2 Trigonometric Functions and Polar Co-ordinates
In this chapter we review trigonometry and periodic functions and discuss their relation to polar
co-ordinates. Some of this will be non-standard.
2.1 Definitions & Measuring Angles
Trigonometric functions date back at least 2000 years. Ancient mathemati-
cians were interested in the relationship between the chord of a circle and
the central angle, often for the purpose of astronomical measurement. It
wasn’t until 1595 that the term trigonometry (literally triangle measure) was
coined, and the functions were considered as coming from triangles.
θ
r
crd θ
Here are several related definitions of sine, cosine and tangent based either on triangles or circles.
Definition 2.1. 1. (a) Given a right triangle with hypotenuse (longest
side) 1 and angle θ, define sin θ and cos θ to be the side lengths
opposite and adjacent to θ.
Define tan θ =
sin θ
cos θ
to be the slope of the hypotenuse.
(b) Given a right triangle with angle θ, hypotenuse r, adjacent x and
opposite y, define
sin θ =
y
r
cos θ =
x
r
tan θ =
y
x
2. (a) ( cos θ, sin θ) are the co-ordinates of a point on the unit circle,
where θ is its polar angle measured counter-clockwise from the
positive x-axis. Provided cos θ = 0, also define tan θ =
sin θ
cos θ
.
(b) Repeat the definition for a circle of radius r with co-ordinates
(r cos θ, r sin θ).
θ
1
cos θ
sin θ
θ
r
x
y
y
x
θ
1
Discuss some of the advantages and weaknesses of these definitions:
• What prerequisites are you assuming in each case?
• Is it easier to think about lengths rather than ratios?
• Where do you need basic facts from Euclidean geometry such as congruent/similar triangles?
• Convince yourself that that the triangle definitions follow from the circle definitions. What is
missing if you try to use the triangle definition to justify the circle version?
• If you were introducing trigonometry for the first time, what would you use?
If you’ve done sufficient calculus you might know of other definitions, for instance using power
(Maclaurin) series. Plainly these are not suitable for grade-school, but have the great benefit of
making the calculus relationship
d
dθ
sin θ = cos θ very simple. Establishing this using the triangle
definition is a somewhat tricky!
21
Measuring Angles
There are two standard ways to measure angles (to sensibly associate a number to each angle).
Degrees A full revolution has 360° and a right-angle 90°. Degree measure dates back to ancient Baby-
lon 2–4000 years ago.
7
Radians The radian measure of an angle is the length of the arc subtending the angle in a circle of
radius 1. Since the circumference of a unit circle is 2π, we have the following identifications.
Degrees Radians sin θ cos θ tan θ
0° 0 0 1 0
30°
π
6
1
2
√
3
2
1
√
3
45°
π
4
1
√
2
1
√
2
1
60°
π
3
√
3
2
1
2
√
3
90°
π
2
1 0 n/a
180° π 0 −1 0
θ
1
0°
30°
60°
90°
120°
150°
180°
210°
240°
270°
300°
330°
π
6
π
3
π
2
2π
3
5π
6
π
7π
6
4π
3
3π
2
5π
3
11π
6
In elementary mathematics, degrees are the most common way to measure angles. Do you know any
other methods?
Exercises 2.1. 1. The identity cos
2
θ + sin
2
θ = 1 is the Pythagorean Theorem in disguise. Why?
2. The word sine is the result of a long list of translations and transliterations from an ancient
Sanskrit term meaning half-chord. For the chord picture on page 21, how does the length of the
chord crd θ relate to modern trigonometric functions?
3. It is conventional not to state units when using radians since they are effectively a ratio and
therefore unitless. Think this through: if the central angle in a circle of radius r is subtended by
an arc with arc-length ℓ, what is the radian measure of the angle? What facts from Euclidean
geometry justify this observation?
4. Explain how to get the values of sine and cosine in the above table.
(Hint: Draw some triangles and use Pythagoras!)
5. Using the pictures, explain why we have the relations
sin(
π
2
− θ) = cos θ = sin(θ +
π
2
), sin(−θ) = −sin θ, sin( π − θ) = sin θ
(You cannot use multiple-angle formulæ for this!)
7
It is not known why they chose 360, but it fits nicely with their base-60 system of counting (decimals are base-10).
The traditional subdivisions of a degree are also base-60. For instance, 34°12
′
45” is 34 degrees, 12 (arc)minutes and 45
(arc)seconds; converted to decimal notation, this becomes
34°12
′
45” = 34 +
12
60
+
45
60
2
= 34.2125°
The standard hour-minute-second measurement of time has the same origin.
22
2.2 Periodicity, Graphs & Inverses
One advantage of the circle definition is that it makes sketching the graphs of sine and cosine very
easy. Simply draw axes next to a unit circle and transfer the heights across.
−1
0
1
y = sin x
x
1
x
π
2
π
3π
2
2π
By Exercise 2.1.5, the graph of cos x = sin(x +
π
2
) is simply that of sine shifted
π
2
= 90° to the left.
Moreover, the circle definition allows us easily to extend trigonometric functions periodically since we
can measure the polar angle by looping as many times round the origin as we like: for any integer n,
sin(θ + 2nπ) = sin θ, cos(θ + 2nπ) = cos θ
Otherwise said, sine and cosine have period 2π radians (360°).
Sine and cosine are non-invertible unless we choose a domain on which they are 1–1.
−1
1
y
x
π
2
−
π
2
π−π
3π
2
5π
2
2π
y = sin x
f ( x) = sin x is 1–1 on the domain [−
π
2
,
π
2
]
Inverse function f
−1
(x) = arcsin x = sin
−1
x
Domain dom( arcsin) = [−1, 1] = range(sin)
Range range( arcsin) = [−
π
2
,
π
2
] = dom(sin)
This is why your calculator always returns a value in the
interval [−
π
2
,
π
2
] = [−90°, 90°] when you hit the sin
−1
button.
−1
1
y
−1
1
x
y = sin x
y = sin
−1
x
π
2
−
π
2
π
2
−
π
2
Example 2.2. If you know the graphs, then symmetry and periodicity help you solve equations. For
example, if sin θ =
9
10
then all solutions are given by
θ = sin
−1
9
10
+ 2πn or π − sin
−1
9
10
+ 2πn (n is any integer)
−1
1
y
θ
sin
−1
9
10
π
2
−
π
2
π−π
3π
2
−
3π
2
−2π
π − sin
−1
9
10
Alternatively, we could use the circle definition directly: sin θ =
9
10
means we want angles θ corre-
sponding to the intersections of the unit circle with the horizontal line y =
9
10
.
23
Periodic Models Trig functions find applications in modeling precisely because they are periodic.
In general, a function has period T if
f ( x + T) = f (x) for all x
It is easy to find the period of the function f (x) = sin kx just by considering what we have to add to
the input x to increase the argument kx of sine by 2π:
T =
2π
k
=⇒ f (x + T) = sin(kx + 2π) = sin kx = f (x)
We may therefore obtain a simple periodic model regardless of what period is required.
Example 2.3. On a given day, high tide occurs at 2:00 with a water depth of 10 ft, whereas low
tide occurs at 8:12 with a depth of 4 ft. We might model this using a periodic function with period
T = 2 ×6
12
60
=
62
5
hours. For instance
h(t) = 7 + 3 cos
5π
31
(t −4)
where t is measured in hours from midnight might be suitable. In reality, tidal height is very close to
being periodic, but the magnitude of the high and low tides are somewhat variable.
In fact any periodic function may be approximated using trigonometric functions. Indeed if f (x) has
period T and we define constants
a
n
=
Z
T
2
−
T
2
f ( x) cos
2πnx
T
dx, b
n
=
Z
T
2
−
T
2
f ( x) sin
2πnx
T
dx (∗)
then
f ( x) ≈
a
0
2
+ a
1
cos
2πx
T
+ b
1
sin
2πx
T
+ a
2
cos
4πx
T
+ b
2
sin
4πx
T
+ ··· (†)
This is the Fourier series of f (x). It often takes only a small number of terms to obtain a very good
approximation. Modern data-compression algorithms often employ Fourier series. Given a periodic
function f (x), one uses a computer to estimate (say) the first 100 Fourier coefficients (∗) and transmits
these values to the receiver, who recovers an approximation to the original function using (†).
Example 2.4. A square-wave function with period T = 2π is given by
f ( x) =
(
1 if 0 ≤ x < π
−1 if π ≤ x < 2π
extended periodically to the real line. With a little calculus,
it is easily checked that the Fourier coefficients are
−1
1
f (x)
x
π 2π
a
n
=
Z
π
−π
f ( x) cos nx dx = 0, b
n
=
Z
π
−π
f ( x) sin nx dx =
(
4
πn
if n is odd
0 if n is even
Use a graphics tool to see how the first few terms of the series approximate the function.
24
Exercises 2.2. 1. f (x) = sin x is also 1–1 on the interval [
π
2
,
3π
2
]. Sketch the graph of its corresponding
inverse function.
2. Draw the graph for cosine and observe that it is invertible if we restrict the domain to the
interval [0, π]. Draw the graph of cos
−1
.
3. Describe all solutions to the equation cos x = −0.2.
4. Explain why the tangent function has period π; that is tan(θ + nπ) = tan θ. What facts are we
using about sine and cosine and why are they obvious from the definition?
5. Describe all solutions to the equation tan x = 5.
6. (a) Suppose θ = cos
−1
9
41
. Find the exact values for sin θ and tan θ.
(b) What changes if θ = cos
−1
−9
41
?
7. Let f (x) = csc x =
1
sin x
be the cosecant function. Describe a domain on which this function is
1–1 and sketch the graph of its inverse y = f
−1
(x).
8. Use a computer to sketch the curve
y = 2
sin x −
1
2
sin 2x +
1
3
sin 3x −
1
4
sin 4x +
1
5
sin 5x
What simple periodic function do you think this is approximating?
25
2.3 Solving Triangles
Basic trigonometry often involves finding the edges and angles of a triangle given partial data.
Example 2.5. To find the height h of a tall tree, two angles of elevation 45° and 30° are measured a
distance 20 ft apart along a straight line from the base of the trunk.
This is easily attacked by drawing a picture and observing that we have two right-triangles. If the
(unknown) distance from the base of the tree to the nearer measurement is x, then
1
√
3
= tan 30° =
h
x + 20
1 = tan 45° =
h
x
Substituting the second equation into the first returns
h =
20
√
3 −1
≈ 27.32 ft
20
x
45°
30°
In fact there is enough data in the problem to recover everything about the original triangle.
• The second base angle is (180° −45° = 135°).
• The third (summit) angle is 180° − 30° −135° = 15°.
• Two applications of Pythagoras compute the remaining sides of the triangle
p
x
2
+ h
2
=
√
2h =
20
√
2
√
3 −1
≈ 38.64
q
h
2
+ (x + 20)
2
=
p
h
2
+ 3h
2
= 2h =
40
√
3 −1
≈ 54.64
The example is just a disguised version of solving a triangle: computing
all six sides and angles of a triangle given three of them. The Euclidean
triangle congruence theorems tell us which combinations are sufficient
to determine all the others. The example is the ASA congruence: angle-
side-angle data (30°–20–135°) is enough to compute everything else
about the triangle.
c
a
b
A
B
C
When in doubt, you can always attack basic trigonometry problems as we did in the example: create
a right-triangle, then use the definitions of sin/cos/tan and/or Pythagoras.
Example 2.6. Given the SAS (side-angle-side) combination 5–60°–9, find the third side of the triangle.
The altitude h creates two right-triangles, from which
h = 5 sin 60°, x = 5 cos 60°
=⇒ c
2
= (9 −x)
2
+ h
2
= 9
2
+ (x
2
+ h
2
) −18x
= 9
2
+ 5
2
−18 · 5 cos 60° = 61
=⇒ c =
√
61 ≈ 7.81
h
x
c
5
9
60°
Since we now know c and 9 − x =
13
2
the remaining angles could also be easily found.
26
In elementary situations it is typically easier to have students drop the perpendicular as we’ve done.
However, once comfortable with the method, it is helpful to have short-cuts which skip the need to
work with the perpendicular at all.
Theorem 2.7 (Sine and Cosine Rules). For any triangle,
sin A
a
=
sin B
b
=
sin C
c
and c
2
= a
2
+ b
2
−2ab cos C
The cosine rule is just the Pythagorean Theorem with a correction for non-right triangles. Both rules
follow straightforwardly by drawing an altitude as before!
Proof. Consider the picture. We have
h = a sin C = c sin A, x = a cos C, b − x = c cos A
The first equation rearranges to
sin A
a
=
sin C
c
Two applications of Pythagoras give the cosine rule
c
a
b
A
B
C
x
h
c
2
= h
2
+ (b − x)
2
= h
2
+ x
2
+ b
2
−2bx = a
2
+ b
2
−2ab cos C
The remaining part of the sine rule and the other versions of the cosine rule are obtained by choosing
other altitudes.
Here are two examples where we use the rules instead of explicitly drawing an altitude.
Examples 2.8. 1. A triangle has sides 2 and
√
3 − 1, and the angle between them is 120°. Find the
remaining sides and angles.
We apply the cosine rule with a = 2, b =
√
3 −1 and C = 120°
c
2
= a
2
+ b
2
−2ab cos C
= 2
2
+ (
√
3 −1)
2
−2 · 2(
√
3 −1) cos 120°
= 4 + 3 + 1 −2
√
3 + 2(
√
3 −1) = 6
We have an opposite pair (c, C) = (
√
6, 120°), so the sine rule may be used
sin A =
2
√
6
sin 120° =
2
√
3
2
√
6
=
1
√
2
=⇒ A = 45°
We chose the acute angle since A = 180° − B − C = 60° − B < 90°.
The final angle is then B = 180° −45° − 120° = 15°.
c
2
√
3 −1
A
B
120°
You could instead drop a perpendicular, say from the vertex A to the extension of the side of
length 2. Think about why the perpendicular has to be outside the triangle. . .
27
2. A triangle has one side with length 5 and its two adjacent angles are 40° and 65°. Find the
remaining data.
This time the initial data is ASA. Writing c = 5, A = 40° and
B = 65°, the remaining angle is plainly
C = 180° −40° −65° = 75°
This gives us an opposite pair (c, C), so we can apply the sine rule
a = c
sin A
sin C
= 5
sin 40°
sin 75°
≈ 3.327
A second application yields
b = c
sin B
sin c
= 5
sin 65°
sin 75°
≈ 4.691
5
a
b
40°
65°
C
3. (Discuss: courtesy of an 8 year-old contributor) Model Earth as a sphere of radius 3963 miles.
If identical vertical ladders are placed in Irvine, CA and Irvine, Scotland, 5145 miles apart by
great circle, how tall would they have to be for people at the top to ‘see’ each other?
Multiple-angle Formulae
Also useful in the context of basic trigonometry is the ability
to sum angles. The picture provides a simple justification of
sin(α + β) = sin α cos β + cos α sin β
at least when 0 < α + β <
π
2
. If you look carefully, you
should be able to see how the same picture establishes
cos(α + β) = cos α cos β − sin α sin β
α
β
1
α
α
cos β
sin β
sin α cos β
cos α sin β
Exercises 2.3. 1. Find the remaining angles in the triangle in Example 2.6.
2. The other Euclidean congruence theorems are SSS and SAA. Explain how to solve triangles
using these minimal data in two ways:
(a) By drawing an altitude. (b) Using the sine/cosine rules.
3. SSA isn’t a triangle congruence theorem. For instance, there are two non-congruent triangles
with data a = 1, b =
√
3 and A = 30°. Find them.
4. Use the multiple-angle formulae to derive the familiar expressions for sin 2θ and cos 2θ.
5. Find the exact value of sin 105°.
6. (a) Find an expression for tan(α + β) purely in terms of tan α and tan β.
(b) Two wooden wedges with slope
1
4
are placed on top of each other to make a steeper slope.
What is the gradient of the new slope?
7. Carefully explain why the answer to Example 2.8.3 is approximately 1012 miles.
28
2.4 Polar Co-ordinates
Definition 2.1 provides an alternative way to describe points in the plane. If θ is the polar angle of
a point with Cartesian (rectangular) co-ordinates (x, y) , then its polar-coordinates are precisely the
values (r, θ) seen in the definition!
Computing x = r cos θ and y = r sin θ is easy given r and θ.
Example 2.9. A point with polar co-ordinates (r, θ) = (2,
5π
6
) has Cartesian co-ordinates
(x, y) =
2 cos
5π
6
, 2 sin
5π
6
=
−
√
3, 1
Computing polar co-ordinates from Cartesian is harder, requiring some visualization.
1. Every point (x, y) has a unique radius r =
p
x
2
+ y
2
, but not polar angle. If θ is a polar angle,
so is θ + 2πn for any integer n ∈ Z. The origin (x, y) = (0, 0) is even stranger; certainly r = 0,
but any θ is a legitimate polar angle!
2. Whenever x = 0 (away from the y-axis),
(
x = r cos θ
y = r sin θ
=⇒ tan θ =
y
x
however, this doesn’t guarantee that θ = tan
−1
y
x
. Continuing the example shows us why. . .
Example (2.9, cont). If (x, y) = (−
√
3, 1), then the radius is easy
r =
q
(
√
3)
2
+ 1
2
= 2
For the polar angle,
tan θ =
y
x
= −
1
√
3
= tan
−
π
6
=⇒ θ = −
π
6
Arctan has range (−
π
2
,
π
2
), so always returns an angle in quad-
rants 1 or 4. Our point is in the second quadrant (x < 0 < y ) so we
need to adjust, using the fact that tan is π-periodic:
θ = π −
π
6
=
5π
6
= 150°
We could alternatively add any integer multiple of 2π.
π
6
−
5π
6
π
3
−
2π
3
2π
3
−
π
3
5π
6
−
π
6
√
3−
√
3
1
−1
The example wasn’t too tricky since the polar angle was exactly computable. When you have to rely
on a calculator, it is much easier to make a mistake.
Example 2.10. The point (x, y) = (−8, −15) has polar co-ordinates (quadrant 3!)
r =
p
8
2
+ 15
2
= 17, θ = π + tan
−1
15
8
≈ 241.93°
We could summarize with formulæ describing precisely how to compute θ dependent on quadrant
(the signs of x, y), though it is better to get in to the habit of drawing a picture!
29
Curves in Polar Co-ordinates
Polar co-ordinates are well-suited to describing curves that encircle the origin. Indeed circles centered
at the origin with radius a > 0 have the very simple polar form r = a. Converting to rectangular
co-ordinates recovers the the natural parametrization of a circle:
x(θ) = a cos θ, y(θ) = a sin θ
This partly explains why mathematicians call sine and cosine circular functions.
General polar graphs are harder to visualize, though the major reason is lack of familiarity. Have
a little empathy: to graph polar functions, you’ll likely have to follow the same approach as new
students use to sketch Cartesian curves like y = x
2
! Here are a couple of examples.
Examples 2.11. 1. The curve r = θ is relatively easy to plot since r increases at exactly the same rate
as the angle; we therefore have a spiral.
To confirm this, plot several points (θ, θ); we’ve done for θ in multiples of
π
6
(30°) from 0 to 2π.
It is sensible to use ‘polar graph paper’ with concentric circles separated by (say)
π
2
≈ 1.57.
−6
−3
3
6
−6 −3 3 6
π
6
7π
6
π
3
4π
3
2π
3
5π
3
5π
6
11π
6
x
y
r = θ, 0 ≤ θ ≤ 2π
1
2
−2 −1 0 1 2
π
6
π
3
2π
3
5π
6
x
y
r = 2 sin θ, 0 ≤ θ ≤ π
θ
π
6
2π
3
π
2
4π
3
5π
6
π
2 sin θ 1 1.73 1 1.73 1 0
2. The curve r = 2 sin θ is a little easier to work with since we know exact values for sine, assisted
by
√
3 ≈ 1.73.
This looks like a circle! To see this, multiply both sides by r and complete the square:
r
2
= 2r sin θ =⇒ x
2
+ y
2
= 2y =⇒ x
2
+ (y −1)
2
= 1
describes the set of points with distance 1 from the point (0, 1): a circle!
You should think about what happens in both examples if we extend the domain:
• What would r = θ look like if θ were allowed to be negative?
• What happens to r = 2 sin θ when θ > π?
30
Exercises 2.4. 1. Convert the following points to polar co-ordinates.
(a) (−5, 5) (b) (3, −4) (c) (−5
√
3, −15)
(d) (−1, tan 3) (tricky—this is 3 radians!)
2. If a > 0, describe the curve with polar equation r = 2a cos θ.
(Be careful with θ >
π
2
since cosine goes negative. . . )
3. The algebraic trickery in the last example sometimes bears fruit, though you have to be lucky!
By multiplying both sides by 1 −sin θ and converting to rectangular co-ordinates, show that
the polar function
r(θ) =
2a
1 −sin θ
is a parabola in disguise and sketch it when a = 1. How does the graph depend on a?
4. In a similar vein to the previous question, sketch the curve r =
2
2+sin θ
. What type of curve is
this?
5. Try to sketch the following curves.
(a) r = θ(θ − 4) (b) r = (θ −1)
2
+ 1 (c) r = (θ −1)
2
−1
As well as plotting points directly, you should sketch the curve first on rectangular axes (e.g.,
(a) is y = x(x − 4)). What happens to (c) when θ = 1?
Once you’ve tried these, use a grapher to see if you’re right, though see how close you can get
without it!
31