7.4 The Other Trigonometric Functions
A wheelchair ramp that meets the standards of the Americans with Disabilities Act must make an angle with the ground whose tangent is \(\frac{1}{12}\) or less, regardless of its length. A tangent represents a ratio, so this means that for every 1 inch of rise, the ramp must have 12 inches of run. Trigonometric functions allow us to specify the shapes and proportions of objects independent of exact dimensions. We have already defined the sine and cosine functions of an angle. Though sine and cosine are the trigonometric functions most often used, there are four others. Together they make up the set of six trigonometric functions. In this section, we will investigate the remaining functions.
7.4.1 Finding Exact Values of the Trigonometric Functions Secant, Cosecant, Tangent, and Cotangent
We can also define the remaining functions in terms of the unit circle with a point \(\left( {x,y} \right)\) corresponding to an angle of \(t,\) as shown in Figure 1. As with the sine and cosine, we can use the \(\left( {x,y} \right)\) 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 \(t\) is equal to \(\frac{y}{x},x \neq 0.\) Because the \(y\)-value is equal to the sine of \(t,\) and the \(x\)-value is equal to the cosine of \(t,\) the tangent of angle \(t\) can also be defined as \(\frac{\sin\; t}{\cos\; t},\cos\; t \neq 0.\) The tangent function is abbreviated as \(\text{tan}\text{.}\) 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 \(t\) is equal to \(\frac{1}{\cos\; t} = \frac{1}{x},x \neq 0.\) The secant function is abbreviated as \(\text{sec}\text{.}\)
- The cotangent function is the reciprocal of the tangent function. In Figure 1, the cotangent of angle \(t\) is equal to \(\frac{\cos\; t}{\sin\; t} = \frac{x}{y},y \neq 0.\) The cotangent function is abbreviated as \(\text{cot}\text{.}\)
- The cosecant function is the reciprocal of the sine function. In Figure 1, the cosecant of angle \(t\) is equal to \(\frac{1}{\sin\; t} = \frac{1}{y},y \neq 0.\) The cosecant function is abbreviated as \(\text{csc}\text{.}\)
The point \(\left( {- \frac{\sqrt{3}}{2},\frac{1}{2}} \right)\) is on the unit circle, as shown in Figure 2. Find \(\sin\; t,\cos\; t,\tan\; t,\sec\; t,\csc\; t,\) and \(\cot\; t.\)
Figure 2
Solution (click to reveal)
Because we know the \((x,y)\) coordinates of the point on the unit circle indicated by angle \(t,\) we can use those coordinates to find the six functions:
\[\begin{array}{rllccc} {\text{sin~}t} & {= y} & {= \frac{1}{2}} & & & \\ {\text{cos~}t} & {= x} & {= - \frac{\sqrt{3}}{2}} & & & \\ {\text{tan~}t} & {= \frac{y}{x}} & {= \frac{\frac{1}{2}}{- \frac{\sqrt{3}}{2}}} & {= \frac{1}{2}\left( {- \frac{2}{\sqrt{3}}} \right)} & {= - \frac{1}{\sqrt{3}}} & {= - \frac{\sqrt{3}}{3}} \\ {\text{sec~}t} & {= \frac{1}{x}} & {= \frac{1}{- \frac{\sqrt{3}}{2}}} & {= - \frac{2}{\sqrt{3}}} & {= - \frac{2\sqrt{3}}{3}} & \\ {\text{csc~}t} & {= \frac{1}{y}} & {= \frac{1}{\frac{1}{2}}} & {= 2} & & \\ {\text{cot~}t} & {= \frac{x}{y}} & {= \frac{- \frac{\sqrt{3}}{2}}{\frac{1}{2}}} & {= - \frac{\sqrt{3}}{2}\left( \frac{2}{1} \right)} & {= - \sqrt{3}} & \end{array}\]
The point \(\left( {\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}} \right)\) is on the unit circle, as shown in Figure 3. Find \(\sin\; t,\cos\; t,\tan\; t,\sec\; t,\csc\; t,\) and \(\cot\; t.\)
Figure 3
Solution (click to reveal)
\(\sin t = - \frac{\sqrt{2}}{2}\cos t = \frac{\sqrt{2}}{2},\tan t = - 1,sect = \sqrt{2},\csc t = - \sqrt{2},\cot t = - 1\)
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 \(x\) equal to the cosine and \(y\) equal to the sine and then using the definitions of tangent, secant, cosecant, and cotangent. The results are shown in Table 1.
| Angle | \(0\) | \(\frac{\pi}{6},\text{or~30°}\) | \(\frac{\pi}{4},\text{or~45°}\) | \(\frac{\pi}{3},\text{or~60°}\) | \(\frac{\pi}{2},\text{or~90°}\) |
|---|---|---|---|---|---|
| Cosine | 1 | \(\frac{\sqrt{3}}{2}\) | \(\frac{\sqrt{2}}{2}\) | \(\frac{1}{2}\) | 0 |
| Sine | 0 | \(\frac{1}{2}\) | \(\frac{\sqrt{2}}{2}\) | \(\frac{\sqrt{3}}{2}\) | 1 |
| Tangent | 0 | \(\frac{\sqrt{3}}{3}\) | 1 | \(\sqrt{3}\) | Undefined |
| Secant | 1 | \(\frac{2\sqrt{3}}{3}\) | \(\sqrt{2}\) | 2 | Undefined |
| Cosecant | Undefined | 2 | \(\sqrt{2}\) | \(\frac{2\sqrt{3}}{3}\) | 1 |
| Cotangent | Undefined | \(\sqrt{3}\) | 1 | \(\frac{\sqrt{3}}{3}\) | 0 |
Table 1
7.4.2 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 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 “\(\mathbf{A}\),” \(\mathbf{a}\)ll of the six trigonometric functions are positive. In quadrant II, “Smart,” only \(\mathbf{s}\)ine and its reciprocal function, cosecant, are positive. In quadrant III, “Trig,” only \(\mathbf{t}\)angent and its reciprocal function, cotangent, are positive. Finally, in quadrant IV, “Class,” only \(\mathbf{c}\)osine and its reciprocal function, secant, are positive.
Figure 4 The trigonometric functions are each listed in the quadrants in which they are positive.
7.4.3 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 \(f(x) = x^{2},\) 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: \({(4)}^{2} = {(-4)}^{2},{(-5)}^{2} = {(5)}^{2},\) and so on. So \(f(x) = x^{2}\) is an even function, a function such that two inputs that are opposites have the same output. That means \(f\left( {- x} \right) = f(x).\)
Figure 5 The function \(f(x) = x^{2}\) is an even function.
Now consider the function \(f(x) = x^{3},\) 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 \(f(x) = x^{3}\) is an odd function, one such that two inputs that are opposites have outputs that are also opposites. That means \(f\left( {- x} \right) = -f(x).\)
Figure 6 The function \(f(x) = x^{3}\) 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 \(y.\) 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 Table 2.
Figure 7
| \(\begin{array}{rcc} {\text{sin~}t} & = & y \\ {\text{sin}(-t)} & = & {-y} \\ {\text{sin~}t} & \neq & {\text{sin}(-t)} \end{array}\) | \(\begin{array}{rcl} {\text{cos~}t} & = & x \\ {\text{cos}(-t)} & = & x \\ {\text{cos~}t} & = & {\text{cos}(-t)} \end{array}\) | \(\begin{array}{rcl} {\text{tan}(t)} & = & \frac{y}{x} \\ {\text{tan}(-t)} & = & {- \frac{y}{x}} \\ {\text{tan~}t} & \neq & {\text{tan}(-t)} \end{array}\) |
| \(\begin{array}{rcl} {\text{sec~}t} & = & \frac{1}{x} \\ {\text{sec}(-t)} & = & \frac{1}{x} \\ {\text{sec~}t} & = & {\text{sec}(-t)} \end{array}\) | \(\begin{array}{rcl} {\text{csc~}t} & = & \frac{1}{y} \\ {\text{csc}(-t)} & = & \frac{1}{-y} \\ {\text{csc~}t} & \neq & {\text{csc}(-t)} \end{array}\) | \(\begin{array}{rcl} {\text{cot~}t} & = & \frac{x}{y} \\ {\text{cot}(-t)} & = & \frac{x}{- y} \\ {\text{cot~}t} & \neq & {\text{cot}(-t)} \end{array}\) |
Table 2
7.4.4 Recognizing and Using 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 alternate forms of the Pythagorean Identity, \(\cos^{2}t + \sin^{2}t = 1.\) One form is obtained by dividing both sides by \(\cos^{2}t.\)
\[\begin{array}{rcl} {\frac{\cos^{2}t}{\cos^{2}t} + \frac{\sin^{2}t}{\cos^{2}t}} & = & \frac{1}{\cos^{2}t} \\ {1 + \tan^{2}t} & = & {\sec^{2}t} \end{array}\]
The other form is obtained by dividing both sides by \(\sin^{2}t.\)
\[\begin{array}{rcl} {\frac{\cos^{2}t}{\sin^{2}t} + \frac{\sin^{2}t}{\sin^{2}t}} & = & \frac{1}{\sin^{2}t} \\ {\cot^{2}t + 1} & = & {\csc^{2}t} \end{array}\]
If \(\cos(t) = \frac{12}{13}\) and \(t\) is in quadrant IV, as shown in Figure 8, find the values of the other five trigonometric functions.
Figure 8
Solution (click to reveal)
We can find the sine using the Pythagorean Identity, \(\cos^{2}t + \sin^{2}t = 1,\) and the remaining functions by relating them to sine and cosine.
\[\begin{array}{rcl} {\left( \frac{12}{13} \right)^{2} + \sin^{2}t} & = & 1 \\ {\sin^{2}t} & = & {1 - \left( \frac{12}{13} \right)^{2}} \\ {\sin^{2}t} & = & {1 - \frac{144}{169}} \\ {\sin^{2}t} & = & \frac{25}{169} \\ {\text{sin~}t} & = & {\pm \sqrt{\frac{25}{169}}} \\ {\text{sin~}t} & = & {\pm \frac{\sqrt{25}}{\sqrt{169}}} \\ {\text{sin~}t} & = & {\pm \frac{5}{13}} \end{array}\]
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, \(- \frac{5}{13}.\)
The remaining functions can be calculated using identities relating them to sine and cosine.
\[\begin{array}{rlll} {\text{tan~}t} & {= \frac{\text{sin~}t}{\text{cos~}t}} & {= \frac{- \frac{5}{13}}{\frac{12}{13}}} & {= - \frac{5}{12}} \\ {\text{sec~}t} & {= \frac{1}{\text{cos~}t}} & {= \frac{1}{\frac{12}{13}}} & {= \frac{13}{12}} \\ {\text{csc~}t} & {= \frac{1}{\text{sin~}t}} & {= \frac{1}{- \frac{5}{13}}} & {= \frac{- 13}{5}} \\ {\text{cot~}t} & {= \frac{1}{\text{tan~}t}} & {= \frac{1}{- \frac{5}{12}}} & {= - \frac{12}{5}} \end{array}\]
As we discussed at the beginning of the chapter, 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 \(2\pi,\) 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 \(x\) represents the length time, measured in years, and \(f(x)\) represents the number of days in February, then \(f(x + 4) = f(x).\) This pattern repeats over and over through time. In other words, every four years (except for multiples of 100), February typically has 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.
Find the values of the six trigonometric functions of angle \(t\) based on Figure 9.
Figure 9
Solution (click to reveal)
\[\begin{array}{rllc} {\text{sin~}t} & {= y} & {= - \frac{\sqrt{3}}{2}} & \\ {\text{cos~}t} & {= x} & {= - \frac{1}{2}} & \\ {\text{tan~}t} & {= \frac{\text{sin~}t}{\text{cos~}t}} & {= \frac{- \frac{\sqrt{3}}{2}}{- \frac{1}{2}}} & {= \sqrt{3}} \\ {\text{sec~}t} & {= \frac{1}{\text{cos~}t}} & {= \frac{1}{- \frac{1}{2}}} & {= -2} \\ {\text{csc~}t} & {= \frac{1}{\text{sin~}t}} & {= \frac{1}{- \frac{\sqrt{3}}{2}}} & {= - \frac{2\sqrt{3}}{3}} \\ {\text{cot~}t} & {= \frac{1}{\text{tan~}t}} & {= \frac{1}{\sqrt{3}}} & {= \frac{\sqrt{3}}{3}} \end{array}\]
Find the values of the six trigonometric functions of angle \(t\) based on Figure 10.
Figure 10
Solution (click to reveal)
\(\begin{array}{l} {\sin t = - 1,\cos t = 0,\tan t = \text{Undefined}} \\ {\sec t = \text{Undefined,}\csc t = - 1,\cot t = 0} \end{array}\)
7.4.5 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 \(\frac{\pi}{180}\) to convert the degrees to radians. To find the secant of \(30{^\circ},\) we could press
\[\begin{matrix} {\text{(for~a~scientific~calculator):}\;\frac{1}{30\; \times \;\frac{\pi}{180}}\;\text{COS}} \\ \text{or} \\ {\text{(for~a~graphing~calculator):}\;\frac{1}{\cos\left( \frac{30\pi}{180} \right)}} \end{matrix}\]
Section Exercises
Verbal
1. On an interval of \(\left\lbrack {0,2\pi} \right),\) can the sine and cosine values of a radian measure ever be equal? If so, where?
Solution (click to reveal)
Yes, when the reference angle is \(\frac{\pi}{4}\) and the terminal side of the angle is in quadrants I and III. Thus, a \(x = \frac{\pi}{4},\frac{5\pi}{4},\) the sine and cosine values are equal.
2. What would you estimate the cosine of \(\pi\) 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?
Solution (click to reveal)
Substitute the sine of the angle in for \(y\) in the Pythagorean Theorem \(x^{2} + y^{2} = 1.\) Solve for \(x\) and take the negative solution.
4. Describe the secant function.
5. Tangent and cotangent have a period of \(\pi\text{.}\) What does this tell us about the output of these functions?
Solution (click to reveal)
The outputs of tangent and cotangent will repeat every \(\pi\) units.
Algebraic
For the following exercises, find the exact value of each expression.
6. \(\tan\;\frac{\pi}{6}\)
7. \(\sec\;\frac{\pi}{6}\)
Solution (click to reveal)
\(\frac{2\sqrt{3}}{3}\)
8. \(\csc\;\frac{\pi}{6}\)
9. \(\cot\;\frac{\pi}{6}\)
Solution (click to reveal)
\(\sqrt{3}\)
10. \(\tan\;\frac{\pi}{4}\)
11. \(\sec\;\frac{\pi}{4}\)
Solution (click to reveal)
\(\sqrt{2}\)
12. \(\csc\;\frac{\pi}{4}\)
13. \(\cot\;\frac{\pi}{4}\)
Solution (click to reveal)
1
14. \(\tan\;\frac{\pi}{3}\)
15. \(\sec\;\frac{\pi}{3}\)
Solution (click to reveal)
2
16. \(\csc\;\frac{\pi}{3}\)
17. \(\cot\;\frac{\pi}{3}\)
Solution (click to reveal)
\(\frac{\sqrt{3}}{3}\)
For the following exercises, use reference angles to evaluate the expression.
18. \(\tan\;\frac{5\pi}{6}\)
19. \(\sec\;\frac{7\pi}{6}\)
Solution (click to reveal)
\(- \frac{2\sqrt{3}}{3}\)
20. \(\csc\;\frac{11\pi}{6}\)
21. \(\cot\;\frac{13\pi}{6}\)
Solution (click to reveal)
\(\sqrt{3}\)
22. \(\tan\;\frac{7\pi}{4}\)
23. \(\sec\;\frac{3\pi}{4}\)
Solution (click to reveal)
\(- \sqrt{2}\)
24. \(\csc\;\frac{5\pi}{4}\)
25. \(\cot\;\frac{11\pi}{4}\)
Solution (click to reveal)
–1
26. \(\tan\;\frac{8\pi}{3}\)
27. \(\sec\;\frac{4\pi}{3}\)
Solution (click to reveal)
-2
28. \(\csc\;\frac{2\pi}{3}\)
29. \(\cot\;\frac{5\pi}{3}\)
Solution (click to reveal)
\(- \frac{\sqrt{3}}{3}\)
30. \(\tan\; 225{^\circ}\)
31. \(\sec\; 300{^\circ}\)
Solution (click to reveal)
2
32. \(\csc\; 150{^\circ}\)
33. \(\cot\; 240{^\circ}\)
Solution (click to reveal)
\(\frac{\sqrt{3}}{3}\)
34. \(\tan\; 330{^\circ}\)
35. \(\sec\; 120{^\circ}\)
Solution (click to reveal)
–2
36. \(\csc\; 210{^\circ}\)
37. \(\cot\; 315{^\circ}\)
Solution (click to reveal)
–1
38. If \(\text{sin}\; t = \frac{3}{4},\) and \(t\) is in quadrant II, find \(\cos\; t,\sec\; t,\csc\; t,\tan\; t,\) and \(\cot\; t.\)
39. If \(\text{cos}\; t = - \frac{1}{3},\) and \(t\) is in quadrant III, find \(\sin\; t,\sec\; t,\csc\; t,\tan\; t,\) and \(\cot\; t.\)
Solution (click to reveal)
\(\sin\; t = - \frac{2\sqrt{2}}{3}\), \(\sec\; t = - 3\), \(\csc\; t = - \frac{3\sqrt{2}}{4}\), \(\tan\; t = 2\sqrt{2}\), \(\cot\; t = \frac{\sqrt{2}}{4}\)
40. If \(\tan\; t = \frac{12}{5}\), and \(0 \leq t < \frac{\pi}{2}\), find \(\sin\; t,\cos\; t,\sec\; t,\csc\; t,\text{and}\;\cot\; t.\)
41. If \(\sin\; t = \frac{\sqrt{3}}{2}\) and \(\cos\; t = \frac{1}{2},\) find \(\sec\; t,\csc\; t,\tan\; t,\) and \(\cot\; t.\)
Solution (click to reveal)
\(\sec t = 2,\) \(\csc t = \frac{2\sqrt{3}}{3},\) \(\tan t = \sqrt{3},\) \(\cot t = \frac{\sqrt{3}}{3}\)
42. If \(\sin\; 40{^\circ} \approx 0.643\) and \(\cos\; 40{^\circ} \approx 0.766,\) find \(\text{sec}\; 40{^\circ},\text{csc}\; 40{^\circ},\text{tan}\; 40{^\circ},\) and \(\text{cot}\; 40{^\circ}.\)
43. If \(\text{sin}\; t = \frac{\sqrt{2}}{2},\) what is the \(\text{sin}(-t)?\)
Solution (click to reveal)
\(- \frac{\sqrt{2}}{2}\)
44. If \(\text{cos}\; t = \frac{1}{2},\) what is the \(\text{cos}(-t)?\)
45. If \(\text{sec}\; t = 3.1,\) what is the \(\text{sec}(-t)?\)
Solution (click to reveal)
3.1
46. If \(\text{csc}\; t = 0.34,\) what is the \(\text{csc}(-t)?\)
47. If \(\text{tan}\; t = -1.4,\) what is the \(\text{tan}(-t)?\)
Solution (click to reveal)
1.4
48. If \(\text{cot}\; t = 9.23,\) what is the \(\text{cot}(-t)?\)
Graphical
For the following exercises, use the angle in the unit circle to find the value of the each of the six trigonometric functions.
49.
Solution (click to reveal)
\(\sin t = \frac{\sqrt{2}}{2}\), \(\cos t = \frac{\sqrt{2}}{2}\), \(\tan t = 1\), \(\cot t = 1\), \(\sec t = \sqrt{2}\), \(\csc t = \sqrt{2}\)
50.
51.
Solution (click to reveal)
\(\sin t = - \frac{\sqrt{3}}{2}\), \(\cos t = - \frac{1}{2}\)\(\tan t = \sqrt{3}\), \(\cot t = \frac{\sqrt{3}}{3}\), \(\sec t = - 2\), \(\csc t = - \frac{2\sqrt{3}}{3}\)
Technology
For the following exercises, use a graphing calculator to evaluate to three decimal places.
52. \(\csc\;\frac{5\pi}{9}\)
53. \(\cot\;\frac{4\pi}{7}\)
Solution (click to reveal)
–0.228
54. \(\sec\;\frac{\pi}{10}\)
55. \(\tan\;\frac{5\pi}{8}\)
Solution (click to reveal)
–2.414
56. \(\sec\;\frac{3\pi}{4}\)
57. \(\csc\;\frac{\pi}{4}\)
Solution (click to reveal)
1.414
58. \(\text{tan}\; 98{^\circ}\)
59. \(\cot\; 33{^\circ}\)
Solution (click to reveal)
1.540
60. \(\cot\; 140{^\circ}\)
61. \(\sec\; 310{^\circ}\)
Solution (click to reveal)
1.556
Extensions
For the following exercises, use identities to evaluate the expression.
62. If \(\tan(t) \approx 2.7,\) and \(\sin(t) \approx 0.94,\) find \(\cos(t).\)
63. If \(\tan(t) \approx 1.3,\) and \(\cos(t) \approx 0.61,\) find \(\sin(t).\)
Solution (click to reveal)
\(\sin(t) \approx 0.79\)
64. If \(\csc(t) \approx 3.2,\) and \(\cos(t) \approx 0.95,\) find \(\tan(t).\)
65. If \(\cot(t) \approx 0.58,\) and \(\cos(t) \approx 0.5,\) find \(\csc(t).\)
Solution (click to reveal)
\(\csc t \approx 1.16\)
66. Determine whether the function \(f(x) = 2\sin x\;\cos\; x\) is even, odd, or neither.
67. Determine whether the function \(f(x) = 3\sin^{2}x\;\cos\; x + \sec\; x\) is even, odd, or neither.
Solution (click to reveal)
even
68. Determine whether the function \(f(x) = \sin\; x - 2\cos^{2}x\) is even, odd, or neither.
69. Determine whether the function \(f(x) = \csc^{2}x + \sec\; x\) is even, odd, or neither.
Solution (click to reveal)
even
For the following exercises, use identities to simplify the expression.
70. \(\csc\; t\;\tan\; t\)
71. \(\frac{\sec\; t}{\csc\; t}\)
Solution (click to reveal)
\(\frac{\sin\; t}{\cos\; t} = \tan\; t\)
Real-World Applications
72. The amount of sunlight in a certain city can be modeled by the function \(h = 15\cos\left( {\frac{1}{600}d} \right),\) where \(h\) represents the hours of sunlight, and \(d\) is the day of the year. Use the equation to find how many hours of sunlight there are on February 11, 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 \(h = 16\cos\left( {\frac{1}{500}d} \right),\) where \(h\) represents the hours of sunlight, and \(d\) 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.
Solution (click to reveal)
13.77 hours, period: \(1000\pi\)
74. The equation \(P = 20\sin\left( {2\pi t} \right) + 100\) models the blood pressure, \(P,\operatorname{}\) where \(t\) 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, \(h,\) in inches, can be modeled by the equation \(y = 3\sin\; x + 1,\) where \(x\) represents the crank angle. Find the height of the piston when the crank angle is \(55{^\circ}.\)
Solution (click to reveal)
3.46 inches
76. The height of a piston, \(h,\) in inches, can be modeled by the equation \(y = 2\cos\; x + 5,\) where \(x\) represents the crank angle. Find the height of the piston when the crank angle is \(55{^\circ}.\)