Solution
Look at the figure at right. An hour and a half represents 1.5 complete rotations, or
[latex]1.5(360°) = 540°[/latex]
Forty minutes is two-thirds of an hour, so the minute hand rotates through
[latex]\dfrac{2}{3}(360°) = 240°[/latex]
Chapter 4: Trig Functions
Evaluate the function.
Look at the figure at right. An hour and a half represents 1.5 complete rotations, or
[latex]1.5(360°) = 540°[/latex]
Forty minutes is two-thirds of an hour, so the minute hand rotates through
[latex]\dfrac{2}{3}(360°) = 240°[/latex]
The volume control on an amplifier is a dial with ten settings, as shown at right. Through how many degrees would you rotate the dial to increase the volume level from 0 to 7?
[latex]252°[/latex]
The degree measure of an angle depends only on the fraction of a whole rotation between its sides, and not on the location or position of the angle. To compare and analyze angles, we place them in standard position so that the vertex of the angle is located at the origin and its initial side lies on the positive [latex]x[/latex]-axis. The figure below shows several angles placed in standard position.
One-half of a complete revolution is [latex]180°{,}[/latex] and three-quarters of one revolution is [latex]270°{.}[/latex] Thus, for angles between [latex]180°[/latex] and [latex]270°[/latex] in standard position, the terminal side lies in the third quadrant, and for angles between [latex]270°[/latex] and [latex]360°{,}[/latex] the terminal side lies in the fourth quadrant.
a. The angle [latex]\alpha[/latex] is one-fifth of a complete revolution, or
[latex]\dfrac{1}{5}(360°) = 72°[/latex]
In standard position, it is a first-quadrant angle, as shown in figure (a) below.
b. The angle [latex]\beta[/latex] is [latex]\dfrac{11}{12}[/latex] of a complete revolution, or
[latex]\dfrac{11}{12}(360°) = 330°[/latex]
In standard position, it is a fourth-quadrant angle. (See Figure [b].)
a. [latex]120°[/latex]
b. [latex]70°[/latex]
In Chapter 3 we defined the sine, cosine, and tangent for obtuse angles by placing the angle in a Cartesian coordinate system. We can do the same for angles that represent rotations.
If [latex]\theta[/latex] is an angle in standard position, and [latex](x,y)[/latex] is a point on its terminal side, with [latex]r = \sqrt{x^2 + y^2}{,}[/latex] then
[latex]{\sin \theta = \dfrac{y}{r}~~~~~~~~~ \cos \theta = \dfrac{x}{r}~~~~~~~~~\tan \theta = \dfrac{y}{x}}[/latex]
We can choose any point on the terminal side of the angle, and the trig ratios defined by its coordinates will be the same. (Can you explain why?)
Because it is the distance from the origin to [latex]P{,}[/latex] [latex]r[/latex] is always positive. However, [latex]x[/latex] and [latex]y[/latex] can be positive or negative (or zero), depending on the angle [latex]\theta{.}[/latex] For example, in the second quadrant, [latex]x[/latex] is negative but [latex]y[/latex] is positive, so the cosine and the tangent of angles between [latex]90°[/latex] and [latex]180°[/latex] are negative, but their sines are positive.
Give the sign of each of the three trigonometric ratios of the angles.
a. In standard position, the terminal side of an angle of [latex]200°[/latex] lies in the third quadrant. (See figure [a] below.) In the third quadrant, [latex]x \lt 0[/latex] and [latex]y \lt 0{,}[/latex] but [latex]r \gt 0{.}[/latex] Thus, [latex]\sin 200°[/latex] is negative, [latex]\cos 200°[/latex] is negative, and [latex]\tan 200°[/latex] is positive.
b. The terminal side of [latex]300°[/latex] lies in the fourth quadrant, so [latex]x \gt 0[/latex] and [latex]y \lt 0{,}[/latex] and [latex]r \gt 0{.}[/latex] Thus, [latex]\sin 300°[/latex] is negative, [latex]\cos 300°[/latex] is positive, and [latex]\tan 300°[/latex] is negative.
| Quadrant | Degrees | Sine | Cosine | Tangent |
| First | [latex]0° \lt \theta \lt 90°[/latex] | positive | positive | positive |
| Second | [latex]90° \lt \theta \lt 180°[/latex] | |||
| Third | [latex]180° \lt \theta \lt 270°[/latex] | |||
| Fourth | [latex]270° \lt \theta \lt 360°[/latex] |
| Quadrant | Degrees | Sine | Cosine | Tangent |
| First | [latex]0° \lt \theta \lt 90°[/latex] | positive | positive | positive |
| Second | [latex]90° \lt \theta \lt 180°[/latex] | positive | negative | negative |
| Third | [latex]180° \lt \theta \lt 270°[/latex] | negative | negative | positive |
| Fourth | [latex]270° \lt \theta \lt 360°[/latex] | negative | positive | negative |
The [latex]y[/latex]-coordinate of the point [latex]P[/latex] is [latex]-5{,}[/latex] and [latex]r = 6{,}[/latex] so
[latex]sin \theta = \dfrac{y}{r} = \dfrac{-5}{6}[/latex]
To find the [latex]x[/latex]-coordinate of [latex]P{,}[/latex] we use the equation of a circle of radius [latex]6{,}[/latex] [latex]x^2 + y^2 = 36{.}[/latex]
[latex]x^2 + (-5)^2 = 36\ x^2 = 36 - 25 = 11\ x = \pm \sqrt{11}[/latex]
Because [latex]P[/latex] is in the third quadrant, [latex]x = -\sqrt{11}{.}[/latex] Thus,
[latex]\cos \theta = \dfrac{x}{r} = \dfrac{-\sqrt{11}}{6}~~~~ {and} ~~~~\tan \theta = \dfrac{y}{x} = \dfrac{-5}{-\sqrt{11}} = \dfrac{5}{\sqrt{11}}[/latex]
[latex]\sin \theta = \dfrac{\sqrt{7}}{4},~~\cos \theta = \dfrac{-3}{4},~~\tan \theta = \dfrac{-\sqrt{7}}{3}[/latex]
In Section 3.1 we learned that the trig ratios for angles in the second quadrant are the same as the trig ratios for their supplements, except for sign. For example, you can use your calculator to verify that
| [latex]\sin 130° = 0.7660[/latex] | [latex]\sin 50° = 0.7660[/latex] |
| [latex]\cos 130° = -0.6428[/latex] | [latex]\cos 50° = 0.6428[/latex] |
| [latex]\tan 130° = -1.1918[/latex] | [latex]\tan 50° = 1.1918[/latex] |
The trig ratios for [latex]130°[/latex] and [latex]50°[/latex] have the same absolute value because the two triangles formed by the angles are congruent, as shown above.
[latex]\triangle OP^{\prime}Q^{\prime}[/latex] is called a reference triangle for [latex]130°{,}[/latex] and [latex]50°[/latex] is called the reference angle.
The trig ratios for angles between [latex]180°[/latex] and [latex]360°{,}[/latex] whose terminal sides lie in the third and fourth quadrants, are also related to the trig ratios of familiar angles in the first quadrant. We “refer” the angle to a first quadrant angle with a congruent reference triangle.
Reference angles will be important when we need to solve trigonometric equations. There will always be two angles with the same trig ratio (except for quadrantal angles), and they will have the same reference angle.
We can construct reference triangles for angles in any of the four quadrants, and the trig ratios of the angle are the same as the trig ratios of its reference angle, up to sign. Here is how to construct a reference triangle for an angle:
The figure below shows angles [latex]\theta[/latex] in all four quadrants, and the reference angle, [latex]\widetilde{\theta}{,}[/latex] for each. Study the figures and make sure you understand the formula for finding the reference angle in each quadrant.
These observations may help you remember the formulas:
[latex]\sin 20° = 0.3420 \sin 200° = -0.3420\ \cos 20° = 0.9397 \cos 200° = -0.9397\ \tan 20° = 0.3640 \tan 200° = 0.3640[/latex]
Here is a summary of our discussion about reference angles.
The trigonometric ratios of any angle are equal to the ratios of its reference angle, except for sign. The sign of the ratio is determined by the quadrant.
Any acute angle [latex]\theta[/latex] is the reference angle for four angles between [latex]0°[/latex] and [latex]360°{,}[/latex] one in each quadrant. The figure below shows the four angles in standard position whose reference angle is [latex]35°{.}[/latex] Note that each angle is found by measuring [latex]35°[/latex] from the [latex]x[/latex]-axis in the appropriate quadrant and that the four angles together make a “bow tie” shape.
From the figure, you can see that the angles in each quadrant with a given reference angle are computed as follows.
| Quadrant I: [latex]\theta = \widetilde{\theta}[/latex] | Quadrant II: [latex]\theta = 180° - \widetilde{\theta}[/latex] | |
| Quadrant III: [latex]\theta = 180° + \widetilde{\theta}[/latex] | Quadrant IV: [latex]\theta = 360° - \widetilde{\theta}[/latex] |
That angle is [latex]360° - 51.3° = 308.7°{,}[/latex] as shown at right.
Because [latex]360°[/latex] represents one complete revolution, we can add or subtract a multiple of [latex]360°[/latex] to any angle, and the terminal side will arrive at the same position. For example, the angles [latex]70°[/latex] and [latex]430°[/latex] have the same terminal side because [latex]430° = 70° + 360°{.}[/latex] Such angles are called coterminal.
The angle [latex]790°[/latex] is also coterminal with [latex]70°{,}[/latex] because [latex]790° = 70°+ 2(360°){.}[/latex] See the figure below.
Because coterminal angles have the same standard position, their trigonometric ratios are equal. For example, you can verify that, to four decimal places,
[latex]\cos 790° = \cos 70° = 0.3420[/latex]
If the direction of rotation is important, we let positive angles represent rotation in the counterclockwise direction, and negative angles represent rotation in the clockwise direction. For example, the angle [latex]-60°[/latex] shown at right lies in the fourth quadrant. It is coterminal with [latex]-60° + 360° = 300°{.}[/latex]
[latex]462°{,}[/latex] [latex]~-258°[/latex]
In Chapter 3 we saw that there are always two angles between [latex]0°[/latex] and [latex]180°[/latex] with a given sine ratio between 0 and 1.
For example, the two solutions to the equation [latex]~~sin \theta = 0.4226~~[/latex] are [latex]25°[/latex] and [latex]155°{.}[/latex] That is,
[latex]\sin 25° = \sin 155° = 0.4226[/latex]
These two angles are supplementary, so they have the same reference angle, as shown below.
There are also two solutions between [latex]0°[/latex] and [latex]360°[/latex] to the equation [latex]~~sin \theta = -0.4226~~{.}[/latex] They are the angles in the third and fourth quadrants whose reference angle is [latex]25°[/latex]—namely, [latex]205°[/latex] and [latex]355°[/latex], as shown above.
In fact, there are always two angles between [latex]0°[/latex] and [latex]360°[/latex] (except for the quadrantal angles) that have the same sine. There are also two angles that have the same cosine, and two angles that have the same tangent.
Enter 2nd SIN [latex]-0.6428[/latex] to find that
[latex]\sin^{-1}(-0.6428) = -40°[/latex]
The angle [latex]-40°[/latex] is in the fourth quadrant, but it is not between [latex]0°[/latex] and [latex]360°{.}[/latex] We need an angle that is coterminal with [latex]-40°{,}[/latex] so we add [latex]360°{.}[/latex]
[latex]-40° + 360° = 320°[/latex]
Thus, one of the solutions is [latex]320°{.}[/latex] The sine is also negative in the third quadrant, so there should also be a solution in the third quadrant. The reference angle for [latex]320°[/latex] is [latex]40°{,}[/latex] and the third-quadrant angle with reference angle [latex]40°[/latex] is [latex]220°{,}[/latex] as shown at right. You can check that, rounded to four decimal places, both angles satisfy the equation—that is,
[latex]\sin 320° = -0.6428~~~~{and} ~~~~\sin 220° = -0.6428[/latex]
When you ask your calculator to find an angle with a given sine by pressing 2ndSIN, it will give you only one of the many possible answers. The same is true of the 2ndCOS and 2ndTAN keys. You must be careful to choose the appropriate answer for the situation, using reference and coterminal angles.
[latex]338.2°{,}[/latex] [latex]~158.2°[/latex]
In the previous example, we found two solutions of the equation [latex]\sin \theta = -0.6428{.}[/latex] Actually, there are infinitely many solutions — namely, all the angles coterminal with [latex]320°[/latex] or [latex]220°{.}[/latex] Some of these angles are listed below.
Angles coterminal with [latex]~ 220° \text{ are }: ~~ \cdots, ~{-500°}, ~{-140°}, ~580°, ~940°, ~\cdots\ \text{ and } 320° \text{ are }:~~ \cdots, ~{-400°}, ~{-40°}, ~680°, ~1040°, ~\cdots[/latex]
Because we can easily find coterminal solutions by adding or subtracting multiples of [latex]360°{,}[/latex] when solving an equation, we usually list only the solutions between [latex]0°[/latex] and [latex]360°{.}[/latex]
An angle of [latex]210°[/latex] lies in the third quadrant, and its reference angle is
[latex]\widetilde{\theta} = 210° - 180° =30°[/latex]
We draw a reference triangle as shown at right. Because the sides of a 30-60-90 triangle are in the ratio [latex]1:\sqrt{3}:2{,}[/latex] we can choose point [latex]P[/latex] with coordinates [latex](-\sqrt{3}, -1){.}[/latex]
(We must be careful to use the correct signs for the coordinates of point [latex]P{;}[/latex] in the third quadrant both coordinates are negative.) Thus, [latex]x = -\sqrt{3},~ y = -1{,}[/latex] and [latex]r = 2{,}[/latex] so
[latex]\sin \theta = \dfrac{-1}{2},~~~\cos \theta = \dfrac{-\sqrt{3}}{2}, ~~~\tan \theta = \dfrac{-1}{\sqrt{3}}[/latex]
All of the special angles are shown at right. In the Homework Problems, you will calculate the three trigonometric ratios for all the special angles, and you should memorize these values or be able to calculate them quickly.
Let [latex]P[/latex] be a point on a unit circle determined by the terminal side of an angle [latex]\theta[/latex] in standard position. Then the coordinates [latex](x,y)[/latex] of [latex]P[/latex] are given by
[latex]{x = \cos \theta,~~~~~~y = \sin \theta}[/latex]
The coordinates of [latex]P[/latex] are given by
[latex]x = \cos 315°,~~~y = \sin 315°{.}[/latex]
The reference angle for [latex]315°[/latex] is [latex]~ 360° - 315° = 45°{,}[/latex] so the trig ratios of [latex]315°[/latex] are the same as the ratios for [latex]45°{,}[/latex] up to sign.
Because [latex]315°[/latex] is in the fourth quadrant, we have [latex]\cos 315° = \dfrac{1}{\sqrt{2}}[/latex] and [latex]\sin 315° = \dfrac{-1}{\sqrt{2}}{.}[/latex] Thus, the coordinates of [latex]P[/latex] are [latex]\left(\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right){.}[/latex]
Find the sine, cosine, and tangent of the angle [latex]\varphi[/latex] shown at right.
[latex]\sin \varphi = -0.7000{,}[/latex] [latex]~\cos \varphi = -0.7141{,}[/latex] [latex]~\tan \varphi = 0.9802[/latex]
If [latex]\theta[/latex] is an angle in standard position, and [latex](x,y)[/latex] is a point on its terminal side, with [latex]r = \sqrt{x^2 + y^2}{,}[/latex] then
[latex]\sin \theta = \dfrac{y}{r}~~~~~~~~~ \cos \theta = \dfrac{x}{r}~~~~~~~~~ \tan \theta = \dfrac{y}{x}[/latex]
Let [latex]P[/latex] be a point on a unit circle determined by the terminal side of an angle [latex]\theta[/latex] in standard position. Then the coordinates [latex](x,y)[/latex] of [latex]P[/latex] are given by
[latex]x = \cos \theta,~~~~~~y = \sin \theta[/latex]
| Quadrant I: [latex]~~~~~~theta = \widetilde{\theta}[/latex] | Quadrant II: [latex]~~~~theta = 180° - \widetilde{\theta}[/latex] | |
| Quad. III: [latex]~~~~theta = 180° + \widetilde{\theta}[/latex] | Quadrant IV: [latex]~~~~theta = 360° - \widetilde{\theta}[/latex] |