Section 8: Basic Trigonometric Functions
Our goal in this section is to define and develop the six basic trigonometric functions. We
start with two of the more important transcendental functions, namely the sine function and
the cosine function. These are universally denoted by y = sin x and y = cos x respectively.
In order to define sin x and cos x, we first need to agree on how we are going to measure
angles. An angle, θ, consists of two rays emanating from a common point called the vertex.
One ray is called the initial side, the other is the terminal side. Because we can choose where
to put the origin in the Cartesian plane and we can decide where the positive x-axis goes, we
will assume that the vertex of our angle is at the origin and that the initial side is along the
positive x-axis. This is called putting the angle in standard position. Finally, since only the
direction of the rays matters, we can chop off both rays at length one. Our angle theta is now
determined by the two points at the end of these line segments of length 1, both of which lie
on the unit circle whose equation is x2 + y 2 = 1. The point on the initial side is simply (1, 0).
Here we will denote the point on the terminal side by P (θ).
All we need to determine how to measure θ is to know how we generated the angle. That
is we need to know how we go from (1, 0) to P (θ). If we generate θ by moving from (1, 0)
counterclockwise (CCW), we say θ has positive measure. If we generate θ by moving from
(1, 0) clockwise (CW), we say that θ has negative measure. The radian measure of θ is the
length of the arc on the unit circle we travel from (1, 0) to P (θ) when generating θ. We use a
minus sign in front when θ is generated CW.
Since the unit circle has circumference 2π · 1 = 2π, once around the circle CCW is 2π
radians. Halfway around the circle CCW is therefore π radians. One fourth the way around
2π
π
the circle CCW is
= radians. So we see that P (θ) = P (ϕ) any time that θ and ϕ have
4
2
measures which differ by a multiple of 2π radians. So we write P (θ) = P (θ + 2kπ) for all
k ∈ ZZ. Two such angles are called coterminal.
We call the x-coordinate of P (θ) the cosine of θ and denote it by cos θ. The y-coordinate
of P (θ) is the sine of θ and is denoted by sin θ. It is fairly clear that these are functions of
θ where θ, as a length, can be any real number. Therefore both the sine function and cosine
function have domain IR, unless restricted for some reason.
Since all the points on the unit circle have coordinates bounded between 1 and −1, we
have that −1 ≤ sin θ ≤ 1 and −1 ≤ cos θ ≤ 1. By the horizontal line test every one of these
29
Section 8: Basic Trigonometric Functions
values is taken on, so the sine function and the cosine function both have range [−1, 1]. These
serve as our first examples of functions with bounded range. Meaning there is a number M
with f (x) ≤ M , for all x ∈ dom f .
Since P (θ) = P (θ + 2kπ) for all k ∈ ZZ, we have sin(θ) = sin(θ + 2kπ), and cos(θ) =
cos(θ + 2kπ) for all k ∈ ZZ. A function which repeats its values regularly is called periodic.
The period of a periodic function is the length of the smallest continuous input interval between
repetitions. Here we see that sin x and cos x have period 2π. Thus we would not be able to
describe the behavior of sin x as x −→ ∞ except to say that the values oscillate.
Neither the sine function nor the cosine function have any asymptote.
The cosine has maximum value 1 and attains this value at θ = 0 + 2kπ, where k ∈ ZZ. The
cosine has minimum value −1 and attains this at θ = π + 2lπ = (2l + 1)π, where l ∈ ZZ. The
sine function attains maximum value of 1 when θ = π/2 + 2kπ = (4k + 1)π/2, where k ∈ ZZ.
The minimum value of the sine function is −1 and occurs when θ = 3π/2 + 2lπ = (4l + 3)π/2,
where l ∈ ZZ.
For a given point P (θ) on the unit circle there are several other points which are natural
to consider. The first is the antipodal point on the opposite side of the unit circle. This
point has coordinates (− cos θ, − sin θ). It is also P (θ + π). We thus derive the relationship
π
sin(θ + π) = − sin θ. Another point to consider is radians CW from P (θ). It has coordinates
2
(sin θ, − cos θ) and is P (θ − π/2). Thus cos(θ − π/2) = sin θ, which means that the graph of
the sine function is the graph of the cosine function shifted right π/2 units. Equivalently the
graph of the cosine function is the graph of the sine function shifted left π/2 units and we have
that cos θ = sin(θ + π/2).
In short to determine the graphs of these functions, it suffices to graph the sine function
for 0 ≤ θ ≤ π and apply all the relations we found above. This is made even easier when we
use the fact that the unit circle is symmetric about the y-axis, giving us sin θ = sin(π − θ). So
we only need to determine values for 0 ≤ θ ≤ π/2.
Before we do so we remark that the x-intercepts of the sine function have x-coordinates
nπ, where n ∈ ZZ. The x-coordinates of the intercepts of the cosine function are (2m + 1)π/2,
where m ∈ ZZ.
Now sin 0 = 0, and sin π/2 = 1. Also as θ increases from 0 to π/2, the value of the sine
√
function will increase. In the next section it is derived that sin π/6 = 1/2, sin π/4 = 2/2,
30
Section 8: Basic Trigonometric Functions
and sin π/3 =
√
3/2. With just these five values (which you should memorize), we can sketch
a remarkably good graph of z = f (θ) = sin θ. We can also therefore graph its shift left by π/2
units, z = cos θ.
The reciprocal of the sine function is the cosecant function, and is denoted csc x. It has a
vertical asymptote when sin x = 0. It has a relative minimum when sin x = 1, and a relative
maximum when sin x = −1. It is also periodic with period 2π.
The reciprocal of the cosine function is the secant function, and is denoted sec x. It has
period 2π, and is a shift left π/2 units of the cosecant function.
The ratio of the sine function to the cosine function is the tangent function, denoted
tan x = sin x/ cos x. This function is zero when sin x = 0, i.e. x = nπ, where n ∈ ZZ. It has a
vertical asymptote whenever cos x = 0, which is when x = (2m + 1)π/2, where m ∈ ZZ. Since
−/− is +, the tangent function has period π. As the ratio of an odd function and an even
function the tangent function is odd. Finally as x −→ (π/2)− , sin x −→ 1, and cos x −→ 0+ ,
so tan x −→ ∞. On the other hand as x −→ (π/2)+ , tan x −→ −∞. The reciprocal of the
tangent function is the cotangent function, denoted cot x. Collectively these make up the six
basic trigonometric functions. We normally concentrate on sin x, cos x, tan x, and sec x.
We will discuss inverse trigonometric functions, and the difference quotients of trigonometric functions in later sections.
Several last points need to be made here. First of all, for any θ ∈ IR, there is a value r ∈ IR,
called the reference angle, so that 0 ≤ r ≤ π/2, with | sin(θ)| = sin r and cos r = | cos(θ)|. For
r 6= 0, π/2, there is therefore a unique right triangle of hypotenuse 1 which has base cos r and
height sin r. This is the reference triangle for θ. We investigate this connection with right
triangles further in a later section. Second, since the point P (θ) lies on the unit circle we have
that (sin θ)2 + (cos θ)2 = 1, for all θ ∈ IR. Since this follows from the Pythagorean Theorem,
this is called the Pythagorean Identity. This is one of the more important of hundreds of
trigonometric identities.
Exercise: Read sections 9 and 10.
31
Section 9: Special Values of Trigonometric Functions
A crucial theorem in trigonometry is Euclid’s Theorem on similar triangles. The triangle
with vertices at A, B and C is denoted by 4ABC. Two triangles 4ABC and 4DEF are
similar provided that their interior angles have the same measure. So the angle with vertex
at A has the same measure as the angle with vertex at D, etc. We state the theorem without
proof.
Theorem: (Euclid) 4ABC with side lengths a = |AB|, b = |BC|, and c = |AC| is similar to
a
b
c
4DEF with side lengths d = |DE|, e = |EF | and f = |DF | if and only if = = .
d
e
f
This theorem allows us to define the trigonometric functions using P (θ). First we draw
a line which touches the unit circle only at P (θ). Such a line is said to be tangent to the
curve at the point P (θ). (You might be asking how this can be done.) Except when the line
is horizontal or vertical (i.e. θ is a multiple of π/2) it has both an x-axis intercept D, and a
y-intercept E. Call the origin O, and put B = (cos θ, 0). Then 4OP B is similar to 4ODP .
|OD|
|OP |
|OD|
1
Therefore
=
. That is,
=
. So since sec θ and cos θ have the same sign
|OP |
|OB|
1
| cos θ|
the coordinates of D are (sec θ, 0). Similarly the coordinates of E are (0, csc θ).
A vertical line is tangent to the circle only when θ = kπ. For these values of θ the sine
function is zero. Hence the cosecant is undefined. Notice though that the coordinates of D
are still (sec θ, 0). The anagolous result holds true for horizontal tangents to the unit circle.
|P D|
|P B|
|P D|
| sin θ|
To define the tangent we note that
=
. Which is to say
=
. So
|OP |
|OB|
1
| cos θ|
the length from P to D measures the absolute value of the tangent of θ. In fact we see that
the tangent line to the curve at P (θ) is perpendicular to the line, l, through O and P , so the
slope of the tangent line is the negative reciprocal of the slope of l. But the slope of l is simply
sin θ
(rise over run)
. So the value of tan θ tells us how to draw the tangent line to the unit
cos θ
circle at P (θ)! The point is that the definitions we made for trigonometric functions in section
8 did not fall out of a hat, they have geometric significance.
Another application of Euclid’s Theorem is that given any right triangle with an included
angle θ, where 0 < θ < π/2, we can define the six trigonometric functions as ratios of the
lengths of the sides of the triangle. To do this draw the given triangle with the right angle in
the lower right hand corner and θ in the lower left corner. Let the length of the base be a,
the height be b, and the length of the hypotenuse be c. The other triangle is drawn with θ in
32
Section 9: Special Values of Trigonometric Functions
standard position in IR2 with vertices (0, 0), (cos θ, 0), and (cos θ, sin θ). This triangle has base
of length cos θ, height sin θ, and hypotenuse of length 1. By Euclid’s Theorem the ratios of
corresponding side lengths are equal so we have
1
sin θ
cos θ
=
=
c
b
a
So given two sides, for example a and c, we have an equation relating the two sides which can
rewritten as an equation describing their ratio. In this case 1/c = cos θ/a implies a/c = cos θ.
Equivalently c/a = 1/ cos θ = sec θ. The other formulas are b/c = sin θ, which is equivalent to
c/b = csc θ, and b/a = tan θ, which is equivalent to a/b = cot θ.
The famous mnemonic here is “soh-cah-toa” which is an acronym for “sine is opposite
over hypotenuse–cosine is adjacent over hypotenuse– tangent is opposite over adjacent.”
The sum of the measures of the angles of a triangle is 180◦ , which is π radians. So for a
right triangle with included angle θ the other nonright angle ϕ must have measure (π/2) − θ
radians. We call the two angles θ and ϕ complementary. Notice that the value of f (ϕ) = cof (θ)
for any of the six trigonometric functions f , when defined as ratios of lengths of sides. Since
the complement of the complement of θ is simply θ we get the mnemonic device “co-co=no-co”.
The fact that a triangle has three sides explains why there are three pairs of complementary
trigonometric functions.
Finally, to demonstrate the utility of these concepts we derive some special values of the
basic trigonometric functions.
√
First take an isosceles right triangle with side s. The hypotenuse has length s2 + s2 =
√
√
2s2 = s 2. Also the included acute angles must be equal and have measure π/4. Therefore
√
√
√
√
cos π/4 = s/s 2 = 1/ 2 = 2/2 = sin π/4. Also tan π/4 = 1 = cot π/4, and sec π/4 = 2 =
csc π/4. Moreover any value of θ with reference angle π/4 has |f (θ)| = f (π/4), where f is any
one of the six trigonometric functions. So to determine the actual value of f (θ) we simply need
to know whether its positive or negative, which can be done since we know what quadrant the
point P (θ) lies in, so we know the sign of its coordinates sin θ and cos θ. This is sufficient since
all of the other basic trigonometric functions are defined in terms of sine and cosine.
Second take an equilateral triangle of side s, drawn with horizontal base. The interior
angles are all π/3. If we drop a perpendicular from the top vertex to the base, we split the
triangle into two right triangles. The one on the left has base s/2 and hypotenuse s. By the
33
Section 9: Special Values of Trigonometric Functions
√
Pythagorean theorem the height is s 3/2. The complementary angle has measure π/6, so we
√
have cos π/6 = sin π/3 = 1/2, sin π/6 = cos π/3 = 3/2. The values of the other four basic
trigonometric functions are determined as before.
Identical to the first case, we can now find exact values of trigonometric functions for any
angle which has reference angle π/6, or reference angle π/3.
Example 1: Let’s find the exact value of some trigonometric functions at 4π/3. The reference
angle is π/3 and P (4π/3) is in the third quadrant where both x and y coordinates are negative.
√
Therefore cos 4π/3 = − cos π/3 = −1/2, and sin 4π/3 = − sin π/3 = − 3/2. Then for example
√
tan 4π/3 = 3, since the ratio of two negative numbers is positive.
Example 2: If we have a right triangle with base of length 12 and height 5, the Pythagorean
theorem gives 13 as the length of the hypotenuse. In this case if θ denotes the angle adjacent to
the base and hypotenuse, sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cot θ = 12/5, sec θ = 13/12
and csc θ = 13/5. Moreover we know how to find lots points on the unit circle which have a
similar reference triangle.
Example 3: If we have an isosceles right triangle of side 4, then if h denotes the length of the
√
√
hypotenuse we have 4/h = sin(π/4) = 2/2. Here we write 8 = h 2 by cross multiplying and
√
√
can solve for h = 8/ 2 = 4 2.
It seems reasonable that we should be able to perform exercise 3 for any angle, θ, where
0 < θ < π/2. And of course it is possible, if we have inverse trigonometric functions. As with
even power functions, due to their periodic nature trigonometric functions are not one-to-one.
Therefore if we wish to invert them we must restrict their domain to a subset where they are
one-to-one. In order to complete exercises like example 3 we want this restricted domain to
include (0, π/2) and to be of most value it would be as large as possible.
For y = sin x we use the convention that its domain is [−π/2, π/2] when discussing its
inverse function denoted arcsin x. Formally the function is defined by the equations y = arcsin x
iff x = sin y. Which means that we can draw a reference triangle with enclosed angle |y|,
√
opposite side of length |x| and hypotenuse of length 1. The remaining side has length 1 − x2
by the Pythagorean theorem. This construct allows us to compute tan arcsin x = tan y =
√
x/ 1 − x2 .
When we start instead with y = cos x we use the convention that its domain is [0, π] and
34
Section 9: Special Values of Trigonometric Functions
denote its inverse by arccos x.
The conventional domain for y = tan x as an invertible function is (−π/2, π/2). Now we
have y = arctan x iff x = tan y. So we can draw a reference triangle with enclosed angle |y|,
opposite side of length |x| and adjacent side of length 1. The hypotenuse therefore has length
√
√
x2 + 1, and we can write for example cos arctan x = cos y = 1/ x2 + 1. Notice that here the
cosine function always has positive value which agrees with our convention about the domain
of a composition of functions.
One last picture to consider is the one flowing from considering sec x as an invertible
function with inverse denoted arcsec x. Now we suppose that the domain of the secant function
is limited to [0, π/2) ∪ (π/2, π]. By definition y = arcsec x iff x = sec y. So we can draw a
right triangle with included angle |y|, hypotenuse of length x, adjacent side of length 1, and
√
√
opposite side of length x2 − 1. Now for example csc arcsec x = csc y = x/ x2 − 1.
The restricted domain for cot x as an invertible function is (0, π). For y = csc x to be
considered invertible our convention is to limit its domain to [−π/2, 0) ∪ (0, π/2].
Exercises:
1. Find the exact value of each of the following:
π 2π
a) sin
b) tan
3
3
−π
7π
d) csc
e) cos
6
6
3π
g) sin (π)
h) cot
4
j) cos (3π)
k) tan (π)
2. Find the exact value of each of the following:
a) tan arccos (−3/5)
b) sin arccot (1/2)
−5π
c) cos
4
−π
f) sin
3
3π
i) sec
2
π l) csc
6
c) sec arccsc (2/3)
3. Sketch the picture described in the second paragraph of this section. Label the points O,
P , B, D, and E.
4. Sketch and label a triangle in which:
a) x = sin y
b) x = tan y
35
Section 10: General Sinusoids
A general sinusoid is a function of the form y = k + A sin(B(x − c)) or y = k + A cos(B(x − c)),
where k, A, B and c are real numbers. Since the sine function and cosine function are horizontal
shifts of each other we will only consider the form y = k + A sin(B(x − c)). Such a function is a
shift right by c units (left if c < 0), of a compression horizontally by a factor of 1/B (expanded
if B < 1), of a vertical scale by A units, of a shift up by k units (down if k < 0) of the basic
function y = sin x. For such a function |A| is the amplitude. About the central line y = k
the sine wave will fluctuate up |A| units and down |A| units. All function values are bounded
below by k − |A| and above by k + |A|. The bounds are obtained when sin(B(x − c)) = ±1,
which means that B(x − c) = (2n + 1)π/2, where n is an integer. This is equivalent to
x = c + [(2n + 1)π/2B]. The general sinusoid is periodic with period 2π/B.
π
Example 1: Let’s sketch a graph of y = f (x) = 2 − 3 sin(3x − ). First we put the function
2
π
in standard form y = f (x) = 2 − 3 sin(3(x − )). Then it’s a matter of determining where
6
f takes on maximal and minimal values, since half-way between it will take value 2. The
extreme values happen when 3(x −
π
6)
= (2n + 1)π/2. So x −
π
6
= (2n + 1)π/6. Thus
6x − π = (2n + 1)π = 2nπ + π. So 6x = (2n + 2)π, or x = (n + 1)π/3. We see that f (π/3) = −1
and f (2π/3) = 5. So f (π/2) = 2, and the rest is determined by symmetry. Of course we need
to mark our x-axis off in units of π/6 to be able to clearly indicate where f takes on its extreme
values and its central values. The y-axis needs to be marked so that all values between −1
and 5 can be seen. Elongating the x-axis a little gives a sketch which is smoother – more like
the basic sine wave. We can also shorten the y-scale somewhat to acheive the same effect.
6
4
2
−π
− 2π
3
π
3
− π3
-2
36
2π
3
π
Section 10: General Sinusoids
It is also useful to be able to combine two (or more) general sinusoids arithmetically.
Given two sinusoids y = k1 + A1 sin(B1 (x − c1 )) and y = k2 + A2 sin(B2 (x − c2 )), their sum
is is shift up k1 + k2 units of y = A1 sin(B1 (x − c1 )) + A2 sin(B2 (x − c2 )). This function is
bounded since −(|A1 | + |A2 |) ≤ y ≤ |A1 | + |A2 |. The bounds may or may not be met. To
graph this function we find scales for the x-axis and y-axis that allow both simpler functions
to be graphed simultaneously. Then we add ordinates. Ordinate is a fancy name for the
y-coordinate.
Example 2: Sketch a graph of y = 3 sin x + sin(3x). First of all the outputs are bounded
between −4 and 4. Second the function y = 3 sin x has period 2π, and is simply a vertical
scale of the regular sine function. Third the function sin 3x has period 2π/3. So it is a
horizontal compression of the sine wave by a factor of 3. To show 2 periods of each component
function simultaneously we need an interval of x-values of length 4π. We might as well take
x ∈ [−2π, 2π] since the overall function is odd. We do need to mark off multiples of π/6 on the
x-axis in order to see all zeroes, and extreme values of sin 3x. We can also make two copies of
our axes, one for scratch work, the other for the final product (neatness counts).
4
2
π
−π
−2π
2π
-2
-4
Exercises:
1. For each of the following general sinusoidal functions identify the amplitude A, the vertical
shift k, the period p, and the phase shift c. Then sketch and label a graph of the function
on the given interval. In each instance choose scales for your axes which allow all important
details to be clearly shown. At least two periods of the function should be shown.
a) y = 2 − 3 cos(2πx +
π
)
2
37
Section 10: General Sinusoids
b) y = 1 + 2 sin(
x π
− )
2
2
c) y = −3 + sin(3x)
2. Sketch and label a graph of each of the functions below on the specified interval. In each
instance choose scales for your axes which allow all important details to be clearly shown. At
least two periods of the resulting function should be graphed.
a) y = 3 sin πx + 2 sin(2πx)
b) y = 2 cos(2x) + cos(4x)
38
Section 11: Trigonometric Identities
We’ve already seen many basic identities which trigonometric functions satisfy. For example if f is a basic trigonometric function, and g is its co-function, then f (x) = g(π/2−x), as long
as both sides of the equation make sense. These are called the co-function identities. Another
example are what are called the reciprocal identities, such as csc x = 1/ sin x. A third example is the quotient identity, tan x = sin x/ cos x. Finally, consider the Pythagorean identity
sin2 x + cos2 x = 1.
For each of these identities there is usually an equivalent way of stating the identity. For
example we may rewrite the Pythagorean identity as sin2 x = 1−cos2 x, or as cos2 x = 1−sin2 x.
We can even divide the equation through by cos2 x to obtain tan2 x+1 = sec2 x, after appealing
to algebra, the quotient identity, and a reciprocal identity. This latest version of the identity is
not true for all real x, but is true as long as cos2 x 6= 0, which means x 6= (2n+1)π/2. One game
we might play is to continue generating equivalent identities from old using other identities.
The result is a very impressive list of identities, too many for most people to memorize. So
instead of recommending that people memorize every form of an identity, we recommend that
they remember one form of the identity and develop the ability to re-derive the rest. This is
very much like the difference between giving someone a fish, and teaching them how to fish.
To build this skill a common practice is to use an exercise such as the following.
Example 1: Verify that 1 + cot2 x = csc2 x is an identity.
Solution :
cos2 x
sin2 x
sin2 x + cos2 x
=
sin2 x
1
=
sin2 x
= csc2 x
1 + cot2 x = 1 +
(quotient identity)
(algebra)
(Pythagorean identity)
(reciprocal identity)
Notice in our solution we did not start by affirming the conclusion. That is we did not
suppose that the identity was true. We simply started with the left hand side and continued
to apply identities until we arrived at the right hand side. Not only is this process logically
proper, it is clearly reversible. So when we verify an identity we will use this process. Next
39
Section 11: Trigonometric Identities
realize that every step was justified as either algebra or as the application of an identity. While
this is not mandatory it is suggested. Third, observe that we did not show every single step of
algebra. The real challenge is for each individual to learn to take as many steps as they need
to verify the identity. Some people are more adroit than others combining algebraic steps. So
two valid solutions might have a different number of delineated algebraic steps.
Now to demonstrate the power of this process let’s derive one identity and develop a
variety of other identities from it.
Example 2: cos(θ − φ) = cos θ cos φ + sin θ sin φ, for θ, φ ∈ IR.
Proof: Let θ, φ ∈ IR. On the unit circle the square of the distance from P (θ) to P (φ) is the
same as the square of the distance from P (θ − φ) to P (0). Therefore by the distance formula
we have
(cos θ − cos φ)2 + (sin θ − sin φ)2 = (cos(θ − φ) − 1)2 + (sin(θ − φ) − 0)2
After expansion the left hand side is cos2 θ − 2 cos θ cos φ + cos2 φ + sin2 θ − 2 sin θ sin φ + sin2 φ.
By applying the Pythagorean identity twice we get 2 − 2 cos θ cos φ − 2 sin θ sin φ. Meanwhile
the right hand side becomes cos2 (θ −φ)−2 cos(θ −φ)+1+sin2 (θ −φ). Which after applying the
Pythagorean identity once becomes 2 − 2 cos(θ − φ). Cancelling the 20 s and dividing through
by −2 yields the identity. This is called the difference identity for cosine.
From the difference identity for cosine we can derive the sum identity for cosine,
Theorem: cos(θ + φ) = cos θ cos φ − sin θ sin φ.
Proof: Here we simply write cos(θ + φ) = cos(θ − (−φ))
= cos θ cos(−φ) + sin θ sin(−φ)
= cos θ cos φ − sin θ sin φ
since cos(−φ) = cos φ and sin(−φ) = − sin φ.
Similarly we can prove the sum identity for sine.
Theorem: sin(θ + φ) = sin θ cos φ + sin φ cos θ
40
Section 11: Trigonometric Identities
Proof: We use sin((π/2) − x) = cos x, sin(−z) = − sin z and cos(y − (π/2)) = sin y.
π
sin(θ + φ) = cos((θ + φ) − )
2
π
= cos(θ + (φ − ))
2
π
π
= cos θ cos(φ − ) − sin θ sin(φ − )
2
2
π
= cos θ sin φ + sin θ sin( − φ)
2
= cos θ sin φ + sin θ cos φ
The rest is a matter of rearranging terms.
The difference identity for sine is sin(θ − φ) = sin θ cos φ − sin φ cos θ. We leave the verification of this identity as an exercise.
As interesting corollaries we get the double angle formulas which are
2 tan x
cos 2x = cos2 x − sin2 x, sin 2x = 2 sin x cos x, and tan 2x =
. For example
1 − tan2 x
Theorem: cos 2x = cos2 x − sin2 x
Proof: cos(2x) = cos(x + x) = cos x cos x − sin x sin x = cos2 x − sin2 x.
Continuing, we can produce cos 2x = cos2 x − sin2 x = cos2 x + sin2 x − sin2 x − sin2 x =
1 − 2 sin2 x. This is an equivalent version of the double angle formula for cosine. One can also
show that cos 2x = 2 cos2 x − 1.
Each of the last two versions of the double angle formula for cosine can be rewritten. For
1 − cos 2x
example sin2 x =
. Which means that the function sin2 x can be analyzed via the
2
cosine function using shifting and scaling. Or if we wish we can derive one of the half-angle
formulas.
r
x
1 − cos x
Theorem: sin = ±
2
2
Proof: Set y = x/2, so 2y = x. Then sin2 (x/2) = sin2 y = (1 − cos 2y)/2 = (1 − cos x)/2. The
result follows by taking square roots of both sides, realizing that the answer may be positive
or negative.
As a nice application of these formulas we have the following examples.
Example 3: Find the exact value of sin(5π/12).
41
Section 11: Trigonometric Identities
8π
3π
2π
π
5π
= sin(
−
) = sin(
− )
12
12
12
3
4
2π
π
π
2π
= sin
cos − sin cos
3
4
4
3
√
√ √
3 2
21
−
=
2 2
2 2
√
√
6− 2
=
4
7π
Example 4: Find the exact value of cos
.
8
Solution: sin
Solution: Since 7π/8 is in the second
s quadrant
s
√
p
√
1 + cos 7π
1 + 22
7π
2+ 2
4
cos
=−
=−
=−
8
2
2
2
Example 5: If we know that cos θ = 4/5 and 3π/2 ≤ θ ≤ 2π then cos 2θ = 2 cos2 θ − 1 =
2(16/25) − 1 = 7/25. While because θ is in quadrant four the Pythagorean identity implies
sin θ = −3/5. So sin 2θ = 2 sin θ cos θ = −24/25. Therefore tan 2θ = −24/7, and we can also
compute exact values for csc 2θ and cot 2θ.
The previous example can be inverted by giving exact value of cos x, for example, and
quadrantal information about x, to enable using the half-angle formulas to compute exact
values of any of the trigonometric functions at
x
2.
Identities are also important since they may be used to simplify a problem. For example
a3
if we want to analyze the formula z = 2
we put u = a tan θ. We find that a2 + u2 =
(a + u2 )3/2
a2 + a2 tan2 θ = a2 (1 + tan2 θ) = a2 sec2 θ. So (a2 + u2 )3/2 = (a2 sec2 θ)3/2 = a3 sec3 θ,
and z = cos3 θ. That is, trigonometric identities and composition of functions may allow us
to change venue to a friendlier setting. To be sure no one has all trigonometric identities
memorized, and many of them are not initially obvious. However we can for instance go one
step further and write z = (cos 3θ + 3 cos θ)/4 which is a restatement of one of the homework
exercises. This version of z as a function of θ can be visualized using addition of ordinates.
Yet another application of trigonometric identities is in simplifying difference quotients of
sin h
trigonometric functions. For example by graphing we can observe that as h −→ 0,
−→ 1,
h
cos h − 1
while
−→ 0. We then have
h
sin(x + h) − sin x
sin x cos h + sin h cos x − sin x
=
h
h
cos h − 1 sin h = sin x
+ cos x
h
h
42
Section 11: Trigonometric Identities
So as h −→ 0, the difference quotient of the sine function approaches cos x.
As a final demonstration of the power of these identities you may now read appendix E
on rotation of axes to clear cross terms in general quadratic equations in two variables.
Exercises:
1. Verify the following trigonometric identities.
a) sin(θ − φ) = sin θ cos φ − sin φ cos θ
tan θ + tan φ
b) tan(θ + φ) =
1 − tan θ tan φ
c) sin 2x = 2 sin x cos x
2 tan x
d) tan 2x =
1 − tan2 x
1 + cos 2x
e) cos2 x =
2
3
f) cos 3x = 4 cos x − 3 cos x
r
x
1 + cos x
g) cos = ±
2
2
x
1 − cos x
h) tan =
2
sin x
1
i) sin α cos β = (sin(α − β) + sin(α + β))
2
1
j) cos α cos β = (cos(α − β) + cos(α + β))
2
1
k) sin α sin β = (cos(α − β) − cos(α + β))
2
2. Use trigonometric identities to find the exact value of each of the following.
π
12
7π
b) sec
12
5π
c) csc
8
a) cot
−4
and that θ is in quadrant III. Find the exact values of the
5
other five trigonometric functions at θ.
3
4. Given that θ is in quadrant I and that cos 2θ = − , find the exact values of the six
5
trigonometric functions at θ.
3. Using the fact that sin θ =
5. Read appendix E.
43
Section 12: Applications of Trigonometry
Whenever a section in a math book has applications in the title it means “word exercises.”
So now you know what you will mostly face in this section. The reason that many people have
difficulty with word problems is that their solutions tend to be more involved than the “plug
and chug” exercises otherwise found in text books. In fact the next to last part of a word
problem usually is completing a particular type of “plug and chug” exercise. The difficulty is
arriving at that stage. Here follows a short description of what is supposed to happen.
Step 1: Read the problem carefully, several times. Make sure you understand what is being
asked.
Step 2: Draw a picture or diagram.
Step 3: Label all known quantities including units, such as ft., lbs. etc. Usually word problems
do not contain superfluous information. Thus any information given is vital.
Step 4: Label all unknown quantities with distinct variables. In contrast to step 3, some
unknown quantities may be irrelevant. With practice should come some facility in deciding
which unknown quantities you can omit. Any unknown quantity which you are being asked
to find must be labeled.
Step 5: This is the hard part. Write down as many equations/relations/identities as you can
which relate the known and unknown quantities. Some of these will be irrelevant. As before,
with practice will come some ability to determine which equations are to be used. The hint is
that the equations usually have something to do with the title of the section.
Step 6: After eliminating as many variables as possible, hopefully you have reduced the word
problem to a “plug and chug” exercise.
Step 7: When you have solved the appropriate exercise, you should carefully and neatly
re-write your solution eliminating all irrelevant unknown quantities,labels,equations,relations
and identities. The finished product should look like the examples, possibly minus some of the
prose.
Before we get to the fun we need to develop a few more tools.
First consider a triangle drawn with a horizontal base and one vertex labelled B above it.
Label the left end point of the base A, the right end point C. Label the angle with vertex A,
α, the angle with vertex B, β, and the angle with vertex C, γ. Finally label the length of the
44
Section 12: Applications of Trigonometry
side opposite A, a, the length of the side opposite B, b, and the length of the third side c.
Drop a perpendicular from B to the line the base lies in. Call the height of the perpendicular h, and the intersection of the perpendicular with the base P . If the perpendicular coincides
with the side CB we have a right triangle and c2 = a2 + b2 . Otherwise the perpendicular splits
the triangle into two right triangles AP B and BP C.
The length from A to P is |c cos α|. Therefore the length from P to C is b − c cos α. Also
we have c2 = h2 + c2 cos2 α and a2 = h2 + (b − c cos α)2 . From the first equation we have
h2 = c2 − c2 cos2 α so by substitution
a2 = h2 + (b − c cos α)2
= c2 − c2 cos2 α + b2 − 2bc cos α + c2 cos2 α
= c2 + b2 − 2bc cos α
This is the law of cosines. Since the parts of the triangle can be rearranged we also have
b2 = a2 + c2 − 2ac cos β, and c2 = a2 + b2 − 2ab cos γ.
Another tool arises from this scenario since in the triangles above we have h = a sin γ =
sin α
sin γ
sin β
c sin α. Therefore
=
. In fact whatever this common ratio is it must equal
.
a
c
b
This is the law of sines.
Though this law is simpler, and therefore more easy to use, it is more dangerous to use
too. If we are only given a, c and α, then if a < c sin α, there is no triangle with these
properties. If a = c sin α , then we get a unique right triangle where γ is the right angle. When
a ≥ c(≥ c sin α) we again get only one triangle with these properties. For this triangle γ must
be acute. However when c sin α < a < c, there are two angles γ1 and γ2 , where γ1 + γ2 = π
and γ1 is obtuse while γ2 is acute. So when applying the law of sines we must be careful to
consider whether our answer makes sense physically.
Now for the fun. A fairly representative exercise whose solution uses trigonometry is the
following.
Example 1: An executive from a window washing company is putting in a bid for the weekly
washing of the windows of a skyscraper. To set his price he needs to know approximately how
tall the building is. His ex-wife works for the city planning office, so he doesn’t dare call them
to ask for the information. Instead he moves away from the building 101 feet and measures
the angle of elevation from that point to the roof of the building as 73◦ = 73π/180 radians.
45
Section 12: Applications of Trigonometry
Thus the height of the building h satisfies h/101 = tan(73◦ ). So h is about 101 tan(73◦ ) ≈ 330
feet.
Example 2: Fred and George are playing wallyball. George wants to pass the ball to Fred but
a very tall opposing player is in the way. Fred is standing 4 feet away from a wall which George
√
is standing 3 feet from. Fred and George are standing 145 feet apart and George decides
that he wants to make a chest-high carom-pass off the wall to Fred. How far along the wall
between them should George pass the ball.
Solution: Let the point where George is standing be labelled A, and the point where Fred is
standing B. Drop perpendiculars from A and B to the wall and label these points C and D
respectively. Let P be the point on the wall which George should hit. Denote the length from
P to C as x, so the length from P to D is 12 − x. Let θ be the angle with vertex P , initial side
along the wall, and terminal side P A. Basic physics tells us that θ should also be the angle with
x
12 − x
vertex P , one side along the wall, and the other side P B. Therefore = cot θ =
. Thus
3
4
1
4x = 3(12 − x) = 36 − 3x, which means x = 36/7 = 5 feet. By the way x/3 = 12/7 = cot θ,
7
so θ = arccot (12/7) ≈ 30.2◦ .
Example 3: A car leaves a little town on a road traveling due South at 75 mph. Twenty minutes
later a truck heads out of town traveling due East at 55 mph. How far apart in miles are the
vehicles forty-five minutes after the truck departed?
Solution: Let t = 0 correspond to when the truck departed and let us measure t in hours. Let
a(t) denote the distance in miles of the car to the town as a function of t, and b(t) denote the dis75
tance of the truck in miles to the town as a function of time. Then a(t) =
+ 75t = 25 + 75t
3
and b(t) = 55t. By the Pythagorean theorem the distance between them in miles as a function
of time t in hours is
d(t) =
p
(25 + 75t)2 + (55t)2 =
p
p
625 + 3750t + 8650t2 = 5 25 + 150t + 346t2
p
√
√
So d(3/4) = 5 25 + 150(3/4) + 346(9/16) = (5/4) 400 + 1800 + 3114 = (5/4) 5314 ≈ 91.1
miles.
Example 4: A passenger jet leaves an airport at 12:00 pm traveling due west at 430 mph. A
private jet leaves the same airport at 12:30 pm traveling 65◦ N of West at 552 mph. How far
apart in miles are the planes at 2:00 pm?
46
Section 12: Applications of Trigonometry
Solution: Let t denote time in hours and let t = 0 correspond to 12:30 pm. Let a(t) denote
the distance of the passenger plane to the airport in miles. Let b(t) denote the distance of
the private jet to the airport in miles. So similar to the previous example a(t) = 215 + 430t
and b(t) = 552t. By the law of cosines the distance squared between them at time t is
d2 (t) = (215 + 430t)2 + (552t)2 − 2(215 + 430t)(552t) cos(65◦ ). So when t = (3/2), the distance
squared is d2 (3/2) = (4 · 215)2 + (828)2 − 2(860)(828) cos(65◦ ). Therefore
p
d(3/2) = (4 · 215)2 + (828)2 − 2(860)(828) cos(65◦ ) ≈ 907 miles.
Example 5: A drug smuggler is traveling due North at 35 knots in a small boat. A coast guard
cutter is 125 nautical miles away on a heading 22◦ East of North from the smuggler. What
heading should the cutter take to intercept the smuggler, supposing that the cutter will run
at its top speed of 40 knots? How long will it take?
Solution: We could use the law of cosines here, but the law of sines is easier. Let A denote the
position of the smuggler, P denote the position of the cutter and I denote the point of intercept.
Let θ be the angle with vertex P and sides AP and IP . Let t denote time to intercept in hours.
sin θ
sin(22◦ )
Then by the law of sines
=
So sin θ = (35/40) sin(22◦ ) = (7/8) sin(22◦ ). Which
40t
35t
means that θ = arcsin((7/8) sin(22◦ )) ≈ 19◦ . So the cutter should head roughly 49◦ South of
sin(139◦ )
sin(22◦ )
West to intercept the smuggler. The time of intercept can be found using
=
125
35t
125 sin(22◦ )
So t =
≈ 2.04 hours.
35 sin(139◦ )
Example 6: A pilot is trying to get to an airport 362 miles away which lies on a major highway
that runs East and West. The control tower tells them to proceed on a heading 35◦ South
of West. The pilot knows the compass in the plane is not reliable so they make a guess at
the correct direction and proceed at 156 mph. Two hours later the pilot reaches the major
highway. How far from the airport is she?
Solution: Label the airport point A, the starting point of the plane P and the point where
the airplane reaches the highway C. Then the length of AP is 362 miles, and the length
of CP is 2 · 156 = 312 miles. The angle with vertex A and sides AP and AC has measure
35◦ . Since 362 sin(35◦ ) < 312 < 362 this is the ambiguous case of the law of sines. Still
if we let γ denote the angle with vertex C and sides CA and CP , the law of sines implies
sin γ = (362/312) sin(35◦ ) So γ = arcsin((181/156) sin(35◦ )) ≈ 41.7◦ or 138.3◦ . Unless the
pilot has a completely lousy sense of direction, γ is the larger angle. Therefore the distance
47
Section 12: Applications of Trigonometry
sin(35◦ )
sin(6.7◦ )
sin(6.7◦ )
=
. Which means d = 312
≈ 63.4 miles. If
d
312
sin(35◦ )
we choose the other value of γ we find d ≈ 529.4 miles!
d from A to P satisfies
As an example of a possible non-word exercise we have
Example 7: Show that for each A, B ∈ IR there is C, δ ∈ IR so that A sin x + B cos x =
C sin(x + δ).
Solution: Let C satisfy C 2 = A2 + B 2 , and δ = arctan(B/A). Draw a right reference triangle
with sides of length |A|, |B| and |C| and included angle |δ|. Then A = C cos δ and B = C sin δ
so that A sin x + B cos x = C cos δ sin x + C sin δ cos x = C sin(x + δ).
Exercises: Give exact answers if possible.
1. The Butler Office Building (BOB) is 847 feet high. An apartment building stands across
Main Avenue from BOB. Someone standing on the edge of the roof of the apartment building
closest to BOB measures the angle of elevation to the top of BOB as 45◦ . They measure the
angle of depression from their location to the foot of BOB as 60◦ . How far apart are the two
buildings, and how tall is the apartment building?
2. An Army Reserve Squad is dropped off at point B and ordered to march on a heading of
42◦ North of East through a dense forest to a bivouac site at point A, 23 miles away on a
North-South road. During the drop all their compasses are damaged. They take a guess at
their proper direction and march off. Marching at 6 miles per hour they arrive at the road in
1
4 hours. How far are they from the bivouac site? How far off was their estimated heading?
2
3. Two jets depart LAX simultaneously at midnight. The first flies on a heading 32◦ North of
East at 350 mph. The second flies on a heading 13◦ South of East at 467 mph. How far apart
are the planes at 2:20 am?
4. Show that for each A, B ∈ IR there is C, δ ∈ IR so that A sin x + B cos x = C cos(x − δ)
5. John has a tree in his back yard that he wants removed. The tree removal company charges
$25 a foot of height to remove it. John wants to know if he can afford it so he needs to know
the height of the tree. He stands 21 feet away from the tree on level ground and measures the
angle of elevation from himself to the top of the tree as 75◦ . About how much will it cost John
to have the tree removed?
48
Section 13: Polar Coordinates
A polar coordinate system is one in which there is a unique point in the plane called
the pole and a ray through the pole called the polar axis. A point, P , in the plane can be
coordinatized by its distance |r| to the pole and the measure of the angle θ from the polar axis
to the ray through the pole and P . Normally we take the pole to lie at (0, 0) in the Cartesian
coordinate system, and take the polar axis to be the positive x-axis. It is therefore common to
measure θ in radians. Of course if θ and φ are coterminal angles, then the polar coordinates
(r, θ) and (r, φ) describe the same point. So polar coordinates are not unique! In fact if we
take the convention that when r is negative we move in the antipodal direction of θ, then the
polar coordinates (r, θ) and (−r, θ ± π) represent the same point. To make our life easier we
normally restrict the values of θ to lie in the range (−π, π].
With these facts in mind we seek a way to convert polar coordinates and Cartesian (or rectangular) coordinates. The point P (θ) lies on the line through the origin and (r, θ). Therefore
by Euclid’s Theorem on similar triangles the Cartesian coordinates of the point are x = r cos θ,
and y = r sin θ. Very simple!
The conversion from rectangular coordinates to polar coordinates is not so smooth. When
x = 0, the polar coordinates of the point are (y, π/2). When y = 0, the polar coordinates
of the point are (x, 0). Otherwise x 6= 0 and the point with Cartesian coordinates (x, y) is a
vertex of the nondegenerate right triangle whose other two vertices are (0, 0) and (x, 0). By
p
the Pythagorean theorem the distance from (x, y) to the origin is r = x2 + y 2 . Let θ denote
the angle with vertex at the origin, initial side the positive x-axis and terminal side the line
segment between the origin and (x, y). If α is the reference angle for this value of θ, then
we know that tan α = |y|/|x| and in fact tan θ = y/x. So if −π/2 < θ < π/2, we can take
θ = arctan(y/x). In general as long as θ is coterminal with an angle in the domain of the
arctangent function, we have no problem. When θ is not coterminal with such and angle,
θ + (2n + 1)π is for some integer n. We can therefore describe the point with rectangular
coordinates (x, y), by the polar coordinates (−r, θ + (2n + 1)π).
A major application of polar coordinates is that the description of certain graphs in polar
coordinates lets us analyze them as functions, rather than relations. For example, x2 + y 2 = a2
has graph a circle with radius a centered at the origin. In polar coordinates x2 + y 2 = r 2
(squaring the top equation on the right) so we can describe this circle as r = a, a very simple
49
Section 13: Polar Coordinates
formula in polar coordinates.
By convention we do try to write polar equations in the form r = f (θ). So r is the
dependent variable and θ is independent. In order to graph polar equations we use T-tables.
π π π π
The values we normally plug-in for θ are multiples of 0, , , , etc. That is values of θ
6 4 3 2
for which we can compute exact values for r. Standardly we keep the θ values in increasing
order and after connecting our dots we use arrows to display the behavior of the curve as θ
increases. This is called orienting the graph. This is a crucial part of the graph. If a > 0,
the polar equations r = a and r = −a both represent the circle centered at the origin with
radius a. The curve r = a is oriented positively, meaning that the motion along the curve as θ
increases is counter-clockwise. The curve r = −a is oriented negatively, since the motion along
the curve as θ increases is clockwise. These are therefore different curves in polar coordinates,
though they transform to the same curve in rectangular coordinates.
There is therefore some inherent danger when we attempt to convert equations in one
coordinate system to the other. When we convert from polar to rectangular we lose orientation.
When we convert from rectangular to polar we must often choose an orientation. The loose
convention we use is to opt for the positively oriented polar curve.
Example 1: Transform the rectangular equation y = 2x to a polar equation.
Solution: By substitution we have r sin θ = 2r cos θ. So if r 6= 0, sin θ = 2 cos θ. Equivalently
tan θ = 2, or θ = arctan 2. This works when r = 0 too, since we are at (0, θ) which is (0, 0)
in rectangular coordinates. This kind of polar curve is not orientable, since θ is fixed. This is
the polar equivalent of the vertical line in rectangular coordinates which has no slope.
Example 2: Transform the polar equation r = 2 sin θ to a rectangular equation.
Solution: Supposing that r 6= 0, we can multiply both sides by r to obtain r 2 = 2r sin θ. By
substitution we have x2 + y 2 = 2y. By completing the square on y we get x2 + (y − 1)2 = 1.
Thus this polar equation describes a circle of radius 1 tangent to the pole, oriented positively.
Example 3: Transform the polar equation r = sin 2θ to a rectangular equation.
Solution: Again supposing that r 6= 0 we first write r 3 = r 2 sin 2θ = r 2 (2 sin θ cos θ). Which
we rewrite as r 3 = 2(r sin θ)(r cos θ). So by substitution (x2 + y 2 )3/2 = 2xy. Therefore
(x2 + y 2 )3 = 4x2 y 2 . But unless you know about the classification of general expressions of
total degree 6 in x and y, this equation will not help you sketch a graph of the curve. It does
50
Section 13: Polar Coordinates
enable us to realize that the un-oriented curve is symmetric with respect to the y-axis, the
x-axis, the line y = x, and with respect to the origin, since substituting in ±x for x and/or
±y for y leaves the equation unchanged. So if we figure out what the graph looks like in the
first quadrant, we can apply symmetry to determine the rest.
The values of θ we want to use to build a T-table are those for which we can compute
exact values of sin 2θ. So for example 2θ = π/6, 2θ = π/4,, 2θ = π/3, etc. So we build the
T-table
θ 0
r(θ) 0
π/12 √
π/8
1/2
2/2
π/6
√
3/2
π/4
1
Thus the graph of this polar equation is
θ=
π
3
θ=
π
6
1
We don’t always need to transform a polar equation to rectangular form to investigate
symmetry. If replacing θ with −θ leaves a polar equation unchanged, then its oriented graph
will have polar axis (x-axis) symmetry. If replacing θ by π − θ leaves a polar equation unchanged, then the oriented graph will have symmetry with respect to the line θ = π/2 (the
y-axis). Finally if replacing r by −r leaves a polar equation unchanged, then its oriented graph
will have polar (original) symmetry.
Example 3 indicates that these tests for polar symmetry are not as easy to implement
as tests for rectangular symmetry. So there can be value to transforming polar equations to
rectangular ones, even if the rectangular equations cannot be easily graphed.
But the real problem persists, what does the graph of a polar equation look like? Aside
from symmetry many polar equations have graphs which are periodic. This is because the
most common functions to plug θ into are trigonometric. In such cases we are interested to
51
Section 13: Polar Coordinates
know the range of θ values we must run through before the graph begins to repeat itself. In
the example above, whereas sin 2θ has period π viewing it as a rectangular equation relating
r and θ, the polar point (sin 2θ, θ) is antipodal to the polar point (sin(2(θ + π)), θ + π), since
sin 2(θ + π) = sin(2θ + 2π) = sin 2θ. So in this case the period of the polar curve turns out to
be 2π.
Example 4: Sketch a graph of the polar equation r = sin 3θ.
Solution: One can check that this graph will have y-axis symmetry. To determine further
symmetries we plot r versus θ rectangularly.
1
−π
− 2π
3
− π3
π
3
2π
3
π
-1
From this we observe that the polar graph will be symmetric about θ = π/6, about θ = π/2
and about θ = 5π/6. Also we note that the period of the curve will be π since sin(3(θ + π)) =
sin(3θ+3π) = sin 3θ cos 3π+sin 3π cos 3θ = − sin 3θ and therefore (sin 3θ, θ) and (− sin 3θ, θ+π)
are the same point. Moreover the part of the polar curve about π/6 in shape will be the same
as the part about π/2 with negative radii, and will be identical to that about 5π/6, back to
positive radii.
52
Section 13: Polar Coordinates
θ=
π
3
θ=
π
6
1
A general analysis leads one to recognize that a curve with polar equation r = a cos 2kθ,
where k is a whole number, will have a graph which is a rose with 4k petals of length a, period
2π, and x-axis symmetry. The graph of r = a sin 2kθ, is virtually identical except that it has
y-axis symmetry. In contrast the graph of the curve with polar equation r = a cos(2l + 1)θ,
where l is a whole number, will be a rose with 2l + 1 petals of length a, each traversed twice
as θ runs from 0 to 2π. This graph therefore has period π and also has x-axis symmetry. So
rose curves might be analogous to quadratic equations in two variables, in that their behavior
is completely describable by the form of the equation after it is put in standard form.
Another class of polar curves with this property are the limaçons. A limaçon is a polar
curve with equation of the form r = a + b cos θ, or r = a + b sin θ, where a and b are constants.
The second class behave like the first except that they have y-axis symmetry instead of x-axis
symmetry.
For the first class there are three distinct possibilities: 1) a < b, 2) a = b, and 3) a > b.
In each case the curve has period 2π. In the first case r = 0 twice in each period, since r = 0
implies that cos θ = −a/b < −1. Which means θ = arccos(−a/b), and 2π − arccos(−a/b). Also
the sign of r changes at θ and again at 2π − θ. Consequently the graph has an inner loop. We
distinguish this case by saying that the curve is a limaçon with an inner loop. In the second
case r = 0 only once per period, when θ is an odd multiple of π. The result is what we called a
cardioid. In the third case r is never zero and the graph looks rather like an egg, or a blunted
ellipse.
53
Section 13: Polar Coordinates
a<b
a=b
a>b
Another application of polar coordinates is that they allow us to easily define and describe
nondegenerate conic sections. This is a two-step process.
First we define a conic as the set of points in the plane so that the distance from any
point to a fixed point F , called the focus, is e times the distance from the point to a fixed line
l, called the directrix. When e = 1 we have by definition a parabola. It turns out that e is
exactly the eccentricity of the conic as discussed in section 7.
We can then derive that if F is the pole, any conic has equation of the form
r=
ed
,
1 + e cos θ
r=
ed
,
1 − e cos θ
r=
ed
,
1 + e sin θ
or
r=
ed
1 − e sin θ
where d is the distance from F to l.
Exercises:
1. Plot the point with polar coordinates (5, 5π/6). b) Convert the polar coordinates to
rectangular coordinates. c) Give two other pairs of polar coordinates which represent the
same point, one pair with r > 0, the other with r < 0.
2. Convert the rectangular equation to a polar equation
a) y = x
b) x2 + y 2 = 4x
c) y = x2
3. Sketch and label a graph of each polar equation clearly indicating orientation.
a) r = 3 cos 3θ
b) r = −3 cos 3θ
c) r = 1 − 2 sin θ
d) r = −1 − 2 sin θ
e) r = 4 sin θ
4. Find a polar equation for the ellipse with focus at the pole, directrix with rectangular
1
equation x = 2 and eccentricity e = .
2
54
Section 14: Parametric Equations
The last topic for us to treat is parametric equations, so called because both rectangular
coordinates of a point on a curve are described by another quantity called the parameter. We
normally take our third variable to be t and think of t as denoting time. This is the situation
in many physical applications. A parametric curve is the set of all ordered pairs (x, y) where
x = g(t) and y = h(t), where g(t) and h(t) describe x and y respectively in terms of t.
An easy way to build a set of parametric equations is to start with y = f (x) and let
x = g(t) = t be the parameter so that y = f (t) = h(t). Any function can be parametrized in
this manner. There is a change in the graph of the curve in this case. As with polar curves,
every parametric curve is oriented. We place arrow marks on the curve to denote the motion
along the curve as the parameter t increases. Similarly any relation of the form x = k(y) can
be parametrized by y.
Every polar curve of the form r = f (θ) gives rise to a parametric curve parametrized
by θ. The parametric equations are x = r cos θ = f (θ) cos θ = g(θ), and y = r sin θ =
f (θ) sin θ = h(θ). This includes r = a, a circle with radius |a| centered at (0, 0). The standard
parametric equations for this curve are x = a cos θ and y = a sin θ. This works because
x2 + y 2 = a2 cos2 θ + a2 sin2 θ = a2 .
This construction generalizes to x = h + a cos t, y = k + b sin t which are parametric
equations for the ellipse (x − h)2 /a2 + (y − k)2 /b2 = 1. The sign of a and b don’t have to be
positive so there are really four possible sets of parametric equations of this form for this ellipse
x = h ± a cos t, and y = k ± b sin t. Since the crucial fact here is that sin2 t + cos2 t = 1, we can
also parametrize this ellipse via x = h ± a sin t, y = k ± b cos t. These eight sets of parametric
equations parametrize the same curve. They are distinct as is seen by considering which point
corresponds to (x(0), y(0)) and whether the curve is oriented positively or negatively.
If we tweak the Pythagorean identity to be in the form 1 = sec2 t − tan2 t, we can
parametrize the hyperbola (x − h)2 /a2 − (y − k)2 /b2 = 1 via x = h ± a sec t, and y =
k ± b tan t. In this case x = h ± a tan t and y = k ± b sec t parametrize the conjugate hyperbola (y − k)2 /b2 − (x − h)2 /a2 = 1.
e2t + 2 + e−2t
et − e−t
et + e−t
Define sinh t =
and cosh t =
, then cosh2 t =
, and
2
2
4
e2t − 2 + e−2t
sinh2 t =
. So cosh2 t − sinh2 t = 1. We can therefore also use these functions,
4
which are the hyperbolic sine and cosine functions, to parametrize a hyperbola.
55
Section 14: Parametric Equations
Since every parabola is either a function of the form y = k + (1/4c)(x − h)2 , or a relation
of the form x = h + (1/4c)(y − k)2 every parabola is parametrizable.
To summarize, every function is parametrizable, nearly every polar curve is parametrizable, and every conic section is parametrizable.
As we see with ellipses and hyperbolas, there are often many ways to parametrize a curve.
In fact if we have a set of parametric equations for a curve x = g(t) and y = h(t), then we can
reparametrize part or all of the curve by s where t = f (s), and f is a function of s. The new
parametric equations are x = g(t) = g(f (s)), and y = h(t) = h(f (s)).
Example 1: Let x = t and y = x2 = t2 be a set of parametric equations describing the parabola
y = x2 . If we set t = sin s, then x = t = sin s and y = t2 = sin2 s. Which means that the points
on the curve x = sin s, y = sin2 s are on the curve y = x2 . However not every point of y = x2
is on this parametric curve since −1 ≤ x = sin s ≤ 1, and 0 ≤ sin2 s = y ≤ 1. This example is
the first of a kind where we have the motion of the curve oscillating. If we want more or less
of the curve we can use x = c sin u, and y = c2 sin u. This parametric curve describes the path
of a particle that oscillates between (−c, c2 ) and (c, c2 ) along the path y = x2 .
12
10
8
6
4
2
-4
-2
2
4
x = 3 sin t, y = 9 sin2 t
Example 2: Let C denote the part of the ellipse x2 /4 + y 2 /9 = 1 in the first quadrant. We can
find sets of parametric equations for C in various ways.
First we can solve for y in terms of x, since this part of the ellipse satisfies the horizontal
line test. We get y 2 /9 = 1 − (x2 /4) = (4 − x2 )/4, so y 2 = (9/4)(4 − x2 ). Thus we can set x = t
√
and y = (3/2) 4 − t2 , where t ∈ [0, 2].
56
Section 14: Parametric Equations
We can reverse orientation by setting x = t = 2 − s. Now 4 − t2 = 4 − (2 − s)2 =
√
4 − (4 − 4s + s2 ) = 4s − s2 , so y = (3/2) 4s − s2 , and s ∈ [0, 2].
√
We can increase speed as it were by setting x = t = 2s, and y = 3 1 − s2 , where s ∈ [0, 1].
√
We can decrease speed by setting x = t = (1/2)s and y = (3/4) 16 − s2 , where s ∈ [0, 4].
p
√
We can call x = t = 2 sin θ, then y = (3/2) 4 − 4 sin2 θ = (3/2) 4 cos2 θ = 3 cos θ. Now
we take θ ∈ [0, π/2].
Example 3: Let C be the line segment between (1, −2) and (3, 6). The line through these points
has rectangular equation y = 4x − 6. We can therefore parametrize C by x = t, y = 4t − 6,
where t ∈ [1, 3].
If we set t = s + 1, then x = s + 1, and y = 4(s + 1) − 6 = 4s − 2, where s ∈ [0, 2].
If we then set s = 2u we get x = 2u + 1, and y = 8u − 2, where u ∈ [0, 1].
If we set u = 1 − v we reverse the orientation and have x = 3 − 2v, and y = 6 − 8v, where
v ∈ [0, 1].
If we have an object for which we know rectangular or polar equations, we can therefore
parametrize the curve fairly easily. One challenge may be to find the best parameterization
with respect to a given set of criteria. It might be instead that we want a parameterization
with a specific orientation, and with a specified domain for the parameter. The tricks above
work as long as we have at least one parameterization.
A problem that arises is that we won’t always be given the rectangular or polar equations
which a curve, or part of a curve satisfy. Instead we’ll begin with just a set of parametric
equations. In order to plot the graph of a set of parametric equations we can use a three
row/column table. The first row is labeled t and contains inputs. The second row consists of
the outputs x = g(t). The third column consists of the outputs y = h(t). The ordered pairs
(x, y) from the second and third columns belong on the graph. When we connect these dots
it is again important to orient the graph to indicate the behavior as t increases. The number
of values in the table varies from person to person. One should be careful to have a sufficient
number of values. Also the t-values should be carefully chosen so that, if possible, exact values
can be calculated for both x and y.
Even when one is careful mistakes can easily be made. Eliminating the parameter to find
an equation relating x and y is therefore a useful skill. We can then use what we know about
this relation to ensure we’ve got a proper graph.
57
Section 14: Parametric Equations
Example 4: Given the parametric equations x = tan θ + sec θ, and y = tan θ − sec θ, where
−π/2 < θ < π/2 let us sketch a graph of the curve. The first thing we might do is to compute
a three-row T-table where the input values of θ are ±π/6, ±π/4, ±π/3, and 0.
θ
−π/3
√
x(θ) 2 − √3
y(θ) −2 − 3
√−π/4
2−1
√
− 2−1
−π/6
√
3/3
√
− 3
0
1
−1
π/6
√
√3
− 3/3
√π/4
2+1
√
− 2+1
π/3√
2 + √3
−2 + 3
Plotting these points by hand and trying to connect the dots is dangerous. Much safer is to
realize that xy = tan2 θ − sec2 θ = −1 by a variation of the Pythagorean identity. So the points
of the parametric curve lie on the hyperbola xy = −1.
5
-5
5
-5
We may also need to remember that the parametric equations do not need to define the
whole rectangular curve.
√
Example 5: Given x = t and y = 1 − t, describe the curve parametrized.
Solution: To begin with the domain for the parameter is at most [0, ∞). Secondly we can solve
for t as a function of x to get t = x2 . So this set of parametric equations describes the right
half of the downward opening parabola y = 1 − x2 . The orientation has (0, 1) = (x(0), y(0))
and as t increases the point moves toward (∞, −∞).
Exercises:
1. Let x(t) = 1 + 4 sin t and y = −2 + 3 cos t. a) Find a rectangular equation for this relation.
b) Sketch and label a graph of the curve clearly indicating orientation.
2. Let x(t) = t + 2 and y(t) = 3t − 5. a) Find a rectangular equation for this relation. b)
Sketch and label a graph of the curve clearly indicating orientation.
58
Section 14: Parametric Equations
3. Find a set of parametric equations for the line through P = (1, −1) and Q = (−2, 4) so
that P = (x(0), y(0)) and Q = (x(1), y(1)).
4. Find a set of parametric equations for the line through P = (2, 1) and Q = (−2, 2) so that
Q = (x(1), y(1)) and P = (x(−1), y(−1)).
5. Find a set of parametric equations for the ellipse
(x(0), y(0)) and (0, −4) = (x(π/2), y(π/2)).
59
x2
(y + 1)2
+
= 1 so that (4, −1) =
16
9