Inverse Trigonometric Functions

Section 2.4 Inverse Trigonometric Functions

Swells are a crucial navigational tool, providing a consistent means of maintaining a straight course over extended periods. Unlike stars, which may be obscured during the day or on cloudy nights, or winds that can change direction frequently, swells tend to remain relatively constant. Crew members can navigate a straight course by keeping the angle at which the swell passes the canoe constant.

🔗

Even in poor visibility, when the crew may not see the swells, they can feel them. Parts of the vaka (canoe) lift and lower as the swells pass beneath. Skilled navigators use these movements to maintain course, even without directly seeing the waves.

🔗

The lateral (side-to-side) rocking motion of a double-hull canoe is known as roll. This motion occurs when a swell approaches from either the port (left) or starboard (right) side, causing the vaka to initially lift the corresponding hull (port or starboard), followed by the opposite hull, depending on the direction of the swell.

🔗

When a swell approaches from the bow (front) of the canoe, it lifts the front, causing the canoe to tilt backward before tilting forward. This motion is known as pitch. Conversely, if the swell comes from the stern (back), the canoe tilts forward and then backward.

🔗

When a swell approaches the canoe from an angle that is not perpendicular to any side, a combination of pitch and roll occurs. The specific motion experienced depends on the precise angle of the swell. For instance, if the swell comes from an angle and lifts the starboard bow (front right), it may subsequently lift the port bow, followed by the starboard stern, and finally the port stern. However, with a slightly different angle, the sequence may change. After lifting the starboard bow, it could lead to the starboard stern being lifted next, followed by the port bow, and finally the port stern. This twisting motion is referred to as a corkscrew effect due to its combination of motions.

🔗

The video in Figure 2.4.1 demonstrates how vaka Paikea moves as swells pass under the hulls. A change in vaka motion can indicate either a change in the canoe’s direction or a shift in the direction of ocean swells. In such cases, the crew must assess the situation and, when conditions allow, utilize celestial markers, such as the rising and setting of stars, to determine the direction of the swells.

🔗

Figure 2.4.1. As swells pass under vaka Paikea, the canoe pitches, rolls, and corkscrews, depending on the angle of the swell. A navigator can use these movements to keep a straight course. If you are viewing the PDF or a printed copy, you can scan the QR code or follow the “Standalone” link to watch the video online.

🔗

As vaka Paikea sails north in the Cook Islands from Rarotonga to Aitutaki with a heading of

0^{\circ},

the swells are approaching the vaka from the southwest and moving towards the northeast. Referring to Figure 1.2.17, which provides heading angles, we observe that the swells have a heading ranging between

0^{\circ}

(north) and

90^{\circ}

(east).

🔗

Additional observations of the crew reveal that the swells hit Paikea in the following order: 1) port stern (back left); 2) starboard stern (back right); 3) port bow (front left); and 4) starboard bow (front right), as shown in Figure 2.4.2. If Paikea is 14.8 m (50 ft) in length and 6.2 m (20 ft) in width, what is the possible range of headings from which the swells may be approaching?

🔗

Figure 2.4.2. Swells moving towards the northeast pass under vaka Paikea heading.
🔗

To determine the range of headings, we start by considering one boundary, which occurs when the swells are moving directly north with a heading of

0^{\circ} .

To identify the other boundary, we want to find the angle at which the swell intersects both the starboard stern and port bow simultaneously as it passes beneath the vaka. Essentially, we want to find the angle that diagonally traverses the canoe from one corner to another. To simplify this, we represent the vaka as a rectangle and create a triangle by connecting the corners. The angle, denoted as

θ,

is formed between the side adjacent to the 14.8 m length of the canoe and the diagonal. This angle corresponds to the heading of the swells, as shown in Figure 2.4.3.

🔗

Figure 2.4.3. The deck of vaka Paikea can be simplified to a rectangle, and its diagonal represents the threshold between different sequences of motions as swells pass. The angle $θ$ corresponds to the heading of the swells.
🔗

You may notice that for our given angle,

θ,

we have the opposite side of length 6.2 m and the adjacent side of length 14.8 m, which are both related to the tangent function:

\tan θ = \frac{6.2}{14.8} .

🔗

Until now, the angles for the trigonometric problems we encountered were given. However, to solve for the angle

θ,

we cannot simply divide by the letters “tan”, since it is part of the tangent function, which takes an angle as input and provides a ratio of sides as output. To find the angle

θ,

we need to use the inverse tangent function, “tan^{-1}”, which takes the ratio of sides as input and gives the angle as output. To find the angle

θ,

we need to use the inverse function, which takes the ratio of sides as input and provides an angle as output. In this section, we will explore inverse trigonometric functions, including their properties and usage.

🔗

Subsection 2.4.1 Inverse Trigonometric Functions

Recall from algebra that for a function

f

and its inverse function

f^{- 1}

we have:

The Domain of $f^{- 1} =$ Range of $f;$
The Range of $f^{- 1} =$ Domain of $f;$
If $f (a) = b$ then $f^{- 1} (b) = a .$

🔗

In terms of trigonometric functions, for example, if

f (x) = \sin x

then

f^{- 1} (x) = \sin^{- 1} x .

Now consider

\sin (\frac{π}{4}) = \frac{1}{\sqrt{2}},

then

\frac{π}{4} = \sin^{- 1} (\frac{1}{\sqrt{2}}) .

🔗

Remark 2.4.4. Be Careful.

Do not confuse the inverse trigonometric notation with an exponent, in other words,

\sin^{- 1} x \neq \frac{1}{\sin x} .

To avoid this, we will use parentheses around the trigonometric function to denote the power of negative one:

(\sin x)^{- 1} .

🔗

Also recall from algebra that for a function,

f,

to have an inverse,

f^{- 1},

it must be one-to-one, meaning no horizontal line intersects the graph more than once. Since this is not true for trigonometric functions, they do not have inverses. We need functions to be one-to-one because both

\sin \frac{π}{4} = \frac{1}{\sqrt{2}}

and

\sin \frac{3 π}{4} = \frac{1}{\sqrt{2}}

but if we took the inverse of sine, we wouldn’t know whether to use

\sin^{- 1} (\frac{1}{\sqrt{2}}) = \frac{π}{4}

\sin^{- 1} (\frac{1}{\sqrt{2}}) = \frac{3 π}{4} .

To avoid this confusion and to ensure the function is one-to-one, we can put restrictions on the domains of each trigonometric function so they attain all the values in the range only once, making the restricted function one-to-one, and thus having an inverse (see Figure 2.4.5).

🔗

Figure 2.4.5. The graph of $f (x) = \sin x$ does not pass the horizontal line test and is not one-to-one. If we restrict the graph to $- \frac{π}{2} \leq x \leq \frac{π}{2}$ so that each value in the range $[- 1, 1]$ is attained only once, then the restricted function is one-to-one and has an inverse.
🔗

We will restrict the domain of

y = \sin x

to the interval

[- \frac{π}{2}, \frac{π}{2}],

the domain of

y = \cos x

[0, π],

and the domain of

y = \tan x

to the interval

(- \frac{π}{2}, \frac{π}{2}) .

Notice that the domain for each trigonometric function includes one quadrant where the function is positive and one quadrant where it is negative. The domains and the corresponding graphs for sine, cosine, and tangent are shown in Figure 2.4.6, Figure 2.4.7, and Figure 2.4.8, respectively.

🔗

Figure 2.4.6. The domain of $y = \sin x$ (left) and its graph on the restricted domain (right).
🔗

Figure 2.4.7. The domain of $y = \cos x$ (left) and its graph on the restricted domain (right).
🔗

Figure 2.4.8. The domain of $y = \tan x$ (left) and its graph on the restricted domain (right).
🔗

With these restrictions on the domains, we now have trigonometric functions that are one-to-one and so we can define their inverse functions:

🔗

Definition 2.4.9. Inverse Sine.

The inverse sine function is symbolized by

\begin{aligned} y = \sin^{- 1} x & and means & x = \sin y . \end{aligned}

The inverse sine function is also called the arcsine function, and is denoted by

\arcsin x .

🔗

Definition 2.4.10. Inverse Cosine.

The inverse cosine function is symbolized by

\begin{aligned} y = \cos^{- 1} x & and means & x = \cos y . \end{aligned}

The inverse cosine function is also called the arccosine function, and is denoted by

\arccos x .

🔗

Definition 2.4.11. Inverse Tangent.

The inverse tangent function is symbolized by

\begin{aligned} y = \tan^{- 1} x & and means & x = \tan y . \end{aligned}

The inverse tangent function is also called the arctangent function, and is denoted by

\arctan x .

🔗

Definition 2.4.12. Inverse Cosecant, Secant, and Cotangent.

Inverse cosecant, inverse secant, and inverse cotangent functions are not as common as the other trigonometric functions and we will just summarize them.

\begin{aligned} y = \csc^{- 1} x & means & x = \csc y, \\ y = \sec^{- 1} x & means & x = \sec y, \\ y = \cot^{- 1} x & means & x = \cot y . \end{aligned}

🔗

Definition 2.4.13. Domain and Range for Inverse Trigonometric Functions.

The domain and range for each function is:

Function	Domain	Range

$\sin^{- 1} x$	$[- 1, 1]$	$[- \frac{π}{2}, \frac{π}{2}]$
$\cos^{- 1} x$	$[- 1, 1]$	$[0, π]$
$\tan^{- 1} x$	$(- \infty, \infty)$	$(- \frac{π}{2}, \frac{π}{2})$
$\csc^{- 1} x$	$(- \infty, - 1] \cup [1, \infty)$	$[- \frac{π}{2}, 0) \cup (0, \frac{π}{2}]$
$\sec^{- 1} x$	$(- \infty, - 1] \cup [1, \infty)$	$[0, \frac{π}{2}) \cup (\frac{π}{2}, π]$
$\cot^{- 1} x$	$(- \infty, \infty)$	$(0, π)$

🔗

Subsection 2.4.2 Finding the Exact Value of an Inverse Trigonometric Function

Example 2.4.14. Evaluating Inverse Trigonometric Functions.

Find the exact value of:

(a)

\cos^{- 1} (\frac{1}{2})

Solution.

Let

θ = \cos^{- 1} (\frac{1}{2}) .

Evaluating the problem is the same as determining the angle,

θ,

for which

\cos θ = \frac{1}{2} .

Although there are infinitely many values of

θ

that satisfy the equation, such as

θ = \frac{π}{3}

and

θ = \frac{5 π}{3},

there is only one value that lies in the interval

[0, π] .

Thus,

\cos^{- 1} (\frac{1}{2}) = \frac{π}{3} .

🔗

(b)

\tan^{- 1} (\sqrt{3})

Solution.

Let

θ = \tan^{- 1} (\sqrt{3}) .

We must find

θ

that satisfies

\tan θ = \sqrt{3}

while also ensuring

θ

is within the range of the inverse tangent function. Because

\tan \frac{π}{3} = \sqrt{3}

and

- \frac{π}{2} < \frac{π}{3} < \frac{π}{2},

we conclude that

\tan^{- 1} (\sqrt{3}) = \frac{π}{3} .

🔗

(c)

\sin^{- 1} (- \frac{1}{\sqrt{2}})

Solution.

The angle,

θ,

in the interval

[- \frac{π}{2}, \frac{π}{2}]

that satisfies

\sin θ = - \frac{1}{\sqrt{2}}

θ = \sin^{- 1} (- \frac{1}{\sqrt{2}}) = - \frac{π}{4} .

🔗

Subsection 2.4.3 Approximations of Inverse Trigonometric Functions

To evaluate inverse trigonometric functions that do not have special angles as values, we will need to use a calculator.

🔗

Remark 2.4.15. Using a Calculator.

Using a calculator for inverse trigonometric functions: Most calculators will have a special key for the inverse sine, inverse cosine, and inverse tangent functions. Depending on the calculator, we may see the following keys for the inverse trigonometric functions:

Function	Calculator Key
inverse sine	`SIN^-1`, `INVSIN`, `ARCSIN`, or `ASIN`
inverse cosine	`COS^-1`, `INVCOS`, `ARCCOS`, or `ACOS`
inverse tangent	`TAN^-1`, `INVTAN`, `ARCTAN`, or `ATAN`

Often, the inverse trigonometry key can be found by first pressing 2nd or SHIFT, followed by the trigonometry function key. For example, to get the SIN^-1 key, press 2ndSIN or SHIFTSIN.

🔗

Many calculators do not have specific keys for the inverse cosecant, inverse secant, and inverse cotangent functions. Instead, we can use the Reciprocal Identities (Definition 1.4.2) to get:

🔗

Function	Calculator Key
inverse cosecant	`1/SIN^-1`
inverse secant	`1/COS^-1`
inverse cotangent	`1/TAN^-1`

🔗

Example 2.4.16. Finding an Approximate Value of Inverse Trigonometric Functions.

Use a calculator to approximate the value of each expression in radians, rounded to two decimals.

(a)

\cos^{- 1} (- 0.39)

Solution.

First verify the mode of the calculator is in radians. Then, press the following keys COS^-1((-)0.39)ENTER to get

\cos^{- 1} (- 0.39) \approx 1.9714279195 .

If the calculator does not have the COS^-1 key, then use the appropriate key(s) for inverse cosine, such as INVCOS, ARCCOS, or ACOS.

🔗

On some calculators, COS^-1 is pressed first, then ((-)0.39); while other calculators the sequence is reversed with ((-)0.39) pressed first, then COS^-1. Verify with the calculator’s manual.

🔗

To find the other boundary, we want to determine the angle at which the swell simultaneously intersects both the starboard stern and port bow as it passes beneath the vaka. This angle, denoted as

θ,

is formed between the side adjacent to the 14.8 m length of the canoe and the diagonal. Using the tangent function we have:

🔗

\tan θ = \frac{6.2}{14.8} .

🔗

Solving for

θ

using the inverse tangent function we get:

🔗

\tan^{- 1} (\frac{6.2}{14.8}) \approx {22.73}^{\circ} .

🔗

Therefore, the possible range of headings from which the swells may be approaching is between

0^{\circ}

(directly north) and approximately

{22.73}^{\circ} .

🔗

Example 2.4.18. A gift for Mau.

In 1999, after Mau Pialug, also known as Papa Mau, sailed from Hawaiʻi to his home in Satawal, Micronesia aboard the Makaliʻi, the crew of Nā Kālai Waʻa expressed their gratitude to the man who shared his knowledge of navigation by constructing a sister canoe to Makaliʻi. This vessel, named the Alingano Maisu, was a 56-foot long double-hulled voyaging canoe. In 2007, accompanied by the Hōkūleʻa, the Alingano Maisu embarked on her inaugural journey to Satawal, continuing Papa Mau’s legacy of navigation in his home islands.

During this journey, the canoes sailed directly towards Johnston Atoll, using it as a sighting point without making a stop, before proceeding to their first destination in Majuro, Marshall Islands. The island of Majuro is situated 1,108 nautical miles west and 575 nautical miles south of Johnston Atoll.

🔗

What house do we need to sail in and what distance will we need to sail? If the waʻa travels at 5 knots, how many days will it take to reach the destination? Note that 1 knot = 1 nautical mile/hour.

🔗

We will use the tangent function because we are given the side opposite to

θ

(575 NM) and the side adjacent to

θ

(1,108 NM) and express it as

\tan θ = \frac{575}{1, 108} .

🔗

Solution.

Since we were able to write

\tan θ = \frac{575}{1, 108},

we can use a calculator or other technology to evaluate the inverse tangent to find our angle:

θ = \tan^{- 1} (\frac{575}{1, 108}) \approx {27.4}^{\circ} .

Because this angle is in Quadrant III, we find its value on the Unit Circle by adding

180^{\circ} + {27.4}^{\circ} = {207.4}^{\circ} .

Next we refer to the Star Compass with angles (Figure 1.2.4) to conclude we will need to sail towards the House ʻĀina Kona.

🔗

To determine the distance,

d,

we use the Pythagorean Theorem:

d = \sqrt{575^{2} + 1, 108^{2}} \approx 1, 248.3 NM .

Finally, we note that since (speed) = (distance)/(duration), we can rearrange the terms to get (duration) = (distance)/(speed). If we travel at 5 knots (5 NM/hr), we can calculate the duration as

🔗

\begin{aligned} duration = & \frac{1, 248.3 NM}{5 knot} & \frac{distance}{speed} \\ = & \frac{1, 248.3 NM}{5 NM/hr} & 1 knot=1 NM/hr \\ = & 249.7 hours & \frac{NM}{NM/hr} = NM \cdot \frac{hr}{NM} = hr \\ = & 249.7 hours \cdot \frac{day}{24 hours} \\ \approx & 10.4 days . \end{aligned}

🔗

Example 2.4.19. Finding Land.

As Hōkūleʻa is sailing towards Rapa Nui, the navigator uses a process called dead reckoning to determine their position based on the latitude, measured by the stars, and other factors such as the estimated distance traveled, speed, and direction. Once the navigator has determined that they are in the vicinity of land, her attention is now focused on looking for signs of land. One method navigators will use is to look for land-based seabirds such as the manu-o-Kū (fairy tern) and the noio (noddy tern), which go out to sea in the morning to fish and return to land at night. However, Rapa Nui’s seabird population has been reduced so she will look for other signs such as drifting land vegetation; clouds that form over islands; the loom of the island when white sand and still lagoons reflect the sunlight or moonlight upwards; and distinctive patterns of swells bending (refracting) around or reflecting off islands. Land will be spotted when the navigator first sees Maʻunga Terevaka, the tallest point in Rapa Nui, which stands at 1,665 ft. Figure 2.4.20 depicts the relation between the canoe and island (left) as well as what is seen from the deck of the canoe (right).

Figure 2.4.20. Watch as the canoe approaches an island from two different views. On the left, see a side view as the island gradually comes into sight over the horizon. On the right, experience the perspective from the canoe’s deck, observing the island appearing to rise from the water. This dual perspective provides a unique glimpse into the legend of Māui, the demigod who pulled the islands from the ocean with his fish-hook. If you are viewing the PDF or a printed copy, you can scan the QR code or follow the “Standalone” link to watch the video online.

On the deck of Hōkūleʻa, a navigator stands at 9 ft above sea level. If she looks out to the sea, how far is she from horizon? Assume the radius of the earth at Rapa Nui is 20,911,171 feet.
How far is Hōkūleʻa from Maʻunga Terevaka when it first becomes visible over the horizon to someone standing 9ft above sea level?

🔗

Solution.

We begin by assuming that the earth is a sphere with radius $R .$ Standing on Hōkūleʻa, the line from the navigator’s eye to the horizon is tangent to the circle of radius $R .$

Here $h = 9$ ft, represents the height of the navigator’s eye above sea level, $R = 20, 911, 171$ ft is the radius of the earth at Rapa Nui, and $s_{1}$ is the arc length or distance along the surface of the earth from the navigator to the horizon. We have written the distances in feet since the height of the navigator is in feet. Recall from Theorem 1.2.26 the formula for finding the arc length is $s_{1} = 2 π R \cdot (\frac{θ_{1}}{360}) .$ Since we know $R,$ we only need to find $θ_{1} .$ Notice the right triangle.

Since we know the adjacent side and hypotenuse of this triangle, we can use cosine:

$\cos θ_{1} = \frac{R}{R + h} = \frac{20, 911, 171}{20, 911, 171 + 9} = \frac{20, 911, 171}{20, 911, 180} .$

To solve for $θ_{1},$ we use the inverse cosine:

$θ_{1} = \cos^{- 1} (\frac{20, 911, 171}{20, 911, 180}) \approx {0.053}^{\circ} .$

Now we are ready to calculate the arc length, $s_{1} :$

$s_{1} = 2 π \cdot \frac{{0.053}^{\circ}}{360^{\circ}} \cdot 20, 911, 171 feet \approx 19, 343 feet .$

Converting to miles we get

$s_{1} \approx 19, 401 feet \cdot \frac{1 mile}{5, 280 feet} \approx 3.7 miles .$

So the horizon is 3.7 miles from the navigator.
Next, to determine how far the navigator is from Maʻunga Terevaka when it emerges over the horizon, we need to align the top of the mountain with line from the navigator’s eye to the horizon.

Here, $H = 1, 665$ ft is the height of Maʻunga Terevaka. To find the distance from the top of the mountain to the horizon, we will need to determine $s_{2} .$ We begin by redrawing the triangle.

The angle is then given by

$\begin{aligned} θ_{1} & = \cos^{- 1} (\frac{R}{R + H}) \\ = \cos^{- 1} (\frac{20, 911, 171}{20, 911, 173 + 1, 665}) \\ = \cos^{- 1} (\frac{20, 911, 171}{20, 912, 836}) \\ \approx {0.723}^{\circ} . \end{aligned}$

We conclude that the distance from Maʻunga Terevaka to the horizon is

$s_{2} = 2 π \cdot \frac{{0.723}^{\circ}}{360^{\circ}} \cdot (20, 911, 171 feet) \cdot \frac{1 mile}{5, 280 feet} \approx 50.0 miles .$

Therefore, the total distance between the navigator and Maʻunga Terevaka is $s_{1} + s_{2} = 3.7 + 50.0 = 53.7$ miles. Note that this is the distance when the island may first be seen, however, weather conditions may reduce visibility.

🔗

Subsection 2.4.4 Graphs of Inverse Trigonometric Functions

Recall from algebra that:

🔗

The point $(a, b)$ is on the graph of $f$ if and only if the point $(b, a)$ is on the graph of $f^{- 1} .$
The graphs of $f^{- 1}$ and $f$ are reflections of each other about the line $y = x .$

🔗

The graph of each inverse trigonometric function can be obtained by reflecting the graph of the original function about the line

y = x .

🔗

Subsection 2.4.5 Composition of Inverse Trigonometric Functions

Recall from Algebra that if

f

is a one-to-one function with inverse

f^{- 1},

then:

🔗

$f (f^{- 1} (y)) = y$ for every $y$ in the domain of $f^{- 1} .$
$f^{- 1} (f (x)) = x$ for every $x$ in the domain of $f .$

🔗

In terms of trigonometric functions,

f (f^{- 1} (y)) = y

holds for all

y

in the domain, however, we need to be careful when evaluating

f^{- 1} (f (x)) = x

because the domain of

f^{- 1}

is restricted.

🔗

Definition 2.4.21. Properties of Composition of Trigonometric Functions.

$\sin (\sin^{- 1} x) = x$	when	$- 1 \leq x \leq 1$
$\cos (\cos^{- 1} x) = x$	when	$- 1 \leq x \leq 1$
$\tan (\tan^{- 1} x) = x$	when	$- \infty < x < \infty$
$\sin^{- 1} (\sin x) = x$	only when	$- \frac{π}{2} \leq x \leq \frac{π}{2}$
$\cos^{- 1} (\cos x) = x$	only when	$0 \leq x \leq π$
$\tan^{- 1} (\tan x) = x$	only when	$- \frac{π}{2} < x < \frac{π}{2}$

🔗

Example 2.4.22. Composition of a trigonometric function and the inverse of the same trigonometric function.

Find the exact value of each expression:

(a)

\cos^{- 1} (\cos \frac{π}{8})

Solution.

Since

\frac{π}{8}

is in the interval

[0, π],

[- \frac{π}{2}, \frac{π}{2}] .

In order to evaluate the expression, we first need to find an angle,

θ,

such that

- \frac{π}{2} \leq θ \leq \frac{π}{2}

and

\sin \frac{5 π}{7} = \sin θ .

Since

\frac{π}{2} < \frac{5 π}{7} < π,

\frac{5 π}{7}

is in Quadrant II. Recall from Section 1.3 that our reference angle is

θ = π - \frac{5 π}{7} = \frac{2 π}{7} .

🔗

Since

\frac{2 π}{7}

is in the interval

[- \frac{π}{2}, \frac{π}{2}],

\sin^{- 1} (\sin \frac{5 π}{7}) = \sin^{- 1} (\sin \frac{2 π}{7}) = \frac{2 π}{7} .

🔗

(e)

\cos (\cos^{- 1} (- 0.283))

Solution.

Since

- 0.283

is in the interval

[- 1, 1],

\cos (\cos^{- 1} (- 0.283)) = - 0.283 .

🔗

Now we will look at what happens when we need to evaluate the composition of a trigonometric function and the inverse of a different trigonometric function.

🔗

Example 2.4.23. Composition of a trigonometric function and the inverse of a different trigonometric function.

Find the exact value of:

(a)

\cos (\tan^{- 1} (\frac{4}{3}))

Solution.

Let

θ

be an angle in the range of the inverse tangent, that is, let

θ

be in the interval

(- \frac{π}{2}, \frac{π}{2})

such that

θ = \tan^{- 1} (\frac{4}{3}) .

This equation is equivalent to

\tan θ = \frac{4}{3} .

Since

\tan θ > 0,

we know that

θ

must be in Quadrant I

(0 < θ < \frac{π}{2}) .

Let

(x, y)

be a point on the terminal side of

θ .

Then by the trigonometric ratios,

\tan θ = \frac{opposite}{adjacent} = \frac{y}{x}

Since we have

\tan θ = \frac{4}{3},

we let

x = 3

and

y = 4,

as shown in the figure below.

🔗

Evaluating

\cos (\tan^{- 1} (\frac{4}{3}))

is equivalent to evaluating

\cos θ = \frac{adjacent}{hypotenuse} = \frac{3}{r},

where

r = \sqrt{x^{2} + y^{2}} = \sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 .

So,

🔗

\cos (\tan^{- 1} (\frac{4}{3})) = \cos θ = \frac{3}{5} .

🔗

(b)

\sin (\cos^{- 1} (- \frac{3}{8}))

Solution.

Let

θ

be an angle in the range of the inverse cosine, that is, let

θ

be in the interval

[0, π]

such that

θ = \cos^{- 1} (- \frac{3}{8}) .

This equation is equivalent to

\cos θ = - \frac{3}{8} .

Since

\cos θ < 0,

we know that

θ

must be in Quadrant II

(\frac{π}{2} < θ < π) .

Let

(x, y)

be a point on the terminal side of

θ .

Then by the trigonometric ratios,

\cos θ = \frac{adjacent}{hypotenuse} = \frac{x}{r} .

Since we have

\cos θ = - \frac{3}{8},

we get

\frac{x}{r} = - \frac{3}{8}

which gives us either

x = - 3

and

r = 8

x = 3

and

r = - 8 .

Since

θ

is in Quadrant II, we know that

x

is negative, thus

x = - 3

and

r = 8,

as shown in the figure below.

🔗

Evaluating

\sin (\cos^{- 1} (- \frac{3}{8}))

is equivalent to evaluating

\sin θ = \frac{opposite}{hypotenuse} = \frac{y}{8},

where

🔗

\begin{aligned} r^{2} & = x^{2} + y^{2} \\ 8^{2} & = (- 3)^{2} + y^{2} \\ 64 & = 9 + y^{2} \\ y^{2} & = 55 . \end{aligned}

Thus

y = \sqrt{55} .

So,

🔗

\sin (\cos^{- 1} (\frac{- 3}{8})) = \sin θ = \frac{\sqrt{55}}{8} .

Note that

\sin θ

is positive, since sine is positive in Quadrant II.

🔗

Exercises 2.4.6 Exercises

\tan^{- 1} (78.9)

Answer.

1.56

🔗

🔗

Exercise Group.

Utilizing the fact that

x = \frac{x}{1}

and sketching a right triangle, find the exact value of the expression in terms of

x .

63.

\cos (\tan^{- 1} x)

Answer.

\frac{1}{\sqrt{x^{2} + 1}}

🔗

64.

\sin (\cos^{- 1} x)

Answer.

\sqrt{1 - x^{2}}

🔗

65.

\sec (\sin^{- 1} x)

Answer.

\frac{1}{\sqrt{1 - x^{2}}}

🔗

66.

\sin (\sec^{- 1} x)

Answer.

The summit of Mauna Kea is the highest point in Hawaiʻi and sits 13,803 ft above the sea level. If we stand at the summit, we will be able to see the sun rise before someone standing at the sea level just north or south of Mauna Kea (at the same latitude). In fact, we will see the sunrise at the same time as someone at sea level sailing in a waʻa to the east of Mauna Kea. Assume the radius of Earth is 20,917,655 feet.

(a)

How far is the horizon when someone 5 ft tall stands at sea level to watch the sun rise? Round the answer to the nearest tenth of a mile.

Answer.

2.7

miles

🔗

(b)

How far is the horizon when someone 5 ft tall stands on the summit to watch the sunrise? Round the answer to the nearest tenth of a mile.

Answer.

143.9

miles

🔗

(c)

How much faster would someone 5 ft tall standing on the top of the mountain see the sunrise than someone 5 ft tall standing at sea level? Express the answer in minutes, rounded to one decimal place.

Hint.

The ratio of the distance to the horizon (the previous answers) to the circumference of the Earth (

2 π R

) is equal to the ratio of the time it takes to see the sunrise to the 24 hours (1,440 minutes) of Earth’s rotation. Compute the time it takes to see the sunrise from the top of the mountain and the shore, then calculate the difference.

Answer.

8.2 minutes

🔗

72. Māui Capturing the Sun.

According to moʻolelo, or legend, the sun traveled very fast across the sky, leaving people with days so short there was not enough time to carry on with their daily lives. Determined to slow the sun, the demigod Māui climbed to the summit of Haleakalā, which stands at 10,023 feet, to snare the sun. Assume the radius of the Earth at Haleakalā is the same as at Mauna Kea, 20,917,655 feet.

(a)

How far is the horizon when Māui stands at the summit? Assume Māui’s eyes are 7 ft from the ground. Round your answer to the nearest tenth of a mile.

Answer.

122.7 miles

🔗

(b)

How much sooner would Māui see the sun emerge over the horizon compared to someone whose eyes are 7 feet above sea level? Express your answer in minutes, rounded to one decimal place.

Answer.

7.1 minutes

🔗

Prev Top Next