1 Trigonometry -Ideas and Applications

1.1 A second look at graphs

The sine and cosine are basic entities of trigonometry, for the other four functions can bedefined in terms of them. The graphs can be stretched and pulled in various ways and formany applications it is important to be able to express the corresponding functions analyti-cally. As we have seen, the behavior of both sine and the cosine graphs on the interval [0, 2π]is repeated on all subsequent and previous intervals of length 2π. What I mean is that oneach of the intervals [2π, 4π], [4π, 6π], [6π, 8π] · · · and [−2π, 0], [−4π,−2π], [−6π,−4π], · · ·the graphs look identical to the graph over the interval [0, 2π]. The restriction of the sinefunction to the interval [0, 2π] is defined to be the basic cycle of the sine function. Similarlythe basic cycle of the cosine function is its restriction to the interval [0, 2π].

Earlier in studying general properties of functions given an arbitrary function whose graphhad been altered in certain ways, we developed methods for analytically expressing a newfunction whose graph was the altered graph. We need to apply these ideas

1.1.1 Stretching and shrinking vertically, reflecting about the x axis

Figure 1: Graphs of y = 3 sin x in blue, y = sinx in green and y = 22 sinx in magenta

Given either the sine function, y = sinx, or the cosine function, y = cos x, and a constanta,we know that :

• if a > 1, multiplication by a stretches the graph vertically

• if 0 < a < 1, multiplication by a shrinks the graph vertically

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• if a < 0, multiplication by a first reflects the graph about the x axis and then eitherstretches it if a < −1 or shrinks it if −1 < a < 0.

Given an equation y = a sinx or y = a cos x, the number a is said to be the amplitude ofthe corresponding equation.

Figure 2: Graph of y − sinx in green and y = −2 sinx in blue

1.1.2 Translating

To translate the graph of a function vertically up, add a positive constant to the function,and to translate down, subtract a positive constant from the function definition. Thus forinstance, the graph of y = sinx + 2, represents the graph of y = sinx translated two unitsup.

Figure 3: The graph of the cosine in green, its vertical translation in magenta y = cos x+2,and its horizontal translate in blue by π

4 to the right - the graph of y = cos(x− π4 ).

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To translate horizontally we have learned that to translate c units to the right, for c > 0,it is necessary to subtract c from the x variable and to translate c units to the left oneadds c to the x variable. The same holds true for the sine and cosine graphs. Thus, forexample, the graph of y = cos(x− π

4 ) is a translation of π4 units to the right of the graph

of y = cos x. See figure 3.

1.1.3 Expanding and compressing horizontally, reflecting about the y axis

Again from our previous discussions, we know that in general and specifically for y = sinxor y = cos x, given a constant .b

Figure 4: Graphs of y = sinx in green, y = sin 2x in blue, and y = sin x2 in magenta

• if b > 1, multiplication of the x variable by b compress the graph horizontally

• if 0 < b < 1, multiplication of the x variable by b expands the graph horizontally

• if b < 0, multiplication of the x variable by b first reflects the graph about the they axis and then either compresses it or expands it according to whether b < −1 or−1 < b < 0

Given either y = sin bx or y = cos bx, the number2π

bis called the period of the correspond-

ing equation. It represents the length of the basic cycle associated with the equation. Infigure 4 we show, superimposed on the graph of y = sinx in green, the graph of y = sin 2xin blue,whose basic cycle is of length π, and the graph of y = sin x

2 in magenta, whoseperiod is of length 4π

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1.1.4 Harmonic motion

Suppose you throw a stone into still pond. The ripples that gradually radiate from wherethe stone landed have the shape of a moving 3-dimensional sine curve. Figure 5 shows apicture of ripples in water, and figure 5 shows a mathematical model built from the sinefunction.

Figure 5: Ripples in a pond

Figure 6: Mathematical surface that models ripples - graph of z = 0.25 sin(x2 + y2)

In a similar way sound creates ripples or vibrations in the air that travel radially out fromthe source. A tuning fork used by musicians for tuning their instruments creates vibrationsin the air much like the ripples in a pond. Such physical wave phenomena are described interms of their frequency. In these situations we consider the sine or cosine functions to befunctions of time - that is , we imagine the graph of a sine or cosine function passing by usat some fixed speed - think of the movement of ripples in water. The frequency of such acurve is the number of basic cycles that pass per time unit. Usually frequency is measured

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in cycles per second.

When we spoke of the period of a sine or cosine curve we defined it to be the length of abasic cycle. Now if we choose to let the x axis represent time instead of physical distanceand we rewrite our equations as y = a sin bt and y = a cos bt, for some constants a and bgreater than zero, then the period, which we said was 2π

b , now refers to the amount of timeit takes a basic cycle to occur. If we measure time in seconds, then we can say that theperiod is the seconds required to complete one cycle. That is seconds per cycle, seconds

cycle . The

reciprocal of the period then measures cycles per second, cyclessecond . Thus for the functions

y = a sin bt and y = a cos bt this is then

frequency =b

2π

Example 1 : Concert pitch A

Most instruments in an orchestra tune to the so called concert pitch A, a sound withfrequency 440 cycles per second. This means in our equations b

2π = 440, so that, b = 880π.The sine equation describing the vibration of the air is then, y = a sin 880π. The constanta is used to model the volume of the sound. Louder sounds require models with largervalues of a.

Example 2 : Vibrating spring

Figure 7: Motion of a weight attached to a spring

Imagine a weight at rest on a table but attached to a spring which is anchored in the wallsomewhat to the left of the spring, see figure 7 Now pull the weight away from the wall

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stretching the spring. Letting the weight go it springs back to a position to the left of itsrest position and then rebounds to a position to the right of the rest position and so forth.Label the rest position zero, the compressed position to the left as -1 and the stretchedposition to the right +1. Once let go the weight oscillates between the -1 position andthe +1 position, gradually due to friction and the nature of the spring the period of theoscillation will shorten and eventually stop. Lets assume an ideal situation in which thereis no friction and no fatigue in the spring. Lets further assume that the motion has beenmodeled by the equation y = 7.2 sin 3πt. We then know that

• the amplitude a = 7.2

• the period is: 2πb = 2π

3π = 23 seconds

• the frequency is 32 Hz (“Hz” stands for cycles per second)

Figure 8: Graph of y = 7.2 sin 3πt in red with a damped version in blue

The graph looks like In actual fact the motion of the the weight undergoes a dampeningeffect whereby the distance traveled gradually decreases. In figure 8 we see the graph ofy = 7.2 sin 3πt in red along with a damped version in blue. The later has been chosen tobe the the graph of y = e−.2(t−10) sin 3πt.

1.1.5 Inverse trigonometric functions

Lets recall what we know about inverse funcitons. Given a function f : A → B defined onan interval A, where B = f(A) be the set of all numbers f(x) such that x ∈ A. Then wehave said that if f is injective f, there will be a function f(x) such that x ∈ A with theproperty that

• f(f−1(y)

)= y for every y ∈ B

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• f−1(f(x)

)= x for every x ∈ A.

Figure 9: Tangent on (−π2 , π

2 )

Now the problem with finding inverses of the trigonometric functions is that none of themsatisfy the horizontal line test - none of them are injective. However for each of them theyare injective when restricted to certain special intervals.

For instance if we consider the tangent only on the interval A = (−π2 , π

2 ), then the graphlook is as we see it in figure 9. The graph clearly satisfies the horizontal line test - so aninverse exists. The inverse of the tangent is denoted either as arctan or as tan−1 . Sincethe range of the tangent is is the entire real line, in our definition of an inverse functionthe set B = R. We then have the relationships,

• arctan(tan x) = x

• tan(arctan y) = y

Said another way, the arctan of a number y is the angle x whose tangent is y. As for thegraph of the arctan , remember that in general given a function f with an inverse f−1, thegraph of an f−1 is obtained by reflecting the graph of f about the diagonal line y = x. Forthe arctan the graph is as in figure 10. Note just as tan is asymptotic to the vertical linesx = −π

2 and x = π2 so arctan is asymptotic to the horizontal lines y = −π

2 and y = π2 .

The definitions other inverse trigonometric functions depends only on finding a convenientinterval over which the associated trigonometric function is injective.

• The definiton of the inverse cotangent, written arccotan or cot−1, relies on restrictingthe definition of the cotangent to the interval (0, π). The arccotan then has domainthe whole real line and range the interval (0, π). See figures 11 and 12.

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Figure 10: Graph of arctan

• The definition of the inverse sine, written arcsin or sin−1, relies on restricting the sinefunction to the interval [−π

2 , π2 ]. Since the sine has range [−1, 1], the arcsine then

has domain [−1, 1] and range [−π2 , π

2 ]. Figure 13 shows the graph of the sine in blueand the graph of the arcsine in purple. Note that the sine is injective on the interval[−π

2 , π2 ].

• For the inverse cosine, written arc-cosine or cos−1, the domain of the cosine is re-stricted to the interval [0, π]. The arc-cosine then has this as its range. See figures 14and 15.

• For the inverse secant, written arcsec or sec−1, the chosen domain for the secantconsists of the intervals [0, π

2 ) and on (π2 , π] on which the secant is injective. The

inverse secant then has domain the range of the secant, namely the two intervals[0, π

2 ) and (π2 , π], and the range of the inverse secant is the domain of the secant

namely the intervals[0, π2 ) and (π

2 , π]. See figures 16 and 17.

• The arc-cosecant function may be similarly defined by considering the cosecant func-tion restricted to the intervals [−π

2 , 0) and (0, π2 ).

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Example 3 Write each of: (a) cos(arcsin 4x) and (b) tan(arcsin 4x) as a function notinvolving a trigonometric function. What is the domain of the function?

Solution: Begin by setting θ equal to the angle whose sine is 4x. In other words setθ = arcsin 4x. Next draw a right triangle as in figure 18. For convenience the triangleis chosen so that the hypotenuse has length one. If this is true and of sin θ = 4x, the

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Figure 11: Cotangent is injective on (0, π)

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Figure 12: Arccotan is defined on all of R with range (0, π)

Figure 13: Sine in blue and arcsine in purple; sine is injective on [−π2 ,−π

2 ], arcsine definedon [−1, 1] with range [−π

2 ,−π2 ]

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Figure 14: Cosine injective on[0, π]

Figure 15: Arc-cosine with domain[−1, 1] and range [0, π]

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Figure 16: Secant defined on [0, π2 ) and on (π

2 , π]. Range:[1,∞) ∪ (−∞,−1]

Figure 17: Arcsecant defined on [1,∞) ∪ (−∞, 1] with range [0, π2 ) and (π

2 , π].

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Figure 18: Right triangle with θ = arcsin 4x.

opposite side must have length 4x. Using the Pythagorean theorem, the adjacent side iscalculated to have length

√1− (4x)2. We can now read off the answers. First, in order

that√

1− (4x)2 not be negative, we must have (4x)2 < 1 or x2 < 116 , which implies that

−14 ≤ x ≤ 1

4 . Then

cos(arcsin 4x) =√

1− (4x)2

tan(arcsin 4x) =4x√

1− (4x)2

Example 4 A sailor sailing single-handed about the globe encountered a ferocious stormin the south Atlantic. All his electronic equipment was damaged with the exception of ascientific calculator, his compass and his watch which had been set to Greenwich meantime. With the help of his compass he was able to determine with reasonable precision thetime ( that is the local Noon) when the sun was directly south . To determine his latitudehe however need to determine the elevation of the sun at Noon. He was able to determinethat a vertical pole 2 meters high cast a shadow at Noon that was precisely 1.8 meters long.With this information he was able to calculate the elevation of the sun. How was this done?See figure 19.

Solution: The sailor being schooled in trigonometry realized that if θ is the angle ofelevation, it must be true that the tangent of θ must be 2

1.8 = 109 = 1.111 Taking out his

calculator he entered 1.11111 and then pushed the inverse tangent button to find the anglewhose tangent is 1.111. The result was tan−1(1.11111) = 48.01o.

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Figure 19: Find the angle θ

1.2 Further Applications

Figure 20: y = sin(−x)is the equation of the red curve - when moved π to the left it hasequation y = sin(π − θ and corresponds to the blue curve y = sin θ

Trigonometry allows an easy way to calculate the area of any triangle. First we need toobserve that starting with the graph of the sine y = sin θ, reflecting about the x axis givesthe graph of y = − sin θ which we also know from the section on identities is the same asthe graph of y = sin(−θ). Using this equation, if we were to translate the graph π units tothe left, the corresponding equation is then y = sin(−θ + π). The graph we see is preciselythe graph of sine. That is we got back to where we had started. See figure 20. Thus wecan say that

sin θ = sin(π − θ).

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1.2.1 Area of a triangle

Given an arbitrary triangle we have learned that its area can be calculated by the formulaA = base×height

2 . The difficulty with this formula that except for the case of right triangles,the height of the triangle does not correspond to the length of one of the edges. And if thelengths of edges is all the information available, one is stuck - that is, if one does not knowa bit of trigonometry.

Figure 21: A base angle θ may be acute or obtuse - in which case the complimentary angleπ − θ is acute

Given a triangle lets let θ be one of the base angles. As we see in figure 21 the angle may beacute (less than 90o) or obtuse (greater than 90o). In the first case we know that sin θ = h

a

and in the second case we know that sin(π−θ) = ha . But as we have seen, sin(π−θ) = sin θ

so thus, in both cases, we conclude that h = a sin θ. We then have in either case that thearea is 1

2ab sin θ. This is expressed more formally as follows

Result 5 Area of triangle

The area A of a triangle with two sides of length a and b and an included angle θ is

A =12ab sin θ.

1.2.2 Law of sines

The law of sines says that for a given triangle ∆ABC the ratio of the sine of angle to thelength of its opposite side is independent of the chosen angle. In other words, if a, b and care the lengths of the sides opposite the angle A,B and C,

sinA

a=

sinB

b=

sinC

c.

See figure 22.

The law of sines is used in situations where given a triangle it is possible to calculate theratio of the sine of one of the angles to the length of its opposite side. Then as long asone more piece of information is present, the length of another side or the measurement

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Figure 22: Triangle with angles A, B, and C with opposite sides of length a,b, and c

of another angle, all the other dimensions of the triangle can be calculated. For instancesuppose we know the size of angle A and the length a of its opposite side. Then if we knowthe length b of one of the other sides , we have sin A

a = sin Bb so that sinB = b sin A

a . Thus theangle at B has measurement arcsin(b sin A

a ). Knowing B we then know the measurement ofthe angle at C and hence sinC and ultimately the length c of the side opposite C. Hereare some examples.

Example 6 Given a triangle ∆ABC suppose that A = 83o, B = 55o and b = 18. Find thelengths of the other sides.

Solution: Since C = 180− (55 + 83) = 42o, we have

sin 55o

18=

sin 83o

a

andsin 55o

18=

sin 42o

c

Thus a =18 sin 83o

sin 55o≈ 21.8 and c =

18 sin 42o

sin 55o≈ 14.7.

Example 7 A construction company was awarded a contract by the provincial governmentto build a foot bridge across a gully on the west shore of Vancouver Island. Although thecompany knew that the bridge was to go from point A on one side of the gully to point B onthe other, they did not know how long the bridge should be. In order to determine the lengththey set a marker C on the same side as the point B and measured the distance between Aand B to be 10 meters. From the other side, with surveying equipment they were able tomeasure the angel at A, ∠BAC, to be 12o. The also measured the angle at C, ∠BCA, tobe 42o. With this information they were able to find what the length of the bridge shouldbe. What is this length?

Solution: We know from the law of sines that if c is the length of the bridge, thenc

sin 42o = 10sin 12o so that c = 10 sin 42o

sin 12o ≈ 32.18 meters. See figure 23.

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Figure 23: Calculate the length of the bridge from A to B

Using our previous result for the area of a general triangle, the law of sines may be easilyproved

Proof:

Considering figure 22 we can calculate the area in three different ways, depending on whichangle we wish to consider.

Area =12bc sinA =

12ac sin B =

12ab sinC.

Multiplying each term by2

abcgives the result

1.2.3 Law of Cosines

The law of cosines may be considered a generalization of the Pythagorean theorem. Givenan arbitrary triangle as in figure 22, if the angle at C were a right angle then by thePythagorean theorem c2 = a2 + b2. But what if the angle at C is not a right angle? Thelaw of cosines tells us that

c2 = a2 + b2 − 2ab cos C

and in the case that C = 90o we have cos C = 0 and the law of cosines reduces to thePythagorean theorem.

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Figure 24: Two boats one traveling N40oE the other travelingS60oE.

Figure 25: position of boats after one hour

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The law of cosines also lets tells us that the converse of the Pythagorean theorem is true,for we then know that if C is not a right angle, then cos C 6= 0 and c2 6= a2 + b2. Fromour work in logic we know that the contrapositive of this last statement must also be true- namely, if c2 = a2 + b2, then C is a right angle. This last statement is the converse of thePythagorean theorem.

Here is an example to show how the law of cosines may be applied.

Example 8 Two boats leave the same harbor at Noon. Boat A travels at speed of 50 km/hin the direction N45oE and the other, boat B, travels at a speed of 35 km/h in a directionS60oE. What is the distance between them at 1:00 P.M. ?

Solution: We see that the angle between the paths of the two boats is 45o+30o = 75o, andafter 1 hour boat A has traveled 50 km and boat B has traveled 35 km. Letting C stand fortheir common starting point we have a diagram as in figure 25. By the law of cosines thedistance c between the two boats is c =

√502 + 352 − 2(50)(35) cos 75o ≈ 53.1km.

Example 9 Given a regular pentagon whose 5 sides are each of length 1. Let A, B, andC be three vertices as in figure 26 and let D be the mid-point of the line segment BC. Whatis the length b of the line segment AD?

Solution: Using the fact that the angle at each of the vertices of a regular pentagon is 108o,a fact that is not difficult to prove using elementary geometry, by the law of cosines

b2 = 1 +14− 2(1)(

12) cos 108

Thusb =

√1.25− cos 108 ≈ 1.2486

Figure 26: For a regular pentagon with sides of length 1, find length b of the segment AD

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