part 4: trigonometry in part 1 we talked about locating things in two dimensions. we restricted...

Part 4: Trigonometry

In part 1 we talked about locating things in two dimensions. We restricted ourselves in that part to the Cartesian system, (x,y), but mentioned there were other systems.

Many of the other systems employ at least one angle. Therefore, we must talk about what an angle is and what are the measures we can use to quantify those angles.

Trig: Angles

Because space is three dimensional (we’ll talk about this soon), we need angles.

What is an angle?

How do you measure an angle?

θ

Trig: AnglesHow do you measure an angle?

1) in (fractions of) circles (cycles, rotations, revolutions)

2) in degrees - but what is a degree?Why do they break the circle into 360 equal degrees?

3) in radians - but why use a weird number like 2 for a circle? Why does have the weird value of 3.1415926535… ?

Trig: Angles

Full circle = 360o (Comes from year - full cycle of seasons is broken into 365 days; but 365 is awkward number; use “nicer” number of 360.)

Full circle = 2 radians (Comes from definition of angle measured in radians:

= arclength / radius = s / r ) r

s

Trig: basic functionsBesides angles, we can define functions

associated with angles. The trig functions are based on a right triangle:

Trig: basic functions• sin() = opposite/hypotenuse = y/r

• cos() = adjacent/hypotenuse = x/r

• tan() = opposite/adjacent = y/x

• (The hypotenuse is the side opposite the right angle.)

ry

x

Trig: sine function

The easiest way to see what the function, sin(θ), looks like is to look at the unit circle where r=1. Then sin(θ) = y/r = y.

x

ysin(θ)

θ

0.5

1.0

90o 1800

π/2 radians

πradians

Trig: sine function

Notice that sin(θ) is symmetric about the y axis of the circle, so sin(θ) is symmetric about 90o or ½π radians (i.e., sin[90o+θ] = sin[90o-θ]).

x

ysin(θ)

θ

0.5

1.0

90o 1800

π/2 radians

πradians

Trig: sine function

Notice that sin(θ) is negative between 180o and 360o.

x

ysin(θ)

θ

-0.5

-1.0

270o 3600

3π/2 radians

2πradians

1800

πradians

Trig: sine functionNotice that sin(θ) repeats after 360o so that

sin(θ) = sin(θ+360o). Notice that this makes sin(θ) antisymmetric about the origin which makes it an odd function so that sin(θ) = -sin(-θ). Notice sine is antisymmetric about 180o so sin[180o+θ] = -sin[180o-θ].

θ

sin(θ)

180o 360o

Trig: cosine function

The easiest way to see what the function, cos(θ), looks like is to look at the unit circle where r=1. Then cos(θ) = x/r = x.

x

ysin(θ)

θ

0.5

1.0

90o 1800

π/2 radians

πradians-0.5

-1.0


We can also use some symmetries to see what cosine looks like. Note that from the definition of sine and cosine and using the unit circle are: sin(θ) = y/r = y and cos(θ) = x/r = x. Since we are dealing with a circle, the cosine function will have exactly the same shape as the sine function, except that it starts at a different place. Sine is maximum at 90o, while cosine is maximum at 0o. (See next slide for graphs).


Notice also that cos(θ) is symmetric about the x axis on the circle and y axis of the graph, so cosine is an even function:

cos[+θ] = cos[-θ].

θ

sin(θ) cos(θ)

180o 360o

Trig: sine and cosine functionsNotice that since x and y are different by 90o (horizontal

versus vertical on the circle), so sin(θ+90o) = cos(θ) , and sin(θ) = cos(θ-90o) which you can see from the diagram. Also since cosine is an even function, cos(θ) = cos(- θ), cos(θ-90o) = cos(90o- θ) = sin(θ) .

θ

sin(θ) cos(θ)

180o 360o

Trig: sine and cosine functions

cos(90o- θ) = sin(θ)This last relation can also be seen from the

right triangle diagram and the definitions.The hypotenuse is r for both angles θ and 90o-θ. Therefore, we havesin(θ) = y/r = cos(90o-θ) and cos(θ) = x/r = sin(90o-θ).

.x

ry

θ

90o-θ


Our calculators provide us with a nice way of evaluating both sine and cosine functions. However, we need to always be aware of whether we are using degrees or radians for the angle. Know how to check your calculator to see what it is expecting, and know how to change your calculator to make it work with either degrees or radians based on the type of problems you are encountering.

Trig: sine and cosine functionsWhy do we continue to have two different major

units for angles: degrees and radians? As strange as it may seem, the answer is basically one of convenience.

The degree goes way back in history but it is an arbitrary unit that is related to us being on the earth with a year of 365 days. The main advantage is that there are an integer number of degrees in a circle which is an advantage when measuring.

The radian measure does NOT have an integer number in one circle, but it is geometrically defined (and so is used a lot in math courses).


Note that once the angle has made a complete circle (gone past 360o = 2π radians), both the sine and cosine functions start repeating themselves: sin(θ+360o) = sin(θ) = sin(θ+2π radians) cos(θ+360o) = cos(θ) = cos(θ+2π radians)

Note also that both sine and cosine are limited to a range between –1 and 1.

Trig: sine and cosine functionsWe can easily calculate several values of sine and cosine from

basic geometry: for a 45o angle, x = y, so r2 = 1 = x2 + y2 = x2 + x2 = 2x2, or x = SQRT(1/2) = y, so

sin(45o) = √½ = cos(45o). For the 60o angle, cos(60o) = [½r / r] = ½ = sin(30o); sin(60o) = y/r where (with r=1) y = SQRT(r2 – [r½]2) = (SQRT[3]/2) so sin(60o) = √(3)/2 = cos(30o).

x

y=xr

r/245o

60o

r r


Here are a few values for sine and cosine that you should know. It helps you keep track of where you are on the oscillating sine and cosine graphs:

sin(0o) = cos(90o) = 0 0o = 0 radssin(30o) = cos(60o) = 0.500 30o = π/6 radssin(45o) = cos(45o) = 0.707 = √(2)/2 45o = π/4 radssin(60o) = cos(30o) = 0.866 = √(3)/2 60o = π/3 radssin(90o) = cos(0o) = 1.000 90o = π/2 rads

Trig: sine and cosine functionsBecause of symmetries, we can extend the table from the

previous slide: sine is symmetric about 90o, cosine is antisymmetric about 90o.

sin(120o) = sin(60o) = 0.866 = √(3)/2 120o = 2π/3 radssin(135o) = sin(45o) = 0.707 = √(2)/2 135o = 3π/4 radssin(150o) = sin(30o) = 0.500 150o = 5π/6 radssin(180o) = sin(0o) = 0.000 180o = π radscos(120o) = -cos(60o) = -0.500cos(135o) = -sin(45o) = -0.707 = -√(2)/2 cos(150o) = -cos(30o) = -0.866 = -√(3)/2 cos(180o) = -cos(0o) = -1.000


For angles between 180o = π radians and 270o = 3π/2 radians, both sine and cosine simply change sign when 180o = π radians is added:

sin(θ+180o) = -sin(θ) = sin(θ+π radians)

cos(θ+180o) = -cos(θ) = cos(θ+π radians)


For angles between 270o = 3π/2 radians and 360o = 2π radians, we can convert to negative angles between –90o = -π/2 radians and 0o by using the fact that both functions repeat over one cycle (±360o = 2π radians) Sine is an antisymmetric (odd) function and cosine is a symmetric (even) function:

sin(θ) = -sin(-θ) ; sin(θ-360o) = -sin(360o-θ)cos(θ) = cos(-θ) ; cos(θ-360o) = cos(360o -θ)

Trig functions – domain and range

Earlier we defined the terms: domain and range. For linear functions, the domain and range were both usually (-∞, +∞). For quadratic functions, the domain was usually (-∞, +∞), while the range usually went to infinity on one side but was limited to the y value of the vertex on the other side. For exponential functions, the domain was usually (-∞, +∞), but the range was usually (0, +∞). For log functions, the domain was usually (0, +∞) while the range was (-∞, +∞). Note that the domain and range flip between the function and its inverse.


But with the functions of sine and cosine, we can see a real use for the concepts of domain and range.

The normal ranges for both sine and cosine are [-1,1].

While the domain for both sine and cosine can be (-∞, +∞), if we limit the domain to be [0, 90o], both sine and cosine are one-to-one functions and inverse functions should exist.


Limiting the trig functions to a domain of [0, 90o], however, does not let them go through their full range of [-1,+1]. For the sine function, we need a domain of [-90o, +90o], and for the cosine we need a domain of [0, 180o].

Trig functions: slope

We have seen before that the slope of a straight line is constant, the slope of a parabola depends on the x value, and the slope of an exponential depends on the y value. What about the slope of an oscillating function (sine and cosine)?

Trig functions: slopeNotice that near θ = 0o the slope is positive, but that the slope gets

less positive until it reaches zero slope at θ = 90o and then the slope become negative and reaches its steepest negative at about θ = 180o and then it becomes less steep until it again reaches zero slope at θ = 270o and then becomes positive until it reaches its steepest slope at θ = 360o . So we see that the slope of the sine wave also oscillates similar to what the sine function does. However, the slope oscillates 90o out of phase with the function.

θ

sin(θ)

180o 360o

Trig functions: slope

What about the slope of an oscillating function (sine and cosine)?

We had a nice picture on the previous slide, but we can calculate the difference quotient (slope near a point) using the sine function to get a more quantitative answer. You might want to calculate the difference quotients near the following points (with h = Δθ = .01 radians): 0 rads, π/6 rads, π/4 rads, π/3, rads, π/2 rads, 2π /3 rads, 3π/4 rads, π rads, 3π/2 rads, and 2π rads. What does this remind you of? Would we get similar results for the cosine function?

Inverse trig functions

Recall that an inverse “undoes” what the original function did. In general, an inverse function, g(x), operating on the original function, f(x), gives the original value of x back again: g(f(x)) = x. We call the inverse of the sine function the Arcsin(x) or sin-1(x). Note that sin-1[sin(θ)] = θ.

Since the value of sin(θ) is a number between –1 and 1, (the range of the sine function), the domain of the sin-1(x) function is a number between –1 and 1, and the range is an angle between –90o and +90o .

Inverse trig functions

sin(θ) has domain [-90o, 90o] & range [-1, 1]sin-1(x) has domain [-1, 1] & range [-90o, 90o].

In a similar way, cos(θ) has domain [0o, 180o] & range [-1, 1]cos-1(x) has domain [-1, 1] & range [0o, 180o].

To see the different ranges of arcsine and arccosine, try using your calculator to find sin-1(-0.5) and then find cos-1(-0.5).

Inverse trig functionsCan you draw the graph of the arcsine (sin-1)

and the arccosine (cos-1) functions? You need to be careful since the x axis will have to be in numbers and the y axis will have to be in either degrees or radians! Will there be any

asymptotic behavior? Are there anysymmetries?

-90o +90o

+1

-1

sin(θ)

θ

Inverse trig functionsNote that there are no asymptotic behaviors.

Note that the sine function is an odd function so it is symmetric through the origin, so the sin-1 function should also be antisymmetric.

-90o +90o

+1

-1

+90o

-90o

-1 +1

sin(θ)

θ

sin-1(x)

x

Inverse trig functionsThe graphs look the same but the dotted symmetry

line is different if you use radians instead of degrees. From this, we see that if you use radians and have small angles, sin(θ) ≈ θ . For example, sin(0.1 rad) = .0988 ≈ 0.1 .

π/2 +π/2

+1

-1

π/2

-π/2

-1 +1

sin(θ)

θ

sin-1(x)

x

Inverse trig functionsCan you draw the graph of the arcsine and the

arccosine functions? You need to be careful since the x axis will have to be in numbers and the y axis will have to be in either degrees or radians! Any asymptotes or

symmetries?

90o 180o

+1

-1

cos(θ)

θ

Inverse trig functionsAgain no asymptotes, but there is an

antisymmetry through 90o .

90o 180o

+1

-1

+180o

90o

-1 +1

cos(θ)

θ

cos-1(x)

x

Trig: tangent function

While the sine and cosine functions relate the rectangular (Cartesian) coordinates (x,y) to the polar coordinates (r, θ), there is another function that directly relates θ to x and y without going through the radius, r. It is the tangent function: tan(θ) = y/x. This function is directly related to the sine and cosine functions: tan(θ) = y/x = (y/r) / (x/r) = sin(θ) / cos(θ).


tan(θ) = y/x = sin(θ) / cos(θ)

This function is a little more “tricky” than either sine or cosine. Can you see where the “tricky” parts come in?


tan(θ) = y/x = sin(θ) / cos(θ)Since the value of cos(θ) equals zero when

θ = 90o = π/2 radians, the tangent function will approach infinity as θ approaches 90o from below 90o, but since cos(θ) is less than zero (negative) for θ greater than 90o, the tangent function will approach negative infinity as θ approaches 90o from above 90o. This gives asymptotic behavior!


tan(θ) = y/x = sin(θ) / cos(θ)The second “tricky” thing about tangent is

trying to determine a domain for it in which an inverse function can be defined. The normal domain for sine to define the inverse sine function was [-90o, 90o], while the normal domain for the cosine to define the inverse cosine was [0, 180o]. What should the normal domain be for the tangent function to define the inverse tangent function, called arctan(x) or tan-1(x) ?

Trig: arctangent function

One way is simply to plot the tan(θ) function and then plot the tan-1(x) relation:

Trig: arctangent functionAs you can see, the tan(θ) function is one-to-one on

the interval from –90o to +90o, and alternately on the interval from 0o to 180o.

The usual choice for use with the inverse tangent is from –90o to +90o . Using your calculator, see whether tan-1(-0.5) is less than 0o or greater than +90o.

-90o +90o

+3

-3

tan(θ)

θ +180o +270o

Trig: arctangent functionNote that tan(θ) ≈ θ (red curve below almost falls on

dotted line) for angles close to zero as long as the angle is measured in radians. Try checking this on your calculator: tan(.1 rads) ≈ .100335 .1 rads = 5.73o

tan(.3 rads) ≈ .309336 .3 rads = 17.2o

What will the tan-1(x) function (in this range) look like?Note that tangent is anti-symmetric (odd function),and note the asymptoticbehavior near ±90o .

-90o +90o

+3

-3

tan(θ)

θ

Trig: arctangent function

-90o +90o

+3

-3

tan(θ)

θ

Since tan(θ) goes through the origin and has anti-symmetry, tan-1(x) will also go through the origin and have antisymmetry. Since tan(θ) approaches infinity as the angle approaches 90o (vertical asymptote), tan-1(x) will approach 90o as x approaches infinity (horizontal asymptote).

+90o

-90o

+3-3

tan-1(θ)

x

Trig: example of vectors

One of the main uses of the trig functions is in the expression of vectors. A vector is simply a quantity that not only has a magnitude (amount), but it also has a direction.

Trig: VectorsHow do we work with a quantity that needs two (or

more) numbers to specify it (like position does)?

We can work either with a group of numbers sometimes put in parenthesis, or we can work with unit vectors that we add together:

(x,y) or x*x + y*y (where x indicates the x direction and x indicates how far in that direction).

Trig: Vectors

• The individual numbers in the vector are called components of the vector.

• There are two common ways of expressing a vector in two dimensions: rectangular and polar.

Trig: Vectors

• Rectangular (x,y) is often used on city maps. Streets generally run East-West or North-South. The distance East or West along a street give one distance, and the distance along the North-South street gives the second distance - all measured from some generally accepted origin.

Trig: Vectors

• If you are at home (the origin), and travel four blocks East and then three blocks North, you will end up at position A.

• Position A (relative to your house) is then (4 blocks, 3 blocks) where the first number indicates East (+) or West (-), and the second number indicates North (+) or South (-).

Trig: Vectors

• If you had a helicopter or could walk directly there, it would be shorter to actually head straight there.

How do you specify 3the location of pointA this way? 4

Trig: Vectors

• The distance can be calculated by the Pythagorian Theorem:

r = [ x2 + y2 ] = 5 bl.• The angle can be 5 3

calculated using the inverse tangent function: 4

= tan-1 (y/x) = 37o

Trig: Transformation Equations

• These two equations are called the rectangular to polar transformation equations:

r = [ x2 + y2 ]

= tan-1 (y/x) .

• Do these work for all values of x and y, including negative values?


r = [ x2 + y2 ]

• The r equation does work all the time since when you square a positive or negative value, you still end up with a positive value. Thus r will always be positive (or zero), never negative.


= tan-1 (y/x) .• However, the theta equation does depend on the

signs of x and y. From this equation you get the same angle if x and y are both the same sign (both positive or both negative), or if one is positive and one negative - regardless of which one is the positive one.

• How do we work with this?


= tan-1 (y/x)

Some scientific calculators have a built-in transformation button.

However, you should know how to do this the “hard way” regardless of whether your calculator does or does not have that button.


= tan-1 (y/x)

Note: If x is positive, you must be in the first or fourth quadrant (theta between -90o and +90o). Your calculator will always give you the right answer for theta if x>0.

If x is negative, you must be in the second or third quadrant (theta between 900 and 270o). All you have to do if x<0 is add 180o to what your calculator gives you.

Trig: Inverse Transformations

• Can we go the other way? That is, if we know (r,) can we get (x,y) ?

• If we recall our trig functions,

we can relate x to r and : r

x = r cos(), and similarly y

y = r sin(). x

Do these work for all values

of r and ? YES!

Trig: Inverse Transformations

x = r cos(), and y = r sin()

Example: If (r,) = (5 bl, 37o), what do we get for (x,y) ?

x = r cos() = 5 bl * cos(37o) = 4 bl.

y = r sin() = 5 bl * sin(37o) = 3 bl.

[Note that this is the (x,y) we started with to get the (r,) = (5 bl, 37o).]

Computer Homework

You should be able to do the eighth Computer Homework assignment on Vector Addition, Vol. 1, #2.

Regular Homework Set #8(continued on the next slides)

1. Evaluate sin(1o) and sin(1 radian).2. a) Find the value of sin(361o). b) Using

this value as x, find the sin-1(x). c) Why doesn’t x = 361o ? d) Find the value of sin(179o).

3. Which is largest for = 0.1 radians: [, sin(), or tan()] ? Which of the three is largest for = 1 radian?

Regular Homework Set #8(continued from the previous slide)

4. Give at least three angles that have sin() = .25 . Express your angles in both radians and in degrees.

5. On a test without using a calculator, you may be asked to give values for sine, cosine, and tangent for the cases listed on or near slide #20. For those specific cases, make a graph of both sine and cosine and mark these special cases on the graphs.


6. Given these coordinates in rectangular form, (x,y), determine the coordinates in polar form, (r, ): a) (4,4); b) (-4,4); c) 2,-4); d) (-2,-4) .

7. Given a circle of radius 5 meters, what is the arclength on the circle subtended by an angle of 40o ? [Hint: convert the angle to radians and use the definition on or near slide #4.]


8. Given the following points in polar form, (r, ), give the points in rectangular form, (x,y): a) (4, 45o); b) (5, π/6); c) (9, 120o);d) (13, - π/3).

9. For a right triangle, the Pythagorean Theory says: x2 + y2 = r2 . Show thatthis leads to the following:sin2() + cos2() = 1.

x

yr


10) Verify that sin2() + cos2() = 1 works for specific angles of your choice in each of the four quadrants.

part 4: trigonometry in part 1 we talked about locating things in two dimensions. we restricted...

Documents

o slide

sin cos

radians radians

x y sin

sin180 o

sin90 o

o horizontal

sin antisymmetric