Trigonometry - Ideas and Applications

A second look at graphs

Vertically stretching and shrinking the sine or cosine

Given or • multiplying the function definition by number > 1 stretches• multiplying by number between 0 and 1 shrinks

y = sin x y = cos x

y = sin x in greeny = 3sin x in blue

y= 12

sin x in magenta

Multiplying function definition by a number a < 0reflects about x axis and then stretches or shrinks, depending on whether a < -1 or -1 < a < 0

Reflecting

y = sin x in green y = −2sin x in blue

Horizontal or vertical translation

• adding c to the x variable translates c units to the left • subtracting c from x variable translates c units to the right• adding c to the function definition translates c units up• subtracting c from function definition translates c units down

Let c > 0

y = cos x in green

y = cos x −π4

⎛⎝⎜

⎞⎠⎟

in blue

y = cos x + 2 in magenta

Graph of • in green• in blue oscillates twice as fast

• in magenta oscillates twice as slowly

Expanding or compressing horizontally

Let b > 0• if b > 1, multiplication of x variable by b compresses graph horizontally • if 0 < b <1, multiplication of x variable by b stretches graph horizontally

y = sin xy = sin2x

y = sin 12x⎛

⎝⎜⎞⎠⎟

Reflecting about the y axis

If b < 0, multiplication of x variable by b reflects graph about y axis, then• compresses horizontally if b < - 1 or • stretches horizontally if - 1 < b < 0

y = sin x in bluey = sin(-x) in red

y = 2sin −x2

⎛⎝⎜

⎞⎠⎟

in magenta

For or • length of the basic cycle is the period, period =

• the number is the amplitude and measures vertical stretching

If , the period is

Terminology

For or the basic cycle is the behaviour on the interval . This repeats infinitely to the left and to the right.

If , the period is

y = asin x y = acos x

y = asin(bx) y = acos(bx)

y = sin2x

y = sin x2

⎛⎝⎜

⎞⎠⎟

[0, 2π ]

2πb

a

π

2π12= 4π

Harmonic Motion

Sine and cosine functions occur naturally in wave phenomena.

Ripples is a pond

A mathematical model

Other examples:• sound, light, ....• motion of spring• motion of pendulum

For wave phenomena applications, sine and cosine are considered as functions of time. The curve moves past at a certain speed measured in basic cycles per second described as the frequency.

Concert pitch A has a frequency of 440 Hz (short for Herz named after German physicist Heinrich Herz (1857-1894)-http://www.webstationone.com/fecha/hertz.htm)

With time t as variable the period of may be thought of as the time for a basic cycle to pass - seconds per cycle. It’s value is

The frequency measures cycles per second - reciprocal of the period

frequency =

2πb

b2π

y = asin(bt)

Vibrating spring

Imagine spring with one end attached to weight lying on table- the other attached to wall. As spring is stretched and let go, movement of the weight is described by a sine function.

The stretched position (step 4) is labelled +1 and the compressed position (step 2) is labelled -1

Assuming ideal of no friction and no fatigue in spring - suppose equation of motion described by y = 7.2sin 3πt

• the amplitude is 7.2

• the period is

• the frequency is

2πb

=2π3π

=23

32

In reality the motion of the spring is dampened by friction andthe nature of the spring. This effect is modelled by multiplying by a suitably adjusted exponential function.

First modelling attempt, , in red.

Damped version, , in blue.

y = 7.2sin 3πt

y = e−0.2(t−10) 7.2sin 3πt

Inverse Trigonometric Functions

Problem - given a trigonometric function f and a number x find and angle such that . In other words, find a function such that

Difficulty - for inverse of function f to exist f must be injective none of the trigonometric functions are injective

Solution - restrict domain of trigonometric function to one on which the function is injective

θ f θ( ) = x

f f −1(x)( ) = f (θ) = x

f −1 f (θ)( ) = f −1(x) = θ

f −1

Inverse tangent

Restrict domain to the portion of tangent in the interval

Restricted tangent is injective and for each real number x there is such that ; i.e is bijective

Inverse tangent is function defined for all real numbers x and takes values in ; i.e. −π

2,π2( )

θ, −π2 < θ < π 2, tan : −π 2,

π2( )→ℜ

tan−1 :ℜ→ −π2,

π2( )

tan(θ) = x

graph of restricted tangentgraph of inverse tangent ( tan -1 or arctan)

Inverse tangent also called arctan

In particular: tan−1 tanθ( ) = θ and tan tan−1 x( ) = x

that is: . Also called arc-cotan

Restricted domain of cotangent is and is bijective

Inverse cotangent

0, π( ) cot : 0, π( )→ℜ

cot−1 :ℜ→ 0, π( )Inverse cotangent produces angle in for given real number -0, π( )

graph of restricted cotangentgraph of inverse cotangent (cot -1 or arc-cotan)

For: x ∈ℜ & θ ∈(0,π )

cot cot−1(x)( ) = xcot−1 cotθ( ) = θ

Inverse sine

Restricted domain of sine is

Inverse sine produces an angle for each x in [−1, 1]

−π 2,π2⎡

⎣⎤⎦

That is: - also called the arcsin function sin−1 :[−1, 1]→ −π 2,π2⎡

⎣⎤⎦

In particular for

θ ∈ −π 2,π2⎡

⎣⎤⎦ x ∈[−1, 1]

sin−1 sinθ( ) = θsin sin−1 x( ) = x

and

graph of restricted sine in blue and inverse sine in magenta

Inverse cosine

Restricted domain of cosine is 0, π[ ]Inverse cosine produces an angle for each x in [−1, 1]

That is: - also called the arc-cos function.cos−1 : [−1, 1]→ 0, π[ ]In particular for x ∈[−1, 1] and θ ∈ 0, π[ ]

cos cos−1(x)( ) = xcos−1 cosθ( ) = θ

graph of restricted sinegraph of inverse cosine (cos -1 or arc-cos)

Inverse secant (arcsec) produces an angle in for each

Restricted domain of secant is Inverse secant

0, π 2⎡⎣ )∪ π

2, π( ⎤⎦

It takes values in 1, ∞[ )∪ −∞, −1( ]

x ∈ 1, ∞[ )∪ −∞, −1( ]

0, π 2⎡⎣ )∪ π

2, π( ⎤⎦

That is: sec−1 : 0, π 2⎡⎣ )∪ π

2, π( ⎤⎦→ 1, ∞[ )∪ −∞, −1( ]

For and x ∈ 1, ∞[ )∪ −∞, −1( ] θ ∈ 0, π 2⎡⎣ )∪ π

2, π( ⎤⎦

sec sec−1(x)( ) = x sec−1 secθ( ) = θand

graph of inverse secant (sec-1 or arcsec) graph of restricted secant

Example: Calculate cos(arcsin(4x)) and tan(arcsin(4x))

Letting construct right triangle below - choosing so that the hypotenuse has unit length

θ = arcsin(4x)

Then: cos(arcsin(4x)) = 1− 4x2

1= 1− 4x2

tan(arcsin(4x) = 4x1− 4x2

Example: Suppose a pole of length 2 meters cast a shadow 1.8 meters long. What is the angle in diagram below?

We know from the diagram that tanθ =21.8

≈ 1.111

θ = tan−1 tanθ( )

θ ≈ tan−1 1.111( ) ≈ 48.010Therefore since ,

θ

Area of triangle

With the graph below the blue is the graph of the sine its • reflection about the y axis is in red and has equation • reflection about the x axis also in red and has equation

Conclusion:

y = sin −θ( )y = − sinθ

sin −θ( ) = − sinθNow translate the graph of units to the left getting equation

y = sin −θ( ) π

However - translated graph correspondsto blue graph.

Conclusion:

y = sin π −θ( )

sinθ = sin π −θ( )

Since , in both cases

For left triangle below

sinθ =ha

For right triangle sin π −θ( ) = ha

sinθ = sin π −θ( )

sinθ =ha

Conclusion:

Thus for each case: h = asinθ

absinθ2Area =

Law of Sines

Given arbitrary triangle with angles A, B, C and lengths of opposite sides a, b, c respectively

sinAa

=sinBb

=sinCc

Calculate the area in three different ways

Area = 12bcsinA =

12acsinB =

12absinC

Now multiply each term by . The result follows.2abc

Proof:

Example: Given triangle suppose Find the lengths of the other sides.

ΔABC A = 830, B = 550, and b = 18

Solution: Since , by the Law of Sines:C = 180 − (55 + 83) = 42

sin5518

=sin83a

and sin5518

=sin 42c

Therefore: a =18sin83

sin55≈ 21.8 and c = 18sin 42

sin55≈ 14.7

Law of Cosines

Given arbitrary triangle with angles A, B, C and lengths of opposite sides a, b, c respectively, the law of cosines is a generalization ofPythagorean theorem for triangles that are not right triangles

Instead of , the law of cosines states that c2 = a2 + b2

c2 = a2 + b2 − 2abcosC

In the case that C is a right angle, we know that , so thatas it should be

cosC = 0c2 = a2 + b2

Example: Two boats leave the same harbour at Noon. Boat A travels at a speed of 50km/h in the direction N 450 E and the other, boat B , travels at a speed o f 35 km/h in a direction S 600 E. What is the distance between them at 1:00 P. M. ?

Diagrams show that the angle between the paths of the boats is 750. After 1 hour boat A has travelled 50km and boat A 35km. By rule of cosines, distance c between boats satisfies

c = 502 + 352 − 2(50)(35)cos75 ≈ 53.1

c2 = 502 + 352 − 2(50)(35)cos75

And