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?
θ
How do you measure an angle?
Trig: Angles
How 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 2p for a circle? Why does p 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 = 2p radians
(Comes from definition of
angle measured in radians:
q = arclength / radius
=s/r)
s
q
r
Trig: basic functions
Besides angles, we can define functions
associated with angles. The trig functions
are based on a right triangle:
q
Trig: basic functions
•
•
•
•
sin(q) = opposite/hypotenuse = y/r
cos(q) = adjacent/hypotenuse = x/r
tan(q) = opposite/adjacent = y/x
(The hypotenuse is the side opposite the
right angle.)
r
y
q
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.
sin(θ)
y
1.0
0.5
90o
π/2
radians
θ
1800
π
radians
x
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]).
y
sin(θ)
1.0
0.5
90o
π/2
radians
θ
1800
π
radians
x
Trig: sine function
Notice that sin(θ) is negative between 180o
and 360o.
y
sin(θ)
π
radians
1800
-0.5
-1.0
3π/2
radians
270o
2π
radians
θ
3600
x
Trig: sine function
Notice 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].
sin(θ)
180o
360o
θ
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.
sin(θ)
y
1.0
0.5
-0.5
-1.0
90o
π/2
radians
θ
1800
π
radians
x
Trig: cosine function
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).
Trig: cosine function
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].
sin(θ) cos(θ)
180o
360o
θ
Trig: sine and cosine functions
Notice 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) .
sin(θ) cos(θ)
180o
360o
θ
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 have
sin(θ) = y/r = cos(90o-θ)
r
90o-θ
and
cos(θ) = x/r = sin(90o-θ).
θ
.
x
y
Trig: sine and cosine functions
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 functions
Why 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).
Trig: sine and cosine functions
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 functions
We 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).
r
y=x
r
60o
45o
x
r/2
r
Trig: sine and cosine functions
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
sin(30o) = cos(60o) = 0.500
sin(45o) = cos(45o) = 0.707 = √(2)/2
sin(60o) = cos(30o) = 0.866 = √(3)/2
sin(90o) = cos(0o) = 1.000
0o = 0 rads
30o = π/6 rads
45o = π/4 rads
60o = π/3 rads
90o = π/2 rads
Trig: sine and cosine functions
Because 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 rads
sin(135o) = sin(45o) = 0.707 = √(2)/2 135o = 3π/4 rads
sin(150o) = sin(30o) = 0.500
150o = 5π/6 rads
sin(180o) = sin(0o) = 0.000
180o = π rads
cos(120o) = -cos(60o) = -0.500
cos(135o) = -sin(45o) = -0.707 = -√(2)/2
cos(150o) = -cos(30o) = -0.866 = -√(3)/2
cos(180o) = -cos(0o) = -1.000
Trig: sine and cosine functions
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)
Trig: sine and cosine functions
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.
Trig functions –
domain and range
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.
Trig functions –
domain and range
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: slope
Notice 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.
sin(θ)
180o
360o
θ
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 functions
Can 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?
+1
sin(θ)
Are there any
symmetries?
θ
-90o
-1
+90o
Inverse trig functions
Note 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
+1
sin-1(x)
sin(θ)
θ
-90o
-1
+90o
-1
x
-90o
+1
Inverse trig functions
The 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
+1
sin-1(x)
sin(θ)
π/2
θ
-1
+π/2
-1
x
-π/2
+1
Inverse trig functions
Can 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?
+1
cos(θ)
90o
-1
θ
180o
Inverse trig functions
Again no asymptotes, but there is an
antisymmetry through 90o .
+180o
+1
cos-1(x)
cos(θ)
90o
90o
-1
θ
180o
-1
+1 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).
Trig: tangent function
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?
Trig: tangent function
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!
Trig: tangent function
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 function
As 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. +3
tan(θ)
θ
-90o
-3
+90o
+180o
+270o
Trig: arctangent function
Note 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+3
symmetric (odd function),
tan(θ)
and note the asymptotic
behavior near ±90o .
θ
-90o
-3
+90o
Trig: arctangent function
Since tan(θ) goes through the origin and has antisymmetry, 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
+3
tan-1(θ)
tan(θ)
θ
-90o
-3
+90o
-3
+3
-90o
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: Vectors
How 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
3
the location of point
A 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
q = tan-1 (y/x) = 37o
Trig: Transformation Equations
• These two equations are called the
rectangular to polar transformation
equations:
r = [ x2 + y2 ]
q = tan-1 (y/x) .
• Do these work for all values of x and y,
including negative values?
Trig: Transformation Equations
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.
Trig: Transformation Equations
q = 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?
Trig: Transformation Equations
q = 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.
Trig: Transformation Equations
q = 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,q) can we get (x,y) ?
• If we recall our trig functions,
we can relate x to r and q:
r
x = r cos(q), and similarly
y
y = r sin(q).
x
Do these work for all values
of r and q? YES!
Trig: Inverse Transformations
x = r cos(q), and y = r sin(q)
Example: If (r,q) = (5 bl, 37o), what do we
get for (x,y) ?
x = r cos(q) = 5 bl * cos(37o) = 4 bl.
y = r sin(q) = 5 bl * sin(37o) = 3 bl.
[Note that this is the (x,y) we started with to
get the (r,q) = (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 q = 0.1 radians:
[q, sin(q), or tan(q)] ? Which of the
three is largest for q = 1 radian?
Regular Homework Set #8
(continued from the previous slide)
4. Give at least three angles that have
sin(q) = .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.
Regular Homework Set #8
(continued from the previous slide)
6. Given these coordinates in rectangular form,
(x,y), determine the coordinates in polar
form, (r, q): 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.]
Regular Homework Set #8
(continued from the previous slide)
8. Given the following points in polar form,
(r, q), 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 that
this leads to the following:
r
y
sin2(q) + cos2(q) = 1.
q
x
Regular Homework Set #8
(continued from the previous slide)
10) Verify that sin2(q) + cos2(q) = 1 works
for specific angles of your choice in each of
the four quadrants.