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8.2 Trigonometric Functions of Any Angle

Reference Angle • Evaluating Trigonometric Functions • Evaluations on a Calculator • Quadrantal Angles • Negative Angles

In the last section, we saw that the signs of the trigonometric functions depend on the quadrant of $θ .$ $θ .$ In this section, we will investigate an important connection between certain angles in different quadrants. To illustrate this, let’s start by taking the sine of the angles $20 °$ $20 °$ , $160 °$ $160 °$ , $200 °$ $200 °$ , and $340 °$ $340 °$ . These values are shown in Fig. 8.5.

A calculator screen with inputs for sine of 20, sine of 160, sine of 200, and sine of 340. Outputs for the first two inputs are each 0.3420201433, and outputs for the last two are each negative 0.3420201433. — Fig. 8.5

The signs of the function values agree with what we would expect; $\sin θ$ $\sin θ$ is positive in quadrants I and II and negative in quadrants III and IV. The important thing to notice, however, is that the numerical values (disregarding the signs) are the same for all four angles. The reason for this is that these angles have something in common, called a reference angle, which is defined below:

In our example, all four angles have a reference angle of $20 °$ $20 °$ . Figure 8.6 shows the terminal sides of these angles drawn in standard position along with their reference angles. Since the angles share a common reference angle, they will also have the same trigonometric function values, neglecting the sign. Our example demonstrated this for sine, and the same is true for the other five functions. This is because the values of x, y, and r in the definition of the trigonometric functions will be the same, except for the signs of x and y, which will depend on the quadrant.

A line rises through quadrants 3 and 1 at 200 degrees and 20 degrees, respectively, and another falls through quadrants 2 and 4 at 160 degrees and 340 degrees, respectively. Each line has a 20 degree angle with the x-axis. — Fig. 8.6

For this reason, reference angles are very important in trigonometry. To find the reference angle of a given angle, we sketch the angle in standard position and then determine the acute angle between its terminal side and the x-axis. This is illustrated in the following example.

EXAMPLE 1 Finding reference angles

Figure 8.7 shows the reference angles of $150 °$ $150 °$ , $230 °$ $230 °$ , $285 °$ $285 °$ , and $420 °$ $420 °$ . Pay careful attention to how the reference angle is calculated in each of the four quadrants.

Four position angles with reference angles. — Fig. 8.7

Figure 8.7 Full Alternative Text

We have observed that the trigonometric functions of different angles with the same reference angle will be the same, except for possibly the sign. Since the reference angle is an acute angle, it will always have a positive function value. The signs in the other quadrants will follow the rules stated in Section 8.1. This is summarized below:

According to the statement above, $\sin θ = \pm \sin θ_{ref}, \cos θ = \pm \cos θ_{ref}, \tan θ = \pm \tan θ_{ref},$ $\sin θ = \pm \sin θ_{ref}, \cos θ = \pm \cos θ_{ref}, \tan θ = \pm \tan θ_{ref},$ etc. This means that a trigonometric function of any angle can be found by first finding the trigonometric function of its reference angle and then attaching the correct sign, depending on the quadrant. The following example illustrates this process.

EXAMPLE 2 Evaluating using reference angles

Examples for solving trigonometric function of any angle using reference angles.

.2-2 Full Alternative Text

Figure 8.8 Full Alternative Text

A calculator will give values, with the proper signs, of functions like those in Example 2. To find $\csc θ, \sec θ,$ $\csc θ, \sec θ,$ and $\cot θ,$ $\cot θ,$ we must take the reciprocal of the corresponding function. For example, $\sec 304 ° = \frac{1}{\cos 304 °} .$ $\sec 304 ° = \frac{1}{\cos 304 °} .$ The display for the first five lines of Example 2 is shown in Fig. 8.8.

NOTE

[In most examples, we will round off values to four significant digits (as in Example 2). However, if the angle is approximate, use the guidelines in Table 4.1 of Section 4.3 for rounding off values.]

EXAMPLE 3 Quadrant II angle—area of triangle

A formula for finding the area of a triangle, knowing sides a and b and the included $∠ C,$ $∠ C,$ is $A = \frac{1}{2} ab \sin C .$ $A = \frac{1}{2} ab \sin C .$ A surveyor uses this formula to find the area of a triangular tract of land for which $a = 173.2 m, b = 156.3 m,$ $a = 173.2 m, b = 156.3 m,$ and $C = 112.51 ° .$ $C = 112.51 ° .$ See Fig. 8.9.

A triangle with sides ay = 173.2 meters and b = 156.3 meters that meet at angle 112.51 degrees. — Fig. 8.9

\begin{array}{rcl} A & = & \frac{1}{2} (173.2) (156.3) \sin 112.51 ° \\ = & 12, 500 m^{2} rounded to four significant digits \end{array}

$\begin{array}{rcl} A & = & \frac{1}{2} (173.2) (156.3) \sin 112.51 ° \\ = & 12, 500 m^{2} rounded to four significant digits \end{array}$

The calculator automatically uses a positive value for $\sin 112.51 °$ $\sin 112.51 °$ .

Reference angles are especially important when we wish to find an angle that has a given trigonometric function value. This is because the inverse trigonometric functions on a calculator are programmed to give angles within the specific intervals shown below:

$\sin^{- 1} x$ $\sin^{- 1} x$ always returns an angle between $- 90 °$ $- 90 °$ and $90 °$ $90 °$
$\cos^{- 1} x$ $\cos^{- 1} x$ always returns an angle between $0 °$ $0 °$ and $180 °$ $180 °$
$\tan^{- 1} x$ $\tan^{- 1} x$ always returns and angle between $- 90 °$ $- 90 °$ and $90 °$ $90 °$

CAUTION

Sometimes we wish to find an angle that is not within the intervals given above. This is when reference angles must be used.

For example, if we need to find a second quadrant angle for which $\sin θ = 0.853$ $\sin θ = 0.853$ , then taking $\sin^{- 1} (0.853)$ $\sin^{- 1} (0.853)$ will not return the correct angle. In this type of situation, we must use the reference angle to find the correct angle. This is illustrated in the following examples.

EXAMPLE 4 Finding angles given sin`θ`

If $\sin θ = 0.2250,$ $\sin θ = 0.2250,$ we see from the first line of the calculator display shown in Fig. 8.10 that $θ = 13.00 °$ $θ = 13.00 °$ (rounded off).

A calculator screen with input inverse of sine of, 0.225, and output 13.00287816; input inverse of cosine of, 1 divided by negative 2.722, and output 111.5539364. — Fig. 8.10

This result is correct, but remember that there is also a second quadrant angle (where sine is also positive) that has a sine of 0.2550. This angle has a reference angle of $13.00 °$ $13.00 °$ and is therefore given by $180 ° - 13.00 ° = 167.00 °$ $180 ° - 13.00 ° = 167.00 °$ .

If we need only an acute angle, $θ = 13.00 °$ $θ = 13.00 °$ is correct. However, if a second-quadrant angle is required, we see that $θ = 167.00 °$ $θ = 167.00 °$ is the angle (see Fig. 8.11). These can be checked by finding the values of $\sin 13.00 °$ $\sin 13.00 °$ and $\sin 167.00 °$ $\sin 167.00 °$ .

The graphs are first and second quadrant position angles. The quadrant 1 terminal side is 13.00 degrees, and the quadrant 2 terminal side is 167.00 degrees. Between the second terminal side and negative x-axis is 13.00 degrees. — Fig. 8.11

EXAMPLE 5 Finding angles given sec`θ`

For $\sec θ = - 2.722$ $\sec θ = - 2.722$ and $0 ° \leq θ < 360 °$ $0 ° \leq θ < 360 °$ (this means $θ$ $θ$ may equal $0 °$ $0 °$ or be between $0 °$ $0 °$ and $360 °$ $360 °$ ), we see from the second line of the calculator display shown in Fig. 8.10 that $θ = 111.55 °$ $θ = 111.55 °$ (rounded off).

The angle $111.55 °$ $111.55 °$ is the second-quadrant angle, but $\sec θ < 0$ $\sec θ < 0$ in the third quadrant as well. The reference angle is $θ_{ref} = 180 ° - 111.55 ° = 68.45 °,$ $θ_{ref} = 180 ° - 111.55 ° = 68.45 °,$ and the third-quadrant angle is $180 ° + 68.45 ° = 248.45 ° .$ $180 ° + 68.45 ° = 248.45 ° .$ Therefore, the two angles between $0 °$ $0 °$ and $360 °$ $360 °$ for which $\sec θ = - 2.722$ $\sec θ = - 2.722$ are $111.55 °$ $111.55 °$ and $248.45 °$ $248.45 °$ (see Fig. 8.12). These angles can be checked by finding $\sec 111.55 °$ $\sec 111.55 °$ and $\sec 248.45 °$ $\sec 248.45 °$ .

The graphs are second and third quadrant position angles. — Fig. 8.12

Figure 8.12 Full Alternative Text

EXAMPLE 6 Finding angles given tan `θ`

Given that $\tan θ = 2.050$ $\tan θ = 2.050$ and $\cos θ < 0,$ $\cos θ < 0,$ find $θ$ $θ$ for $0 ° \leq θ < 360 ° .$ $0 ° \leq θ < 360 ° .$

Because $\tan θ$ $\tan θ$ is positive and $\cos θ$ $\cos θ$ is negative, $θ$ $θ$ must be a third-quadrant angle. A calculator will display an angle of $64.00 °$ $64.00 °$ (rounded off) for $\tan^{- 1} (2.050) .$ $\tan^{- 1} (2.050) .$ However, because we need a third-quadrant angle, we must add $64.00 °$ $64.00 °$ (the reference angle) to $180 °$ $180 °$ . Thus, the required angle is $244.00 °$ $244.00 °$ (see Fig. 8.13). Check by finding $\tan 244.00 °$ $\tan 244.00 °$ .

A position angle in quadrant 3 at 244.00 degrees has a reference angle of 64.00 degrees with the negative x-axis. — Fig. 8.13

If $\tan θ = - 2.050$ $\tan θ = - 2.050$ and $\cos θ < 0,$ $\cos θ < 0,$ the calculator will display an angle of $- 64.00 °$ $- 64.00 °$ for $\tan^{- 1} (- 2.050) .$ $\tan^{- 1} (- 2.050) .$ We would then have to recognize that the reference angle is $64.00 °$ $64.00 °$ and subtract it from $180 °$ $180 °$ to get $116.00 °$ $116.00 °$ , the required second-quadrant angle. This can be checked by finding $\tan 116.00 °$ $\tan 116.00 °$ .

The calculator gives the reference angle (disregarding any minus signs) in all cases except when $\cos θ$ $\cos θ$ is negative. Then it can be used to determine the angle required by the problem.

NOTE

[To avoid confusion from the angle displayed by the calculator, a good procedure is to find the reference angle first.]

We can find the reference angle by entering the absolute value of the function. The displayed angle will be the reference angle. The required angle $θ$ $θ$ is then found by using the reference angle along with the quadrant of $θ$ $θ$ as shown below:

\begin{array}{l} θ = θ_{ref} & (first quadrant) \\ θ = 180 ° - θ_{ref} & (second quadrant) \\ θ = 180 ° + θ_{ref} & (third quadrant) \\ θ = 360 ° - θ_{ref} & (fourth quadrant) \end{array}

$\begin{array}{l} θ = θ_{ref} & (first quadrant) \\ θ = 180 ° - θ_{ref} & (second quadrant) \\ θ = 180 ° + θ_{ref} & (third quadrant) \\ θ = 360 ° - θ_{ref} & (fourth quadrant) \end{array}$ (8.3)

EXAMPLE 7 Using reference angle

Given that $\cos θ = - 0.1298,$ $\cos θ = - 0.1298,$ find $θ$ $θ$ for $0 ° \leq θ < 360 ° .$ $0 ° \leq θ < 360 ° .$

Because $\cos θ$ $\cos θ$ is negative, $θ$ $θ$ is either a second-quadrant angle or a third-quadrant angle. Using 0.1298, the calculator tells us that the reference angle is $82.54 °$ $82.54 °$ .

For the second-quadrant angle, we subtract $82.54 °$ $82.54 °$ from $180 °$ $180 °$ to get $97.46 °$ $97.46 °$ . For the third-quadrant angle, we add $82.54 °$ $82.54 °$ to $180 °$ $180 °$ to get $262.54 °$ $262.54 °$ . See Fig. 8.14. Therefore, the two solutions are $97.46 °$ $97.46 °$ and $262.54 °$ $262.54 °$ .

Figure 8.14 Full Alternative Text

In Section 4.2, we showed how the unit circle could be used to evaluate the trigonometric functions. We now restate this as a definition, which is a special case of the general definition in Eq. (8.1) with the restriction that $r = 1.$ $r = 1.$

There are many situations when the unit circle definition is useful, including evaluating trigonometric functions of quadrantal angles, which have their terminal side along one of the axes (see Fig. 8.16). The sine of $θ$ $θ$ is simply the y-coordinate on the unit circle, and the cosine of $θ$ $θ$ is the x-coordinate. The other functions are found by taking ratios, and if the denominator is zero, the function is undefined.

Figure 8.16 Full Alternative Text

EXAMPLE 8 Quadrantal angles and the unit circle

Find the six trigonometric functions of $0 °$ $0 °$ .

As shown in Fig. 8.16(a), $θ = 0 °$ $θ = 0 °$ intersects the unit circle at (1, 0). Therefore,

$\begin{array}{rclrcl} \sin 0 ° & = & 0 & \csc 0 ° & = & \frac{1}{0} (undefined) \\ \cos 0 ° & = & 1 & \sec 0 ° & = & \frac{1}{1} = 1 \\ \tan 0 ° & = & \frac{0}{1} = 0 & \cot 0 ° & = & \frac{1}{0} (undefined) \end{array}$ $\begin{array}{rclrcl} \sin 0 ° & = & 0 & \csc 0 ° & = & \frac{1}{0} (undefined) \\ \cos 0 ° & = & 1 & \sec 0 ° & = & \frac{1}{1} = 1 \\ \tan 0 ° & = & \frac{0}{1} = 0 & \cot 0 ° & = & \frac{1}{0} (undefined) \end{array}$
Find the six trigonometric functions of $270 °$ $270 °$ .

As shown in Fig. 8.16(d), $θ = 270 °$ $θ = 270 °$ intersects the unit circle at $(0, - 1) .$ $(0, - 1) .$ Thus,

$\begin{array}{rclrcl} \sin 270 ° & = & - 1 & \csc 270 ° & = & \frac{1}{- 1} = - 1 \\ \cos 270 ° & = & 0 & \sec 270 ° & = & \frac{1}{0} (undefined) \\ \tan 270 ° & = & \frac{- 1}{0} (undefined) & \cot 270 ° & = & \frac{0}{- 1} = 0 \end{array}$ $\begin{array}{rclrcl} \sin 270 ° & = & - 1 & \csc 270 ° & = & \frac{1}{- 1} = - 1 \\ \cos 270 ° & = & 0 & \sec 270 ° & = & \frac{1}{0} (undefined) \\ \tan 270 ° & = & \frac{- 1}{0} (undefined) & \cot 270 ° & = & \frac{0}{- 1} = 0 \end{array}$

To evaluate functions of negative angles, we can use functions of corresponding positive angles, if we use the correct sign. In Fig. 8.17, note that $\sin θ = y / r,$ $\sin θ = y / r,$ and $\sin (- θ) = - y / r,$ $\sin (- θ) = - y / r,$ which means $\sin (- θ) = - \sin θ .$ $\sin (- θ) = - \sin θ .$ In the same way, we can get all of the relations between functions of $- θ$ $- θ$ and the functions of $θ .$ $θ .$ Therefore,

Counterclockwise position angle theta with length r passes through (x, y), and clockwise position angle negative theta with length r passes through (x, negative y). A dashed vertical segment connects (x, y) and (x, negative y). — Fig. 8.17

\begin{array}{l} \sin (- θ) = - \sin θ & \cos (- θ) = \cos θ & \tan (- θ) = - \tan θ \\ \csc (- θ) = - \csc θ & \sec (- θ) = \sec θ & \cot (- θ) = - \cot θ \end{array}

$\begin{array}{l} \sin (- θ) = - \sin θ & \cos (- θ) = \cos θ & \tan (- θ) = - \tan θ \\ \csc (- θ) = - \csc θ & \sec (- θ) = \sec θ & \cot (- θ) = - \cot θ \end{array}$ (8.5)

EXAMPLE 9 Negative angles

\begin{array}{l} \sin (- 60 °) = - \sin 60 ° = - 0.8660 & \cos (- 60 °) = \cos 60 ° = 0.5000 \\ \tan (- 60 °) = - \tan 60 ° = - 1.732 & \cot (- 60 °) = - \cot 60 ° = - 0.5774 \\ \sec (- 60 °) = \sec 60 ° = 2.000 & \csc (- 60 °) = - \csc 60 ° = - 1.155 \end{array}

$\begin{array}{l} \sin (- 60 °) = - \sin 60 ° = - 0.8660 & \cos (- 60 °) = \cos 60 ° = 0.5000 \\ \tan (- 60 °) = - \tan 60 ° = - 1.732 & \cot (- 60 °) = - \cot 60 ° = - 0.5774 \\ \sec (- 60 °) = \sec 60 ° = 2.000 & \csc (- 60 °) = - \csc 60 ° = - 1.155 \end{array}$

EXERCISES 8.2

In Exercises 1 and 2, make the given changes in the indicated examples of this section and then solve the resulting problems.

In Example 2, add $40 °$ $40 °$ to each angle, express in terms of the same function of a positive acute angle, and then evaluate.
In Example 7, change the $-$ $-$ to $+$ $+$ and then find $θ .$ $θ .$

In Exercises 3–8, express the given trigonometric function in terms of the same function of the reference angle.

$\sin 155 °$ $\sin 155 °$ , $\cos 220 °$ $\cos 220 °$
$\tan 91 °$ $\tan 91 °$ , $\sec 345 °$ $\sec 345 °$
$\tan 105 °$ $\tan 105 °$ , $\csc 328 °$ $\csc 328 °$
$\cos 190 °$ $\cos 190 °$ , $\tan 290 °$ $\tan 290 °$
$\sec 425 °, \sin (- 520 °)$ $\sec 425 °, \sin (- 520 °)$
$\tan 920 °, \csc (- 550 °)$ $\tan 920 °, \csc (- 550 °)$

In Exercises 9–42, the given angles are approximate. In Exercises 9–16, find the values of the given trigonometric functions by finding the reference angle and attaching the proper sign.

$\sin 195 °$ $\sin 195 °$
$\tan 311 °$ $\tan 311 °$
$\cos 106.3 °$ $\cos 106.3 °$
$\sin 93.4 °$ $\sin 93.4 °$
$\sec 328.33 °$ $\sec 328.33 °$
$\cot 516.53 °$ $\cot 516.53 °$
$\tan (- 109.1 °)$ $\tan (- 109.1 °)$
$\csc (- 108.4 °)$ $\csc (- 108.4 °)$

In Exercises 17–24, find the values of the given trigonometric functions directly from a calculator.

$\cos (- 62.7 °)$ $\cos (- 62.7 °)$
$\cos 141.4 °$ $\cos 141.4 °$
$\sin 310.36 °$ $\sin 310.36 °$
$\tan 242.68 °$ $\tan 242.68 °$
$\csc 194.82 °$ $\csc 194.82 °$
$\sec 441.08 °$ $\sec 441.08 °$
$\tan 148.25 °$ $\tan 148.25 °$
$\sin (- 215.5 °)$ $\sin (- 215.5 °)$

In Exercises 25–38, find

θ

$θ$ for

0 ° \leq θ < 360 ° .

$0 ° \leq θ < 360 ° .$

$\sin θ = - 0.8480$ $\sin θ = - 0.8480$
$\tan θ = - 1.830$ $\tan θ = - 1.830$
$\cos θ = 0.4003$ $\cos θ = 0.4003$
$\sec θ = - 1.637$ $\sec θ = - 1.637$
$\cot θ = - 0.012$ $\cot θ = - 0.012$
$\csc θ = - 8.09$ $\csc θ = - 8.09$
$\sin θ = 0.870, \cos θ < 0$ $\sin θ = 0.870, \cos θ < 0$
$\tan θ = 0.932, \sin θ < 0$ $\tan θ = 0.932, \sin θ < 0$
$\cos θ = - 0.12, \tan θ > 0$ $\cos θ = - 0.12, \tan θ > 0$
$\sin θ = - 0.192, \tan θ < 0$ $\sin θ = - 0.192, \tan θ < 0$
$\csc θ = - 1.366, \cos θ > 0$ $\csc θ = - 1.366, \cos θ > 0$
$\cos θ = 0.0726, \sin θ < 0$ $\cos θ = 0.0726, \sin θ < 0$
$\sec θ = 2.047, \cot θ < 0$ $\sec θ = 2.047, \cot θ < 0$
$\cot θ = - 0.3256, \csc θ > 0$ $\cot θ = - 0.3256, \csc θ > 0$

In Exercises 39–42, find the exact value of each expression without the use of a calculator. (Hint: Start by expressing each quantity in terms of its reference angle.)

$\cos 60 ° + \cos 70 ° + \cos 110 °$ $\cos 60 ° + \cos 70 ° + \cos 110 °$
$\sin 200 ° - \sin 150 ° + \sin 160 °$ $\sin 200 ° - \sin 150 ° + \sin 160 °$
$\tan 40 ° + \tan 135 ° - \tan 220 °$ $\tan 40 ° + \tan 135 ° - \tan 220 °$
$\sec 130 ° - \sec 230 ° + \sec 300 °$ $\sec 130 ° - \sec 230 ° + \sec 300 °$

In Exercises 43–46, determine the function that satisfies the given conditions.

Find $\tan θ$ $\tan θ$ when $\sin θ = - 0.5736$ $\sin θ = - 0.5736$ and $\cos θ > 0.$ $\cos θ > 0.$
Find $\sin θ$ $\sin θ$ when $\cos θ = 0.422$ $\cos θ = 0.422$ and $\tan θ < 0.$ $\tan θ < 0.$
Find $\cos θ$ $\cos θ$ when $\tan θ = - 0.809$ $\tan θ = - 0.809$ and $\csc θ > 0.$ $\csc θ > 0.$
Find $\cot θ$ $\cot θ$ when $\sec θ = 6.122$ $\sec θ = 6.122$ and $\sin θ < 0.$ $\sin θ < 0.$

In Exercises 47–50, insert the proper sign,

>

$>$ or

<

$<$ or

=,

$=,$ between the given expressions. Explain your answers.

$\sin 90 ° 2 \sin 45 °$ $\sin 90 ° 2 \sin 45 °$
$\cos 360 ° 2 \cos 180 °$ $\cos 360 ° 2 \cos 180 °$
$\tan 180 ° \tan 0 °$ $\tan 180 ° \tan 0 °$
$\sin 270 ° 3 \sin 90 °$ $\sin 270 ° 3 \sin 90 °$

In Exercises 51–56, solve the given problems. In Exercises 53 and 54, assume

0 ° < θ < 90 ° .

$0 ° < θ < 90 ° .$ (Hint: Review cofunctions on page 125.)

Using the fact that $\sin 75 ° = 0.9659,$ $\sin 75 ° = 0.9659,$ evaluate $\cos 195 °$ $\cos 195 °$ .
Using the fact that $\cot 20 ° = 2.747,$ $\cot 20 ° = 2.747,$ evaluate $\tan 290 °$ $\tan 290 °$ .
Express $\tan (270 ° - θ)$ $\tan (270 ° - θ)$ in terms of the $\cot θ .$ $\cot θ .$
Express $\cos (90 ° + θ)$ $\cos (90 ° + θ)$ in terms of $\sin θ .$ $\sin θ .$
For a triangle with angles A, B, and C, evaluate $tan A + \tan (B + C) .$ $tan A + \tan (B + C) .$
(a) Is $\sin 180 ° = 2 \sin 90 ° ?$ $\sin 180 ° = 2 \sin 90 ° ?$ (b) Is $\sin 360 ° = 2 \sin 180 ° ?$ $\sin 360 ° = 2 \sin 180 ° ?$

In Exercises 57–60, evaluate the given expressions.

The current i in an alternating-current circuit is given by $i = i_{m} \sin θ,$ $i = i_{m} \sin θ,$ where $i_{m}$ $i_{m}$ is the maximum current in the circuit. Find i if $i_{m} = 0.0259 A$ $i_{m} = 0.0259 A$ and $θ = 495.2 ° .$ $θ = 495.2 ° .$
The force F that a rope exerts on a crate is related to force $F_{x}$ $F_{x}$ directed along the x-axis by $F = F_{x} \sec θ,$ $F = F_{x} \sec θ,$ where $θ$ $θ$ is the standard-position angle for F. See Fig. 8.18. Find F if $F_{x} = - 365 N$ $F_{x} = - 365 N$ and $θ = 127.0 ° .$ $θ = 127.0 ° .$

Fig. 8.18
For the slider mechanism shown in Fig. 8.19, $y \sin α = x \sin β .$ $y \sin α = x \sin β .$ Find y if $x = 6.78$ $x = 6.78$ in., $α = 31.3 °,$ $α = 31.3 °,$ and $β = 104.7 ° .$ $β = 104.7 ° .$

Fig. 8.19
A laser follows the path shown in Fig. 8.20. The angle $θ$ $θ$ is related to the distances a, b, and c by $2 a b \cos θ = a^{2} + b^{2} - c^{2} .$ $2 a b \cos θ = a^{2} + b^{2} - c^{2} .$ Find $θ$ $θ$ if $a = 12.9 cm, b = 15.3 cm,$ $a = 12.9 cm, b = 15.3 cm,$ and $c = 24.5 cm .$ $c = 24.5 cm .$

Fig. 8.20

Answers to Practice Exercises

$\sin 15 °; - \csc 15 °$ $\sin 15 °; - \csc 15 °$
$- \sec 15 °; \cos 15 °$ $- \sec 15 °; \cos 15 °$
$- \tan 15 °; \cot 15 °$ $- \tan 15 °; \cot 15 °$
$55.0 °$ $55.0 °$ , $305.0 °$ $305.0 °$
$- 1,$ $- 1,$ undefined

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Table of Contents for 8.2 Trigonometric Functions of Any Angle

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Table of Contents for
8.2 Trigonometric Functions of Any Angle