\frac{17}{8} and 0<t<π0<t<\pi , find the values of the other five functions. Solution
As we discussed in the chapter opening, a function that repeats its values in regular intervals is known as a periodic function. The trigonometric functions are periodic. For the four trigonometric functions, sine, cosine, cosecant and secant, a revolution of one circle, or 2π2\pi , will result in the same outputs for these functions. And for tangent and cotangent, only a half a revolution will result in the same outputs. Other functions can also be periodic. For example, the lengths of months repeat every four years. If xx represents the length time, measured in years, and f(x)f\left(x\right)\\ represents the number of days in February, then f(x+4)=f(x)f\left(x+4\right)=f\left(x\right)\\. This pattern repeats over and over through time. In other words, every four years, February is guaranteed to have the same number of days as it did 4 years earlier. The positive number 4 is the smallest positive number that satisfies this condition and is called the period. A period is the shortest interval over which a function completes one full cycle—in this example, the period is 4 and represents the time it takes for us to be certain February has the same number of days.

A General Note: Period of a Function

The period PP of a repeating function ff is the number representing the interval such that f(x+P)=f(x)f\left(x+P\right)=f\left(x\right)\\ for any value of xx. The period of the cosine, sine, secant, and cosecant functions is 2π2\pi . The period of the tangent and cotangent functions is π\pi .

Example 8: Finding the Values of Trigonometric Functions

Find the values of the six trigonometric functions of angle tt based on Figure 9.
Graph of circle with angle of t inscribed. Point of (1/2, negative square root of 3 over 2) is at intersection of terminal side of angle and edge of circle. Figure 9
 

Solution

sin⁡t=y=−32cos⁡t=x=−12tan⁡t=sin⁡tcos⁡t=−32−12=3sec⁡t=1cos⁡t=1−12=−2csc⁡t=1sin⁡t=1−32=−233cot⁡t=1tan⁡t=13=33\begin{array}{l}\sin t=y=-\frac{\sqrt{3}}{2}\\ \cos t=x=-\frac{1}{2}\\ \tan t=\frac{\sin t}{\cos t}=\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=\sqrt{3}\\ \sec t=\frac{1}{\cos t}=\frac{1}{-\frac{1}{2}}=-2\\ \csc t=\frac{1}{\sin t}=\frac{1}{-\frac{\sqrt{3}}{2}}=-\frac{2\sqrt{3}}{3}\\ \cot t=\frac{1}{\tan t}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\end{array}

Try It 8

Find the values of the six trigonometric functions of angle tt based on Figure 10.
Graph of circle with angle of t inscribed. Point of (0, -1) is at intersection of terminal side of angle and edge of circle. Figure 10
Solution

Example 9: Finding the Value of Trigonometric Functions

If sin⁡(t)=−32\sin \left(t\right)=-\frac{\sqrt{3}}{2} and cos(t)=12\text{cos}\left(t\right)=\frac{1}{2}, find sec(t),csc(t),tan(t), cot(t)\text{sec}\left(t\right),\text{csc}\left(t\right),\text{tan}\left(t\right),\text{ cot}\left(t\right).

Solution

sec⁡t=1cos⁡t=112=2csc⁡t=1sin⁡t=1−32−233tan⁡t=sin⁡tcos⁡t=−3212=−3cot⁡t=1tan⁡t=1−3=−33\begin{array}{l}\begin{array}{l}\\ \sec t=\frac{1}{\cos t}=\frac{1}{\frac{1}{2}}=2\end{array}\hfill \\ \csc t=\frac{1}{\sin t}=\frac{1}{-\frac{\sqrt{3}}{2}}-\frac{2\sqrt{3}}{3}\hfill \\ \tan t=\frac{\sin t}{\cos t}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}\hfill \\ \cot t=\frac{1}{\tan t}=\frac{1}{-\sqrt{3}}=-\frac{\sqrt{3}}{3}\hfill \end{array}

Try It 9

If sin⁡(t)=22\sin \left(t\right)=\frac{\sqrt{2}}{2} and cos⁡(t)=22\cos \left(t\right)=\frac{\sqrt{2}}{2}, find sec(t),csc(t),tan(t), and cot(t)\text{sec}\left(t\right),\text{csc}\left(t\right),\text{tan}\left(t\right),\text{ and cot}\left(t\right). Solution
https://youtu.be/4BR_qUZ5jK0

Evaluating Trigonometric Functions with a Calculator

We have learned how to evaluate the six trigonometric functions for the common first-quadrant angles and to use them as reference angles for angles in other quadrants. To evaluate trigonometric functions of other angles, we use a scientific or graphing calculator or computer software. If the calculator has a degree mode and a radian mode, confirm the correct mode is chosen before making a calculation. Evaluating a tangent function with a scientific calculator as opposed to a graphing calculator or computer algebra system is like evaluating a sine or cosine: Enter the value and press the TAN key. For the reciprocal functions, there may not be any dedicated keys that say CSC, SEC, or COT. In that case, the function must be evaluated as the reciprocal of a sine, cosine, or tangent. If we need to work with degrees and our calculator or software does not have a degree mode, we can enter the degrees multiplied by the conversion factor π180\frac{\pi }{180}\\ to convert the degrees to radians. To find the secant of 30∘30^\circ , we could press
(for a scientific calculator):130×π180COS\text{(for a scientific calculator):}\frac{1}{30\times \frac{\pi }{180}}\text{COS}\\
or
(for a graphing calculator):1cos⁡(30π180)\text{(for a graphing calculator):}\frac{1}{\cos \left(\frac{30\pi }{180}\right)}\\

How To: Given an angle measure in radians, use a scientific calculator to find the cosecant.

  1. If the calculator has degree mode and radian mode, set it to radian mode.
  2. Enter: 1 /1\text{ /}\\
  3. Enter the value of the angle inside parentheses.
  4. Press the SIN key.
  5. Press the = key.

How To: Given an angle measure in radians, use a graphing utility/calculator to find the cosecant.

  1. If the graphing utility has degree mode and radian mode, set it to radian mode.
  2. Enter: 1 /1\text{ /}\\
  3. Press the SIN key.
  4. Enter the value of the angle inside parentheses.
  5. Press the ENTER key.

Example 10: Evaluating the Secant Using Technology

Evaluate the cosecant of 5π7\frac{5\pi }{7}.

Solution

For a scientific calculator, enter information as follows:
1 / ( 5 × π / 7 ) SIN =\text{1 / ( 5 }\times \text{ }\pi \text{ / 7 ) SIN =}\\
csc⁡(5π7)≈1.279\csc \left(\frac{5\pi }{7}\right)\approx 1.279\\

Try It 10

Evaluate the cotangent of −π8-\frac{\pi }{8}\\. Solution
https://youtu.be/rhRi_IuE_18

Key Equations

Tangent function tan⁡t=sin⁡tcos⁡t\tan t=\frac{\sin t}{\cos t}
Secant function sec⁡t=1cos⁡t\sec t=\frac{1}{\cos t}
Cosecant function csc⁡t=1sin⁡t\csc t=\frac{1}{\sin t}
Cotangent function cot⁡t=1tan⁡t=cos⁡tsin⁡t\cot t=\frac{1}{\tan t}=\frac{\cos t}{\sin t}

Key Concepts

  • The tangent of an angle is the ratio of the y-value to the x-value of the corresponding point on the unit circle.
  • The secant, cotangent, and cosecant are all reciprocals of other functions. The secant is the reciprocal of the cosine function, the cotangent is the reciprocal of the tangent function, and the cosecant is the reciprocal of the sine function.
  • The six trigonometric functions can be found from a point on the unit circle.
  • Trigonometric functions can also be found from an angle.
  • Trigonometric functions of angles outside the first quadrant can be determined using reference angles.
  • A function is said to be even if f(−x)=f(x)f\left(-x\right)=f\left(x\right) and odd if f(−x)=−f(x)f\left(-x\right)=-f\left(x\right).
  • Cosine and secant are even; sine, tangent, cosecant, and cotangent are odd.
  • Even and odd properties can be used to evaluate trigonometric functions.
  • The Pythagorean Identity makes it possible to find a cosine from a sine or a sine from a cosine.
  • Identities can be used to evaluate trigonometric functions.
  • Fundamental identities such as the Pythagorean Identity can be manipulated algebraically to produce new identities.
  • The trigonometric functions repeat at regular intervals.
  • The period PP of a repeating function ff is the smallest interval such that f(x+P)=f(x)f\left(x+P\right)=f\left(x\right) for any value of xx.
  • The values of trigonometric functions of special angles can be found by mathematical analysis.
  • To evaluate trigonometric functions of other angles, we can use a calculator or computer software.

Glossary

cosecant
the reciprocal of the sine function: on the unit circle, csc⁡t=1y,y≠0\csc t=\frac{1}{y},y\ne 0
cotangent
the reciprocal of the tangent function: on the unit circle, cot⁡t=xy,y≠0\cot t=\frac{x}{y},y\ne 0
identities
statements that are true for all values of the input on which they are defined
period
the smallest interval PP of a repeating function ff such that f(x+P)=f(x)f\left(x+P\right)=f\left(x\right)
secant
the reciprocal of the cosine function: on the unit circle, sec⁡t=1x,x≠0\sec t=\frac{1}{x},x\ne 0
tangent
the quotient of the sine and cosine: on the unit circle, tan⁡t=yx,x≠0\tan t=\frac{y}{x},x\ne 0

Section Exercises

1. On an interval of [0,2π)\left[0,2\pi \right), can the sine and cosine values of a radian measure ever be equal? If so, where? 2. What would you estimate the cosine of π\pi degrees to be? Explain your reasoning. 3. For any angle in quadrant II, if you knew the sine of the angle, how could you determine the cosine of the angle? 4. Describe the secant function. 5. Tangent and cotangent have a period of π\pi . What does this tell us about the output of these functions? For the following exercises, find the exact value of each expression. 6. tan⁡π6\tan \frac{\pi }{6} 7. sec⁡π6\sec \frac{\pi }{6} 8. csc⁡π6\csc \frac{\pi }{6} 9. cot⁡π6\cot \frac{\pi }{6} 10. tan⁡π4\tan \frac{\pi }{4} 11. sec⁡π4\sec \frac{\pi }{4} 12. csc⁡π4\csc \frac{\pi }{4} 13. cot⁡π4\cot \frac{\pi }{4} 14. tan⁡π3\tan \frac{\pi }{3} 15. sec⁡π3\sec \frac{\pi }{3} 16. csc⁡π3\csc \frac{\pi }{3} 17. cot⁡π3\cot \frac{\pi }{3} For the following exercises, use reference angles to evaluate the expression. 18. tan⁡5π6\tan \frac{5\pi }{6} 19. sec⁡7π6\sec \frac{7\pi }{6} 20. csc⁡11π6\csc \frac{11\pi }{6} 21. cot⁡13π6\cot \frac{13\pi }{6} 22. tan⁡7π4\tan \frac{7\pi }{4} 23. sec⁡3π4\sec \frac{3\pi }{4} 24. csc⁡5π4\csc \frac{5\pi }{4} 25. cot⁡11π4\cot \frac{11\pi }{4} 26. tan⁡8π3\tan \frac{8\pi }{3} 27. sec⁡4π3\sec \frac{4\pi }{3} 28. csc⁡2π3\csc \frac{2\pi }{3} 29. cot⁡5π3\cot \frac{5\pi }{3} 30. tan⁡225∘\tan 225^\circ 31. sec⁡300∘\sec 300^\circ 32. csc⁡150∘\csc 150^\circ 33. cot⁡240∘\cot 240^\circ 34. tan⁡330∘\tan 330^\circ 35. sec⁡120∘\sec 120^\circ 36. csc⁡210∘\csc 210^\circ 37. cot⁡315∘\cot 315^\circ 38. If sint=34\text{sin}t=\frac{3}{4}, and tt is in quadrant II, find cos⁡t,sec⁡t,csc⁡t,tan⁡t,cot⁡t\cos t,\sec t,\csc t,\tan t,\cot t. 39. If cost=−13\text{cos}t=-\frac{1}{3}, and tt is in quadrant III, find sin⁡t,sec⁡t,csc⁡t,tan⁡t,cot⁡t\sin t,\sec t,\csc t,\tan t,\cot t. 40. If tan⁡t=125\tan t=\frac{12}{5}, and 0≤t<π20\le t<\frac{\pi }{2}, find sin⁡t,cos⁡t,sec⁡t,csc⁡t\sin t,\cos t,\sec t,\csc t, and cot⁡t\cot t. 41. If sin⁡t=32\sin t=\frac{\sqrt{3}}{2} and cos⁡t=12\cos t=\frac{1}{2}, find sec⁡t,csc⁡t,tan⁡t\sec t,\csc t,\tan t, and cot⁡t\cot t. 42. If sin⁡40∘≈0.643cos⁡40∘≈0.766sec40∘,csc40∘,tan40∘,andcot40∘\sin 40^\circ \approx 0.643\cos 40^\circ \approx 0.766\text{sec}40^\circ ,\text{csc}40^\circ ,\text{tan}40^\circ ,\text{and}\text{cot}40^\circ . 43. If sint=22\text{sin}t=\frac{\sqrt{2}}{2}, what is the sin(−t)?\text{sin}\left(-t\right)? 44. If cost=12\text{cos}t=\frac{1}{2}, what is the cos(−t)?\text{cos}\left(-t\right)? 45. If sect=3.1\text{sec}t=3.1, what is the sec(−t)?\text{sec}\left(-t\right)? 46. If csct=0.34\text{csc}t=0.34, what is the csc(−t)?\text{csc}\left(-t\right)? 47. If tant=−1.4\text{tan}t=-1.4, what is the tan(−t)?\text{tan}\left(-t\right)? 48. If cott=9.23\text{cot}t=9.23, what is the cot(−t)?\text{cot}\left(-t\right)? For the following exercises, use the angle in the unit circle to find the value of the each of the six trigonometric functions. 49. Graph of circle with angle of t inscribed. Point of (square root of 2 over 2, square root of 2 over 2) is at intersection of terminal side of angle and edge of circle. 50. Graph of circle with angle of t inscribed. Point of (square root of 3 over 2, 1/2) is at intersection of terminal side of angle and edge of circle. 51. Graph of circle with angle of t inscribed. Point of (-1/2, negative square root of 3 over 2) is at intersection of terminal side of angle and edge of circle. For the following exercises, use a graphing calculator to evaluate. 52. csc⁡5π9\csc \frac{5\pi }{9} 53. cot⁡4π7\cot \frac{4\pi }{7} 54. sec⁡π10\sec \frac{\pi }{10} 55. tan⁡5π8\tan \frac{5\pi }{8} 56. sec⁡3π4\sec \frac{3\pi }{4} 57. csc⁡π4\csc \frac{\pi }{4} 58. tan98∘\text{tan}98^\circ 59. cot⁡33∘\cot 33^\circ 60. cot⁡140∘\cot 140^\circ 61. sec⁡310∘\sec 310^\circ For the following exercises, use identities to evaluate the expression. 62. If tan⁡(t)≈2.7\tan \left(t\right)\approx 2.7, and sin⁡(t)≈0.94\sin \left(t\right)\approx 0.94, find cos⁡(t)\cos \left(t\right). 63. If tan⁡(t)≈1.3\tan \left(t\right)\approx 1.3, and cos⁡(t)≈0.61\cos \left(t\right)\approx 0.61, find sin⁡(t)\sin \left(t\right). 64. If csc⁡(t)≈3.2\csc \left(t\right)\approx 3.2, and cos⁡(t)≈0.95\cos \left(t\right)\approx 0.95, find tan⁡(t)\tan \left(t\right). 65. If cot⁡(t)≈0.58\cot \left(t\right)\approx 0.58, and cos⁡(t)≈0.5\cos \left(t\right)\approx 0.5, find csc⁡(t)\csc \left(t\right). 66. Determine whether the function f(x)=2sin⁡xcos⁡xf\left(x\right)=2\sin x\cos x is even, odd, or neither. 67. Determine whether the function f(x)=3sin⁡2xcos⁡x+sec⁡xf\left(x\right)=3{\sin }^{2}x\cos x+\sec x is even, odd, or neither. 68. Determine whether the function f(x)=sin⁡x−2cos⁡2xf\left(x\right)=\sin x - 2{\cos }^{2}x is even, odd, or neither. 69. Determine whether the function f(x)=csc⁡2x+sec⁡xf\left(x\right)={\csc }^{2}x+\sec x is even, odd, or neither. For the following exercises, use identities to simplify the expression. 70. csc⁡ttan⁡t\csc t\tan t 71. sec⁡tcsc⁡t\frac{\sec t}{\csc t} 72. The amount of sunlight in a certain city can be modeled by the function h=15cos⁡(1600d)h=15\cos \left(\frac{1}{600}d\right), where hh represents the hours of sunlight, and dd is the day of the year. Use the equation to find how many hours of sunlight there are on February 10, the 42nd day of the year. State the period of the function. 73. The amount of sunlight in a certain city can be modeled by the function h=16cos⁡(1500d)h=16\cos \left(\frac{1}{500}d\right), where hh represents the hours of sunlight, and dd is the day of the year. Use the equation to find how many hours of sunlight there are on September 24, the 267th day of the year. State the period of the function. 74. The equation P=20sin⁡(2πt)+100P=20\sin \left(2\pi t\right)+100 models the blood pressure, PP, where tt represents time in seconds. (a) Find the blood pressure after 15 seconds. (b) What are the maximum and minimum blood pressures? 75. The height of a piston, hh, in inches, can be modeled by the equation y=2cos⁡x+6y=2\cos x+6, where xx represents the crank angle. Find the height of the piston when the crank angle is 55∘55^\circ . 76. The height of a piston, hh, in inches, can be modeled by the equation y=2cos⁡x+5y=2\cos x+5, where xx represents the crank angle. Find the height of the piston when the crank angle is 55∘55^\circ .

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