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 side-to-side rocking motion of the 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 as well. 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 the 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.

As the vaka Paikea sails north in the Cook Islands from Rarotonga to Aitutaki with a heading of \(0^{\circ}\text{,}\) 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 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?

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 \(0^{\circ}\text{.}\) 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 \(\theta\text{,}\) 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.

You may notice that for our given angle, \(\theta\text{,}\) 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:

Until now, the angles for the trigonometric problems we encountered were given. However, to solve for the angle \(\theta\text{,}\) we cannot simply divide by “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 \(\theta\text{,}\) 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.

Subsection2.4.1Inverse 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 \(f^{-1}(x)=\sin^{-1}x\text{.}\) Now consider \(\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\text{,}\) then \(\frac{\pi}{4}=\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)\text{.}\)

Remark2.4.4.Be Careful.

Do not confuse the inverse trigonometric notations with an exponent, in other words, \(\sin^{-1}x\neq\frac{1}{\sin x}\text{.}\) To avoid this, we will use parentheses around the trigonometric function to denote the power of negative one: \((\sin x)^{-1}\text{.}\)

Also recall from algebra that for a function, \(f\text{,}\) to have an inverse, \(f^{-1}\text{,}\) 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 \(\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}\) and \(\sin\frac{3\pi}{4}=\frac{\sqrt{2}}{2}\) but if we took the inverse of sine, would we wouldn’t know to use \(\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)=\frac{\pi}{4}\) or \(\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)=\frac{3\pi}{4}\text{.}\) 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 function one-to-one, and thus have an inverse (see Figure 2.4.5).

We will restrict the domain of \(y=\sin x\) to the interval \([-\frac{\pi}{2},\frac{\pi}{2}]\text{,}\) the domain of \(y=\cos x\) to \([0,\pi]\text{,}\) and the domain of \(y=\tan x\) to the interval \((-\frac{\pi}{2},\frac{\pi}{2})\text{.}\) 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.

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

The inverse tangent function is also called the arctangent function, and is denoted by \(\arctan x\text{.}\)

Definition2.4.12.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.

Let \(\theta=\cos^{-1}\left(\frac{1}{2}\right)\text{.}\) Then evaluating the problem is the same as determining the angle, \(\theta\text{,}\) for which

Although there are infinite values of \(\theta\) that satisfy the equation, such as \(\theta=\frac{\pi}{3}\) and \(\theta=\frac{5\pi}{3}\text{,}\) there is only one value that lies in the interval \([0,\pi]\text{.}\) Thus, \(\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\text{.}\)

(b)

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

Solution.

Let \(\theta=\tan^{-1}\sqrt{3}\text{.}\) Then we must find \(\theta\) that satisfies \(\tan\theta=\sqrt{3}\) as well as satisfies the range of \(\tan^{-1}\text{.}\) Because \(\tan\frac{\pi}{3}=\sqrt{3}\) and \(-\frac{\pi}{2}\lt\frac{\pi}{3}\lt\frac{\pi}{2}\text{,}\) we conclude that \(\tan^{-1}\sqrt{3}=\frac{\pi}{3}\text{.}\)

(c)

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

Solution.

The angle, \(\theta\text{,}\) in the interval \([-\frac{\pi}{2},\frac{\pi}{2}]\) that satisfies \(\sin\theta=-\frac{\sqrt{2}}{2}\) is \(\theta=\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)=-\frac{\pi}{4}\text{.}\)

Subsection2.4.3Approximations of Inverse Trigonometric Functions

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

Remark2.4.15.Using a Calculator.

Using a calculator for inverse trigonometric functions: Most calculators will have a special key for the inverse sine, cosine, and tangent functions. Depending on your calculator, you may see the following keys for inverse 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, secant, and cotangent functions. Instead, you can 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

Example2.4.16.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

If your 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 your calculator’s manual.

(b)

\(\tan^{-1}12\)

Solution.

First verify the mode of the calculator is in radians. Then, using the appropriate key for inverse tangent, press the following keys TAN^-1(12)ENTER to get

First verify the mode of the calculator is in radians. Then, using the appropriate key for inverse sine, press the following keys SIN^-1(0.8)ENTER to get

At the start of this section, we discussed the swells passing through 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). Given that 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?

Solution.

If the swells are moving directly north, the heading will be \(0^{\circ}\text{.}\)

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 \(\theta\text{,}\) is formed between the side adjacent to the 14.8 m length of the canoe and the diagonal. Using the tangent function we have:

Therefore, the possible range of headings from which the swells may be approaching is between \(0^{\circ}\) (directly north) and approximately \(67.27^{\circ}\text{.}\)

Example2.4.18.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 its 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 you need to sail in and what distance will you 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 \(\theta\) (575 NM) and the side adjacent to \(\theta\) (1,100 NM) and express it as

Since we were able to write \(\tan\theta=\frac{575}{1,108}\text{,}\) we can use a calculator or other technology to evaluate the inverse tangent to find our angle:

Because this angle is in Quadrant III, we find its value on the Unit Circle by adding \(180^{\circ}+27.4^{\circ}=207.4^{\circ}\text{.}\) 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, \(d\text{,}\) we use the Pythagorean Theorem:

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

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 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 sun or moon upwards; and distinctive patterns of swells bending (refracting) around and/or reflecting off islands. Land will be spotted when the navigator first sees Maunga 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).

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 Maunga 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 \(R\text{.}\) Standing on Hōkūleʻa, the line from the navigator’s eye to the horizon is tangent to the circle of radius \(R\text{.}\)

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 1.2.25 the formula for finding the arc length is \(s_1=2\pi R\cdot\left(\frac{\theta_1}{360}\right)\text{.}\) Since we know \(R\text{,}\) we only need to find \(\theta_1\text{.}\) Notice this forms a right triangle.

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

Next, to determine how far the navigator is from Maunga 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 Maunga Terevaka. To find the distance from the top of the mountain to the horizon, we will need to determine \(s_2\text{.}\) We begin by redrawing the triangle.

Therefore, the total distance between the navigator and Maunga Terevaka is \(s_1+s_2=3.7+50.0=53.7\) miles. Please note that this is the distance when the island may first be seen, however, weather conditions may reduce visibility.

Subsection2.4.4Graphs 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 \(f^{-1}\text{.}\)

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 \(y=x\text{.}\)

Subsection2.4.5Composition of Inverse Trigonometric Functions

Recall Algebra that if \(f\) is a one-to-one function with inverse \(f^{-1}\text{,}\) 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, \(f(f^{-1}(y))=y\) will work 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.

Definition2.4.21.Properties of Composition of Trigonometric Functions.

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

when

\(-1\leq x\leq1\)

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

when

\(-1\leq x\leq1\)

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

when

\(-\infty\leq x\leq\infty\)

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

only when

\(-\frac{\pi}{2}\leq x\leq\frac{\pi}{2}\)

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

only when

\(0\leq x\leq\pi\)

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

only when

\(-\frac{\pi}{2}\leq x\leq\frac{\pi}{2}\)

Example2.4.22.Composition of trigonometric function and the inverse of the same trigonometric function.

Find the exact value of each expression

(a)

\(\cos^{-1}\left(\cos\frac{\pi}{8}\right)\)

Solution.

Since \(\frac{\pi}{8}\) is in the interval \([0,\pi]\text{,}\) then from the properties of compositions of inverse functions, we get

Note that \(\frac{5\pi}{7}\) is not in the interval \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\text{.}\) In order to evaluate the expression, we first need to find an angle, \(\theta\text{,}\) such that \(-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\) and \(\sin\frac{5\pi}{7}=\sin\theta\text{.}\) Since \(\frac{\pi}{2}<\frac{5\pi}{7}<\pi\) means \(\frac{5\pi}{7}\) is in Quadrant II. Recall from Section 1.3 that our reference angle is \(\theta=\pi-\frac{5\pi}{7}=\frac{2\pi}{7}\text{.}\)

Since \(\frac{2\pi}{7}\) is in the interval \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\text{,}\)

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.

Example2.4.23.Composition of trigonometric function and the inverse of a different trigonometric function.

Let \(\theta\) be an angle in the range of \(\tan^{-1}\text{,}\) that is, let \(\theta\) be in the interval \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) such that \(\theta=\tan^{-1}\left(\frac{4}{3}\right)\text{.}\) This equation is equivalent to \(\tan\theta=\frac{4}{3}\text{.}\) Since \(\tan\theta>0\text{,}\) we know that \(\theta\) must be in Quadrant I \(\left(0<\theta<\frac{\pi}{2}\right)\text{.}\) Let \((x,y)\) be a point on the terminal side of \(\theta\text{,}\) then by the trigonometric ratios,

Let \(\theta\) be an angle in the range of \(\cos^{-1}\text{,}\) that is, let \(\theta\) be in the interval \(\left[0,\pi\right]\) such that \(\theta=\cos^{-1}\left(-\frac{3}{8}\right)\text{.}\) This equation is equivalent to \(\cos\theta=-\frac{3}{8}\text{.}\) Since \(\cos\theta<0\text{,}\) we know that \(\theta\) must be in Quadrant II \(\left(\frac{\pi}{2}\leq\theta\leq\pi\right)\text{.}\) Let \((x,y)\) be a point on the terminal side of \(\theta\text{,}\) then by the trigonometric ratios,

Since we have \(\cos\theta=-\frac{3}{8}\text{,}\) we get \(\frac{x}{r}=-\frac{3}{8}\) which gives us either \(x=-3\) and \(r=8\) or \(x=3\) and \(r=-8\text{.}\) Since \(\theta\) is in Quadrant II, we know that \(x\) is negative, thus \(x=-3\) and \(r=8\text{,}\) as shown in the figure below.

Evaluating \(\sin\left(\cos^{-1}\left(-\frac{3}{8}\right)\right)\) is equivalent to evaluating

In Exercise Group 1.4.7.1–8, we determined the values of the six trigonometric functions for each triangle. Now, use a calculator to find the value of \(\theta\) in degrees, rounding to two decimal places.

The summit of Mauna Kea is the highest point in Hawaiʻi and sits 13,803 ft above the sea level. If you stand at the summit, you 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, you 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 the 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 your answer to the nearest tenth of a mile.

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 your answer in minutes, rounded to one decimal place.

The ratio of the distance to the horizon (your previous answers) to the circumference of the Earth (\(2\pi R\)) is equal to the ratio of the time it takes to see the sunrise to the 24 hours (1,440 minutes) in 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.

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 Maui 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 Mauna Kea at 20,917,655 feet.

(a)

How far is the horizon when Maui stands at the summit? Assume Maui’s eyes are 5 ft from the ground. Round your answer to the nearest tenth of a mile.

How much faster would Maui see the sun emerge over the horizon than someone 5 ft tall and standing at the seashore? Espress your answer in minutes, rounded to one decimal place.