Angles
Trigonometric Angles
A trigonometric angle is determined by rotating a ray about its endpoint, called the vertex of the angle. The starting position of the ray is called the initial side and the ending position is the terminal side. (See Fig. 22-1.)
If the displacement of the ray from its starting position is in the counterclockwise direction, the angle is assigned a positive measure, if in the clockwise direction, a negative measure. A zero angle corresponds to zero displacement; the initial and terminal sides of a zero angle are coincident.
Angles in Standard Position
An angle is in standard position in a Cartesian coordinate system if its vertex is at the origin and its initial side is the positive x-axis. Angles in standard position are categorized by their terminal sides: If the terminal side falls along an axis, the angle is called a quadrantal angle; if the terminal side is in quadrant n, the angle is referred to as a quadrant n angle (see Figs. 22-2 to 22-5).
| Positive Quadrantal Angle | Negative Quadrantal Angle | Positive Quadrant IV Angle | Negative Quadrant II Angle |
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Radian Measurement of Angles
In calculus, angles are normally measured in radian measure. One radian is defined as the measure of an angle that, if placed with vertex at the center of a circle, subtends (intersects) an arc of length equal to the radius of the circle. In Fig. 22-6, angle $ $ has measure 1 radian.
Since the circumference of a circle of radius r has length $ 2r $ , a positive angle of one full revolution corresponds to an arc length of $ 2r $ and thus has measure $ 2$ radians.
EXAMPLE 22.1 Draw examples of angles of measures $ $ , $ $ , and $ $ radians.
Arc Length and Radian Measure
In a circle of radius r, an angle of radian measure $ $ subtends an arc of length $ s = r$ .
EXAMPLE 22.2 Determine the radius of a circle in which a central angle of 3 radians subtends an arc of length 30 cm. Since $ = 3 $ and $ s = 30 , cm $ , $ 30 , cm = 3r $ ; hence $ r = 10 , cm $ .
Degree Measure
In applications, angles are commonly measured in degrees ( $ ^{} $ ). A positive angle of one full revolution has measure $ 360^{} $ . Thus, $ 2$ radians = $ 360^{} $ , or
\[ 180^{\circ}=\pi\;r a d i a n s \]
To transform radian measure into degrees, use this relation in the form $ 180^{}/= 1 $ radian and multiply the radian measure by $ 180^{}/$ . To transform degree measure into radians, use the relation in the form $ 1^{} = /180 $ radians and multiply the degree measure by $ /180^{} $ . The following table summarizes the measure of common angles:
| Degree Measure | 0° | 30° | 45° | 60° | 90° | 120° | 135° | 150° | 180° | 270° | 360° |
| Radian Measure | 0 | $ \(</td><td>\) \(</td><td>\) \(</td><td>\) \(</td><td>\) \(</td><td>\) \(</td><td>\) \(</td><td>\) \(</td><td>\) $ | 2 $ $ |
EXAMPLE 22.3 (a) Transform $ 210^{} $ into radians. (b) Transform $ 6$ radians into degrees.
- $ 210^{} = 210^{} $ radians $ = $ radians; (b) $ 6$ radians $ = 6 = 1080^{} $
Degrees, Minutes, and Seconds
If measurements smaller than a degree are required, the degree may be subdivided into decimal fractions. Alternatively, a degree is subdivided into minutes (‘) and seconds (’‘). Thus, $ 1^{} = 60’ $ and $ 1’ = 60’’ $ ; hence, $ 1^{} = 3600’’ $ .
EXAMPLE 22.4 Transform $ 35^{}24’36’’ $ into decimal degrees.
\[ 35^{\circ}24^{\prime}36^{\prime \prime}=\left(35+\frac{24}{60}+\frac{36}{3600}\right)^{\circ}=35.41^{\circ} \]
Terminology for Special Angles
An angle of measure between 0 and $ /2 $ radians (between $ 0^{} $ and $ 90^{} $ ) is called an acute angle. An angle of measure $ /2 $ radians ( $ 90^{} $ ) is called a right angle. An angle of measure between $ /2 $ and $ $ radians (between $ 90^{} $ and $ 180^{} $ ) is called an obtuse angle. An angle of measure $ $ radians ( $ 180^{} $ ) is called a straight angle. An angle is normally referred to by giving its measure; thus $ = 30^{} $ means that $ $ has a measure of $ 30^{} $ .
Complementary and Supplementary Angles
If \(\alpha\) and \(\beta\) are two angles such that \(\alpha + \beta = \pi/2\), \(\alpha\) and \(\beta\) are called complementary angles. If \(\alpha\) and \(\beta\) are two angles such that \(\alpha + \beta = \pi\), \(\alpha\) and \(\beta\) are called supplementary angles.
EXAMPLE 22.5 Find an angle complementary to $ $ if (a) $ = /3 $ ; (b) $ = 37^{}15’ $ .
The complementary angle to θ is $ - = - = $ .
The complementary angle to $ $ is $ 90{}-{}-37{}15{}=89{}60{}-37{}15{}=52{}45{} $
Coterminal Angles
Two angles in standard position are coterminal if they have the same terminal side. There are an infinite number of angles coterminal with a given angle. To find an angle coterminal with a given angle, add or subtract $ 2$ (if the angle is measured in radians) or $ 360^{} $ (if the angle is measured in degrees).
EXAMPLE 22.6 Find two angles coterminal with (a) 2 radians; (b) $ -60^{} $
Coterminal with 2 radians are $ 2 + 2$ and $ 2 - 2$ radians, as well as many other angles.
Coterminal with $ -60^{} $ are $ -60^{} + 360^{} = 300^{} $ and $ -60^{} - 360^{} = -420^{} $ , as well as many other angles.
Trigonometric Functions of Angles
If $ $ is an angle with radian measure t, then the value of each trigonometric function of $ $ is its value at the real number t.
EXAMPLE 22.7 Find (a) cos 90°; (b) tan 135°.
\[ \left(a\right)\cos90^{\circ}=\cos\frac{\pi}{2}=0;\\\left(b\right)\tan135^{\circ}=\tan\left(135^{\circ}\cdot\frac{\pi}{180^{\circ}}\right)=\tan\frac{3\pi}{4}=-1 \]
Trigonometric Functions of Angles as Ratios
Let $ $ be an angle in standard position, and $ P(x, y) $ be any point except the origin on the terminal side of $ $ . If $ r = $ is the distance from P to the origin, then the six trigonometric functions of $ $ are given by:
\[ \sin\theta=\frac{y}{r} \]
\[ \csc\theta=\frac{r}{y}\quad(ify\neq0) \]
\[ \cos\theta=\frac{x}{r} \]
\[ \sec\theta=\frac{r}{x}\quad(if x\neq0) \]
\[ \tan\theta=\frac{y}{x}\quad(if x\neq0) \]
\[ \cot\theta=\frac{x}{y}\quad(ify\neq0) \]
EXAMPLE 22.8 Let $ $ be an angle in standard position with $ P(-3, 4) $ a point on the terminal side of $ $ (see Fig. 22-8). Find the six trigonometric functions of $ $ .
\[ x=-3,y=4,r=\sqrt{x^{2}+y}=\sqrt{(-3)^{2}+4^{2}}=5;hence \]
\[ \sin\theta=\frac{y}{r}=\frac{4}{5} \]
\[ \cos\theta=\frac{x}{r}=\frac{-3}{5}=-\frac{3}{5} \]
\[ \tan\theta=\frac{y}{x}=\frac{4}{-3}=-\frac{4}{3} \]
\[ \csc\theta=\frac{r}{y}=\frac{5}{4}\qquad\sec\theta=\frac{r}{x}=\frac{5}{-3}=-\frac{5}{3} \]
\[ \cot\theta=\frac{x}{y}=\frac{-3}{4}=-\frac{3}{4} \]
Trigonometric Functions of Acute Angles
If $ $ is an acute angle, it can be regarded as an angle of a right triangle. If $ $ is placed in standard position, and the sides of the right triangle are named as hypotenuse (hyp), opposite (opp), and adjacent (adj), the lengths of the adjacent and opposite sides are the x- and y-coordinates, respectively, of a point on the terminal side of the angle. The length of the hypotenuse is $ r = $ . (See Fig. 22-9.)
For an acute angle \(\theta\), the trigonometric functions of \(\theta\) are then as follows:
\[ \sin\theta=\frac{y}{r}=\frac{opp}{hyp} \]
\[ \csc\theta=\frac{r}{y}=\frac{hyp}{opp} \]
\[ \cos\theta=\frac{x}{r}=\frac{adj}{hyp} \]
\[ \sec\theta=\frac{r}{x}=\frac{hyp}{adj} \]
\[ \tan\theta=\frac{y}{x}=\frac{opp}{adj} \]
\[ \cot\theta=\frac{x}{y}=\frac{adj}{opp} \]
EXAMPLE 22.9 Find the six trigonometric functions of $ $ as shown in Fig. 22-10.
For $ $ as shown, opp = 5, adj = 12, hyp = 13, hence
\[ \sin\theta=\frac{opp}{hyp}=\frac{5}{13} \]
\[ \cos\theta=\frac{adj}{hyp}=\frac{12}{13} \]
\[ \tan\theta=\frac{opp}{adj}=\frac{5}{12} \]
\[ \csc\theta=\frac{hyp}{opp}=\frac{13}{5} \]
\[ \sec\theta=\frac{hyp}{adj}=\frac{13}{12} \]
\[ \cot\theta=\frac{adj}{opp}=\frac{12}{5} \]
Reference Angles
The reference angle for $ $ , a nonquadrantal angle in standard position, is the acute angle $ _{R} $ between the x-axis and the terminal side of $ $ . Fig. 22-11 shows angles and reference angles for cases $ 0 < < 2$ . To find reference angles for other nonquadrantal angles, first add or subtract multiples of $ 2$ to obtain an angle coterminal with $ $ that satisfies $ 0 < < 2$ .
| Quadrant I | Quadrant II | Quadrant III | Quadrant IV |
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| θR = θ | θR = π - θ = 180° - θ | θR = θ - π = θ - 180° | θR = 2π - θ = 360° - θ |
Trigonometric Functions of Angles in Terms of Reference Angles
For any nonquadrantal angle $ $ , each trigonometric function of $ $ has the same absolute value as the same trigonometric function of $ {R} $ . To find a trigonometric function of $ $ , find the function of $ {R} $ , then apply the correct sign for the quadrant of $ $ .
EXAMPLE 22.10 Find cos\(\frac{3\pi}{4}\).
The reference angle for $ $ , a second quadrant angle, is $ - = $ . In quadrant II, the sign of the cosine function is negative. Hence, $ = - = - $ .
SOLVED PROBLEMS
22.1. List all angles coterminal with (a) $ 40^{} $ ; (b) $ $ radians.
To find angles coterminal with $ 40^{} $ , add or subtract any integer multiple of $ 360^{} $ . Thus, $ 400^{} $ and $ -320^{} $ are examples of angles coterminal with $ 40^{} $ , and all angles coterminal with $ 40^{} $ can be expressed as $ 40^{} + n360^{} $ , where n is any integer.
To find angles coterminal with $ 2/3 $ , add or subtract any integer multiple of $ 2$ . Thus, $ 8/3 $ and $ -4/3 $ are examples of angles coterminal with $ 2/3 $ and all angles coterminal with $ 2/3 $ can be expressed as $ 2/3 + 2n $ , where n is any integer.
22.2. Find the trigonometric functions of (a) $ 180^{} $ ; (b) $ -360^{} $ .
\(180^{\circ} = \pi\) radians; hence, \(\sin 180^{\circ} = \sin \pi = 0\), \(\cos 180^{\circ} = \cos \pi = -1\), \(\tan 180^{\circ} = \tan \pi = 0\), \(\cot 180^{\circ} = \cot \pi\) is undefined, \(\sec 180^{\circ} = \sec \pi = -1\), \(\csc 180^{\circ} = \csc \pi\) is undefined.
$ -360^{} = -2$ radians, hence, $ (-360^{}) = (-2) = 0 $ , $ (-360^{}) = (-2) = 1 $ ,
$ (-360^{}) = (-2) = 0 $ , $ (-360^{}) = (-2) $ is undefined, $ (-360^{}) = (-2) = 1 $ ,
$ (-360^{}) = (-2) $ is undefined.
22.3. Find an angle supplementary to $ $ if (a) $ = /3 $ ; (b) $ = 37^{}15’ $ .
Supplementary to $ $ is $ - = $ .
Supplementary to 37°15’ is 180° − 37°15’ = 179°60’ − 37°15’ = 142°45’.
22.4. Transform 5 radians into degrees, minutes, and seconds.
First note that 5 radians = $ 5 = ^{} $ . To transform this into degrees and minutes, write
\[ 286.4789^{\circ}=286^{\circ}+\frac{4789^{\circ}}{10000}=286^{\circ}+\frac{4789^{\circ}}{10000}\cdot\frac{60^{\prime}}{1^{\circ}}=286^{\circ}+28.734^{\prime} \]
To transform this into degrees, minutes, and seconds, write
\[ 286^{\circ}+28.734^{\prime}=286^{\circ}+28^{\prime}+\frac{734^{\prime}}{1000}=286^{\circ}+28^{\prime}+\frac{734^{\prime}}{1000}\cdot\frac{60^{\prime \prime}}{1^{\prime}}=286^{\circ}28^{\prime}44.031^{\prime \prime} \]
22.5. Transform $ 424^{}34’24’’ $ into radians.
First note that $ 424^{}34’24’’ $ = $ (424 + + )^{} $ ≈ $ 424.57333^{} $ . To transform this into radians, write
\[ 424.57333^{\circ}=424.57333^{\circ}\cdot\frac{\pi}{180^{\circ}}\approx7.41radians. \]
22.6. (a) Derive the relationship $ s = r$ . (b) Find the angle in radians subtended by an arc of length 5 cm on a circle of radius 3 cm. (c) Find the linear distance traveled by a point on the rim of a bicycle wheel of radius 26 in as the wheel makes 10 rotations.
- Draw two circles of radius r, as shown in Fig. 22-12.
From plane geometry it is known that the ratio of the arc lengths equals the ratio of the angles. Thus
\[ \frac{s}{s_{1}}=\frac{\theta}{\theta_{1}} \]
Take $ {1}=1 $ radian, then $ s{1}=r $ ; hence, $ = $ , that is, $ s=r$ .
Use $ s = r$ with $ s = 5 , cm $ , $ r = 3 , cm $ , then $ 5 = 3$ ; thus $ = $ radians.
First note that 10 rotations represents an angle of $ 10 = 20$ radians. Hence,
\[ s=r\theta=26in\cdot20\pi radians=520\pi in\approx136ft. \]
22.7. Show that the definitions of the trigonometric functions as ratios are consistent with the definitions of the trigonometric functions of angles.
Let $ $ be a nonquadrantal angle in standard position. Choose an arbitrary point $ Q(x, y) $ on the terminal side of $ $ . (See Fig. 22-13).
Then $ r = $ . Let $ P(x_{1}, y_{1}) $ be a point on the terminal side of $ $ with $ = 1 $ . Then P lies on the unit circle and $ = y_{1} $ . Drop perpendicular lines from P and Q to the x axis at A and B, respectively. Then triangles OAP and OBQ are similar, and hence ratios of corresponding sides are equal; thus
\[ \frac{\left|y\right|}{r}=\frac{\left|y_{1}\right|}{1} \]
Since y and $ y_{1} $ have the same sign, it follows that
\[ y_{1}=\frac{y}{r} \]
Thus \(\sin\theta = y_{1} = \frac{y}{r}\) and the two definitions are consistent for the sine function. The proof is easily extended to the other trigonometric functions and to quadrantal angles.
22.8. If $ $ is in standard position and $ (-20,21) $ lies on its terminal side, find the trigonometric functions of $ $ .
\[ x=-20\text{and}y=21\text{hence}r=\sqrt{x^{2}+y^{2}}=\sqrt{(-20)^{2}+21^{2}}=29. \]
\[ \sin\theta=\frac{y}{r}=\frac{21}{29} \]
\[ \cos\theta=\frac{x}{r}=\frac{-20}{29}=-\frac{20}{29} \]
\[ \tan\theta=\frac{y}{x}=\frac{21}{-20}=-\frac{21}{20} \]
\[ \csc\theta=\frac{r}{y}=\frac{29}{21} \]
\[ \sec\theta=\frac{r}{x}=\frac{29}{-20}=-\frac{29}{20} \]
\[ \cot\theta=\frac{x}{y}=\frac{-20}{21}=-\frac{20}{21} \]
22.9. If $ $ is in standard position and its terminal side lies in quadrant I on the line y = 2x, find the trigonometric functions of $ $ .
To find the trigonometric functions of \(\theta\), any point on the terminal side of \(\theta\) may be chosen; let \(x=1\), then \(y=2\) and \(r=\sqrt{x^{2}+y^{2}}=\sqrt{1^{2}+2^{2}}=\sqrt{5}\). Therefore
\[ \sin\theta=\frac{y}{r}=\frac{2}{\sqrt{5}} \]
\[ \cos\theta=\frac{x}{r}=\frac{1}{\sqrt{5}} \]
\[ \tan\theta=\frac{y}{x}=\frac{2}{1}=2 \]
\[ \csc\theta=\frac{r}{y}=\frac{\sqrt{5}}{2} \]
\[ \sec\theta=\frac{r}{x}=\frac{\sqrt{5}}{1}=\sqrt{5} \]
\[ \cot\theta=\frac{x}{y}=\frac{1}{2} \]
22.10. If \(\theta\) is an acute angle, find the other trigonometric functions of \(\theta\), given
$ = $ ; (b) $ = $ .
Draw a figure. In the right triangle, take opp = 3 and hyp = 5. Then the third side is found from the Pythagorean theorem: $ adj = = 4 $ . See Fig. 22-14.
Hence.
\[ \sin\theta=\frac{opp}{hyp}=\frac{3}{5} \]
\[ \cos\theta=\frac{adj}{hyp}=\frac{4}{5} \]
\[ \tan\theta=\frac{opp}{adj}=\frac{3}{4} \]
\[ \csc\theta=\frac{hyp}{opp}=\frac{5}{3} \]
\[ \sec\theta=\frac{hyp}{adj}=\frac{5}{4} \]
\[ \cot\theta=\frac{adj}{opp}=\frac{4}{3} \]
- Draw a figure. In the right triangle, take opp = 2 and adj = 3. Then the third side is found from the Pythagorean theorem: $ hyp = = $ . See Fig. 22-15.
Hence.
\[ \sin\theta=\frac{opp}{hyp}=\frac{2}{\sqrt{13}}\qquad\cos\theta=\frac{adj}{hyp}=\frac{3}{\sqrt{13}}\qquad\tan\theta=\frac{opp}{adj}=\frac{2}{3} \]
\[ \csc\theta=\frac{\mathrm{h y p}}{\mathrm{o p p}}=\frac{\sqrt{13}}{2}\qquad\sec\theta=\frac{\mathrm{h y p}}{\mathrm{a d j}}=\frac{\sqrt{13}}{3}\qquad\cot\theta=\frac{\mathrm{a d j}}{\mathrm{o p p}}=\frac{3}{2} \]
22.11. Find the trigonometric functions of $ 30^{} $ , $ 45^{} $ , and $ 60^{} $ .
To find the trigonometric functions of \(30^{\circ}\) and \(60^{\circ}\), draw a \(30-60^{\circ}\) right triangle (Fig. 22-16). Since the side opposite the \(30^{\circ}\) angle is one-half the hypotenuse, for \(30^{\circ}\) take opp = 1, hyp = 2. Then from the Pythagorean theorem, adj = \(\sqrt{3}\).
Hence
\[ \sin30^{\circ}=\frac{opp}{hyp}=\frac{1}{2}\quad\cos30^{\circ}=\frac{adj}{hyp}=\frac{\sqrt{3}}{2}\quad\tan30^{\circ}=\frac{opp}{adj}=\frac{1}{\sqrt{3}} \]
\[ \csc30^{\circ}=\frac{hyp}{opp}=\frac{2}{1}=2\qquad\sec30^{\circ}=\frac{hyp}{adj}=\frac{2}{\sqrt{3}}\qquad cot30^{\circ}=\frac{adj}{opp}=\frac{\sqrt{3}}{1}=\sqrt{3} \]
Fig. 22-16 can also be used to determine the trigonometric functions of \(60^{\circ}\), even though the \(60^{\circ}\) angle is not in standard position. Take \(\mathrm{opp} = \sqrt{3}\), adj = 1, and hyp = 2, then
\[ \sin60^{\circ}=\frac{opp}{hyp}=\frac{\sqrt{3}}{2}\qquad\cos60^{\circ}=\frac{adj}{hyp}=\frac{1}{2}\qquad\qquad\tan60^{\circ}=\frac{opp}{adj}=\frac{\sqrt{3}}{1}=\sqrt{3} \]
\[ \csc60^{\circ}=\frac{hyp}{opp}=\frac{2}{\sqrt{3}}\qquad\sec60^{\circ}=\frac{hyp}{adj}=\frac{2}{1}=2 \]
\[ \cot60^{\circ}=\frac{adj}{opp}=\frac{1}{\sqrt{3}} \]
To find the trigonometric functions of $ 45^{} $ , draw an isosceles right triangle (Fig. 22-17). Take opp = 1, adj = 1, and hyp = $ $ , then
\[ \sin45^{\circ}=\frac{opp}{hyp}=\frac{1}{\sqrt{2}} \]
\[ \cos45^{\circ}=\frac{adj}{hyp}=\frac{1}{\sqrt{2}} \]
\[ \tan45^{\circ}=\frac{opp}{adj}=\frac{1}{1}=1 \]
\[ \csc45^{\circ}=\frac{hyp}{opp}=\frac{\sqrt{2}}{1}=\sqrt{2}\qquad\sec45^{\circ}=\frac{hyp}{adj}=\frac{\sqrt{2}}{1}=\sqrt{2}\qquad cot45^{\circ}=\frac{adj}{opp}=\frac{1}{1}=1 \]
22.12. Form a table of the trigonometric functions of $ 0, , , $ , and $ $ radians.
The trigonometric functions of 0 and \(\pi/2\) radians are the same as the functions of the real numbers 0 and \(\pi/2\), respectively, calculated in Problems 20.7 and 20.8. The trigonometric functions of \(\pi/6\), \(\pi/4\), and \(\pi/3\) are the same as the functions of \(30^{\circ}\), \(45^{\circ}\), and \(60^{\circ}\), calculated in Problem 22.11. Summarizing yields the following table (U stands for undefined):
| θ (radians) | θ (degrees) | sinθ | cosθ | tanθ | cotθ | secθ | cscθ |
| 0 | 0° | 0 | 1 | 0 | U | 1 | U |
| π/6 | 30° | 1/2 | √3/2 | 1/√3 | √3 | 2/√3 | 2 |
| π/4 | 45° | 1/√2 | 1/√2 | 1 | 1 | √2 | √2 |
| π/3 | 60° | √3/2 | 1/2 | √3 | 1/√3 | 2 | 2/√3 |
| π/2 | 90° | 1 | 0 | U | 0 | U | 1 |
22.13. Show that for any nonquadrantal angle $ $ , each trigonometric function of $ $ has the same absolute value as the same trigonometric function of its reference angle $ _{R} $ .
The four possible positions of $ $ and $ _{R} $ are shown in Fig. 22-18.
In each case, let $ P(x, y) $ be a point on the terminal side of $ $ and draw a line from P perpendicular to the x-axis at A. In triangle $ OAP, _{R} $ is an acute angle with $ opp = |y| $ , $ adj = |x| $ , and $ hyp = = r $ .
\[ \left|\sin\theta\right|=\left|\frac{y}{r}\right|=\frac{\left|y\right|}{r}=\sin\theta_{R}\qquad\left|\cos\theta\right|=\left|\frac{x}{r}\right|=\frac{\left|x\right|}{r}=\cos\theta_{R}\qquad\left|\tan\theta\right|=\left|\frac{y}{x}\right|=\frac{\left|y\right|}{\left|x\right|}=\tan\theta_{R} \]
and similarly for the other trigonometric functions.
22.14. Find the reference angle for (a) $ 480^{} $ ; (b) $ - $ radians.
First note that $ 480{}-360{}=120^{} $ is an angle between $ 0^{} $ and $ 360^{} $ coterminal with $ 480^{} $ . Since $ 90{}<120{}<180^{} $ , $ 120^{} $ is a second quadrant angle. Hence the reference angle for $ 120^{} $ and therefore for $ 480^{} $ is $ 180{}-120{}=60^{} $ .
First note that $ - + 2= $ is an angle between 0 and $ 2$ radians coterminal with $ - $ . Since $ < < $ , $ $ is a third quadrant angle. Hence the reference angle for $ $ and therefore for $ - $ is $ - = $ .
22.15. Find the trigonometric functions for (a) $ 480^{} $ ; (b) $ - $ radians.
- To find the trigonometric functions of an angle, find the functions of its reference angle and attach the correct sign for the quadrant. $ 480^{} $ is a second quadrant angle. In quadrant II, sine and coscant are positive, and the other trigonometric functions are negative. Using the reference angle found in the previous problem yields:
\[ \begin{aligned}\sin480^{\circ}&=\sin60^{\circ}=\frac{\sqrt{3}}{2}\quad&\cos480^{\circ}&=-\cos60^{\circ}=-\frac{1}{2}\quad&\tan480^{\circ}&=-\tan60^{\circ}=-\sqrt{3}\\ csc480^{\circ}&=\csc60^{\circ}=\frac{2}{\sqrt{3}}\quad&\sec480^{\circ}&=-\sec60^{\circ}=-2\quad&\cot480^{\circ}&=-\cot60^{\circ}=-\frac{1}{\sqrt{3}}\end{aligned} \]
- $ -3/4 $ is a third quadrant angle. In quadrant III, tangent and cotangent are positive, and the other trigonometric functions are negative. Using the reference angle found in the previous problem yields:
\[ \begin{aligned}&\sin\left(-\frac{3\pi}{4}\right)=-\sin\frac{\pi}{4}=-\frac{1}{\sqrt{2}}\qquad\cos\left(-\frac{3\pi}{4}\right)=-\cos\frac{\pi}{4}=-\frac{1}{\sqrt{2}}\qquad\tan\left(-\frac{3\pi}{4}\right)=\tan\frac{\pi}{4}=1\\ &\quad\csc\left(-\frac{3\pi}{4}\right)=-\csc\frac{\pi}{4}=-\sqrt{2}\qquad\sec\left(-\frac{3\pi}{4}\right)=-\sec\frac{\pi}{4}=-\sqrt{2}\qquad\cot\left(-\frac{3\pi}{4}\right)=\cot\frac{\pi}{4}=1\\ \end{aligned} \]
22.16. Find all angles $ , 0 < 2$ , such that (a) $ = $ ; (b) $ = - $ .
The sine function is increasing on the interval from 0 to $ /2 $ ; thus it is a one-to-one function on this interval. For values of a in the interval $ 0 a $ , the notation $ t = ^{-1} a $ is used to denote the unique value t in the interval $ 0 t /2 $ such that $ t = a $ . (See Chapter 25 for a fuller discussion of inverse trigonometric functions.)
From the table in Problem 22.12, $ = $ ; thus, $ =^{-1} $ . Since the sine function is positive in quadrants I and II, there is also an angle $ $ in quadrant II with reference angle $ $ and $ = $ . This angle must be $ -= $ . Then $ $ and $ $ are the two required angles.
From the table in Problem 22.12, $ = $ , thus, $ =^{-1} $ . Since the sine function is negative in quadrants III and IV, the required angles are the angles $ {1} $ and $ {2} $ in these quadrants with reference angle $ $ and $ {1}={2}=- $ . In quadrant III, $ {1}-= $ ; thus, $ {1}= $ . In quadrant IV, $ 2-{2}= $ ; thus, $ {2}= $ .
22.17. Find all angles $ , 0 < 2$ , such that (a) $ = $ ; (b) $ = - $ .
The cosine function is decreasing on the interval from 0 to $ /2 $ ; thus it is a one-to-one function on this interval. For values of a in the interval $ 0 a $ , the notation $ t = ^{-1} a $ is used to denote the unique value t in the interval $ 0 t /2 $ such that $ t = a $ .
From the table in Problem 22.12, $ = $ , thus, $ =^{-1} $ . Since the cosine function is positive in quadrants I and IV, there is also an angle $ $ in quadrant IV with reference angle $ $ and $ = $ . This angle must satisfy $ 2-= $ ; thus, $ = $ . Thus $ $ and $ $ are the two required angles.
From the table in Problem 22.12, $ = $ ; thus $ =^{-1} $ . Since the cosine function is negative in quadrants II and III, the required angles are the angles $ {1} $ and $ {2} $ in these quadrants with reference angle $ $ and $ {1}={2}=- $ . In quadrant II, $ -{1}= $ ; thus $ {1}= $ . In quadrant III, $ {2}-= $ ; thus, $ {2}= $ .
22.18. Find all angles $ $ , $ 0^{} < 360^{} $ , such that (a) $ = $ ; (b) $ = -1 $ .
The tangent function is increasing on the interval from 0 to $ /2 $ , thus it is a one-to-one function on this interval. For nonnegative values of a, the notation $ t = ^{-1} a $ is used to denote the unique nonnegative value t such that $ t = a $ .
From the table in Problem 22.12, $ = $ , thus $ =^{-1} $ . Hence $ 60^{} $ is one required angle. Since the tangent function is positive in quadrants I and III, there is also an angle $ $ in quadrant III with reference angle $ 60^{} $ and $ = $ . In quadrant III, $ {}=60{} $ ; thus $ ^{} $ . The required angles are $ 60^{} $ and $ 240^{} $ .
From the table in Problem 22.12, $ =1 $ ; thus $ =^{-1}1 $ . Since the tangent function is negative in quadrants II and IV, the required angles are the angle $ {1} $ and $ {2} $ in these quadrants with reference angle $ =45^{} $ . In quadrant II, $ 180{}-_{1}=45{} $ ; thus $ {1}=135^{} $ . In quadrant IV, $ 360{}-_{2}=45{} $ ; thus $ {2}=315^{} $ .
22.19. Use a scientific calculator to find approximate values for (a) sin 42°; (b) cos 238°;
- $ (-61.5^{}) $ ; (d) $ ^{}25’ $
In using a scientific calculator for trigonometric calculations, it is crucial to make certain that the correct mode (degree mode or radian mode) is selected. Refer to the calculator manual for instructions for choosing the mode. In this problem, put the calculator in degree mode. (a) $ ^{} = 0.6691 $ ; (b) $ ^{} = -0.5299 $ ; (c) $ (-61.5^{}) = -1.8418 $ ; (d) secant cannot be calculated directly on a calculator; use a trigonometric identity:
\[ \sec\left(341^{\circ}25^{\prime}\right)=\frac{1}{\cos\left(341^{\circ}25^{\prime}\right)}=\frac{1}{\cos\left(341+25/60\right)^{\circ}}=1.055 \]
22.20. Use a scientific calculator to find approximate values for (a) sin 3; (b) cos(−5.3);
- $ (2.356) $ ; (d) $ (12.3) $ .
See comments in previous problem. In this problem, put the calculator in radian mode. (a) $ = 0.1411 $ ; (b) $ (-5.3) = 0.5544 $ ; (c) $ (2.356) = -1.0004 $ ; (d) cotangent cannot be calculated directly on a calculator; use a trigonometric identity:
\[ cot(12.3)=\frac{1}{\tan\left(12.3\right)}=-3.6650 \]
22.21. Use a scientific calculator to find approximate values for all angles $ $ , $ 0 < 2$ , such that
- $ = 0.7543 $ (b) $ = -4.412 $
Put the calculator into a radian mode.
First find $ ^{-1}0.7543 = 0.8546 $ . Since the sine function is positive in quadrants I and II, there is also an angle $ $ in quadrant II with reference angle 0.8546 and $ = 0.7543 $ . This angle must be $ - 0.8546 = 2.2870 $ . Then 0.8546 and 2.2870 are the two required angles.
First find $ ^{-1}4.412=1.3479 $ . Since the tangent function is negative in quadrants II and IV, the required angles are the angles $ {1} $ and $ {2} $ in these quadrants with reference angle 1.3479. In quadrant II, $ -{1}=1.3479 $ ; thus $ {1}=1.7937 $ . In quadrant IV, $ 2-{2}=1.3479 $ , thus $ {2}=4.9353 $ .
22.22. Use a scientific calculator to find approximate values for all angles $ $ , $ 0^{} < 360^{} $ , such that
- $ = 0.8455 $ ; (b) $ = -3 $ ; (c) $ = 0.333 $ .
Put the calculator into degree mode.
First find $ ^{-1}0.8455 = 0.5633 = 32.27^{} $ . Since the cosine function is positive in quadrants I and IV, there is also an angle $ $ in quadrant IV with reference angle $ 32.27^{} $ and $ = 0.8455 $ . This angle must satisfy $ 360^{} - = 32.27^{} $ ; thus $ = 327.73^{} $ . Then $ 32.27^{} $ and $ 327.73^{} $ are the two required angles.
Cosecant cannot be calculated directly on a calculator; use a trigonometric identity. $ = -3 $ is equivalent to $ 1/() = -3 $ , thus, $ = - $ .
First find $ {-1}=0.3398=19.47{} $ . Since the sine function is negative in quadrants III and IV, the required angles are the angles $ {1} $ and $ {2} $ in these quadrants with reference angle $ 19.47^{} $ and $ {1}={2}=- $ . In quadrant III, $ {1}-180{}=19.47{} $ ; thus, $ {1}=199.47^{} $ . In quadrant IV, $ 360{}-_{2}=19.47{} $ ; thus $ _{2}=340.53^{} $ .
- There is no angle that satisfies $ = 0.333 $ since 0.333 is not in the range of the secant function. A calculator will return an error message.
SUPPLEMENTARY PROBLEMS
22.23. List all angles coterminal with (a) θ radians; (b) θ degrees.
Ans. (a) $ + 2n $ , n any integer; (b) $ + n360^{} $ , n any integer
22.24. Find the trigonometric functions of $ 270^{} $ .
Ans. \(\sin 270^{\circ} = -1\), \(\cos 270^{\circ} = 0\), \(\tan 270^{\circ}\) is undefined, \(\cot 270^{\circ} = 0\), \(\sec 270^{\circ}\) is undefined, \(\csc 270^{\circ} = -1\)
22.25. Complete the proof in Problem 22.7 that the definitions of the trigonometric functions as ratios are consistent with the definitions of the trigonometric functions of angles.
22.26. Find (a) sin 120°; (b) cos \(\frac{5\pi}{6}\); (c) tan \((-45^{\circ})\); (d) cot \(\frac{7\pi}{6}\); (e) sec 240°; (f) csc \(\frac{2\pi}{3}\).
\[ Ans.\quad(a)\frac{\sqrt{3}}{2};(b)-\frac{\sqrt{3}}{2};(c)-1;(d)\sqrt{3};(e)-2;(f)\frac{2}{\sqrt{3}} \]
22.27. Find (a) \(\sin\frac{7\pi}{4}\); (b) \(\cos450^{\circ}\); (c) \(\tan\frac{8\pi}{3}\); (d) \(\cot(-720^{\circ})\); (e) \(\sec\frac{17\pi}{6}\); (f) \(\csc(-510^{\circ})\).
Ans. (a) $ - $ ; (b) 0; (c) $ - $ ; (d) undefined; (e) $ - $ ; (f) -2
22.28. If θ is in standard position and (−1, −4) lies on its terminal side, find the trigonometric functions of θ.
\[ Ans.\quad\sin\theta=-\frac{4}{\sqrt{17}},\cos\theta=-\frac{1}{\sqrt{17}},\tan\theta=4,\cot\theta=\frac{1}{4},\sec\theta=-\sqrt{17},csc\theta=-\frac{\sqrt{17}}{4} \]
22.29. If \(\theta\) is an acute angle, find the other trigonometric functions of \(\theta\), given
\[ \begin{aligned}(a)\sin\theta=\frac{12}{13};(b)\cos\theta=\frac{5}{7};(c)\tan\theta=\frac{1}{\sqrt{2}}.\end{aligned} \]
Ans. (a) $ = $ , $ = $ , $ = $ , $ = $ , $ = $
22.30. If $ $ is in standard position and its terminal side lies in quadrant II on the line $ x + 3y = 0 $ , find the trigonometric functions of $ $ .
Ans. $ =,=-,=-,,=-,= $
22.31. Find approximate values for all angles $ $ , $ 0 < 2$ , such that
- sinθ = 0.1188; (b) tanθ = 8.7601; (c) secθ = −2.3.
Ans. (a) 0.1191, 3.0225; (b) 1.4571, 4.5987; (c) 2.0206, 4.2626
22.32. Find approximate values for all angles $ $ , $ 0^{} < 360^{} $ , such that
- $ = 0.0507 $ ; (b) $ = 62 $ ; (c) $ = -5.2 $ .
Ans. (a) 87.09°, 272.91°; (b) 0.92°, 180.92°; (c) 191.09°, 348.91°
22.33. The angular speed $ $ of a point moving in a circle is defined as the quotient $ /t $ , where $ $ is the angle in radians through which the point travels in time t.
Find the angular speed of a point that moves through an angle of 4 radians in 6 seconds.
Find the angular speed of a point on the rim of a wheel that travels at 60 rpm (revolutions per minute).
Show that the linear speed v of a point moving in a circle is related to the angular speed by the formula $ v = r$ .
A car is moving at the rate of 60 miles per hour, and the diameter of each wheel is 2.5 feet. Find the angular speed of the wheels.
Ans. (a) $ $ rad/sec (b) $ 120$ rad/min (d) 4224 rad/min