Trigonometry:
A Crash Review
An in-depth
study of Basic trigonemtry
written
by
Kenneth
Boyd
Unfortunately,
for much of this subject, there is no substitute for brute memorization.
This course is often used as pre-calculus. "Algebra masters"
will have a much smaller memorization load than other students.
[I mean College Algebra, not the higher-math version.] Much
of this subject is highly dependent on either calculators+ or
tables [slide rules have been passé since the electronic calculator,
but could be used.] Traditionally, this subject relies on some
familiarity with "College Algebra", but not an intensive mastery.
Note: in
many places, I will mention algebraic identities. For serious
students who are either algebra-deficient or have to use the
material under time pressure, I highly recommend working through
"all of the plausible forms" and memorizing them. This usually
means algebraic isolation of various variables in the equations,
and figuring out how to use multiple equations in one problem.
Currently, I am not providing the homework that is required
to learn the material. I would like to partially remedy this
deficiency. [No timetables!]
Note: There
are several kludge notations lurking in this crash review:
-
SQRT
for "square root" (correctly
notated for 2, 3, 5, and 6)
-
pi
for p (correctly notated outside
of popup text for pictures)
I will remove
instances of these kludge notations, without advance warning,
when it is convenient in my schedule to do so.

Trigonometry
Survival 101 -- basic material [not coordinate geometry]
What
are these trigonometric [trig] functions, anyway?
What
does a negative angle (measure) mean?
Why
should I know the generic right triangle?
What
are the reference triangles?
Why
should I know the unit circle?
How
does "wrapping the real number line" around the unit circle
work?
What
are the triangle area formulae?
What
is the law of sines?
What
is the law of cosines?
What
are the angle sum, difference, and/or doubling formulae?
What
are the angle halving formulae?
What
are the trig(onometric) sum and difference formulae?
What
are the trig(onometric) product formulae?

Trigonometry
Survival 201 -- useful material, but not in your standard Trigonometry
course
Reference values -- 15° and 75° i.e. p/12
and 5p/12 radians i.e. 100/6 and 500/6 gradients
Reference
values -- 18°, 36°, 54°, & 72° i.e. p/10, p/5, 3p/10, &
2p/5 radians i.e. 20, 40, 60, & 80 gradients
Instant
numerical trigonometric tables for sin, cos, and tan near 0

Formula
Lists
Angle
classifications
Angle
relations
Arc
length from central angle and radius
Cofunction
relations
Supplementary angle
relations
Algebraically
even and odd trigonometric functions
Basic
reference tables
Trigonometric
function evaluation from sides of a right triangle
Trigonometric
function signs from quadrants of the unit circle
Trigonometric
functions that are multiplicative inverses of each other
Trigonometric
functions in terms of sin(X) and cos(X)
Pythagorean
Theorem
Pythagorean
Identities

A
Vastly Incomplete Selection of Web Resources
Trig FAQ, J. David Eisenberg
Short
Course in Trigonometry, David E. Joyce
Introduction
to Trigonometry, Johan. Claeys [and a possible mirror]
Physics
Dept., University of Guelph
Trigonometry
Overview, Kenny Felder
Basic
Trigonometry, Purdue
Web
Graphing Calculator
Web
evaluator
for sin, cos, tan, and arctan
What
are these trigonometric [trig] functions anyway?
Their names
are:
These functions
have a single argument, and take angles. By the heuristic of
"wrapping the real number line around the unit circle," we can
think of these functions as taking real numbers as arguments.
Note that in "stratosphere" math, the above approach is very
painful; other techniques are used to define these functions.
These techniques are the ones which calculators are programmed
with -- but they're not appropriate for this section.
Incidentally,
there are three angle units listed on my HP-48SX: degrees, radians,
and gradients. [Gradients are not commonly used in the U.S.A.]
To rebuild the conversion formulae, remember:
1 circle
= 360 degrees = 2p radians = 400 gradients.
-
We inherited
degrees from Babylonia. There are 60 minutes in a degree,
and 60 seconds in a minute. [When dealing with the Earth's
rotation in astronomy, the Earth rotates "about" 1 angular
degree in 4 solar-day minutes. The Earth rotates 1 degree
in 4 sidereal-day minutes.] Also, a nautical mile is "about"
1 arc-minute on the Earth's surface. This would be exact
if the Earth was a perfect sphere. A common notation for,
say, 3 degrees 4 minutes 5 seconds is 3°4'5".
-
Radians
are the "natural" angle unit when dealing with "stratosphere"
math. The defining formulae for sin and cos are much simpler
then. Radians also behave better if you have to deal with
calculus [deriviatives and integrals].
-
I don't
know much about the history of gradients.
A right
angle is 1/4th of a circle; i.e. 90°, p/2 radians, or 100 gradients.
I will use the common symbol
for a right angle, in the rest of this crash review. An angle
larger than 0, but smaller than a
, is called an acute
angle. An angle larger than a
, but smaller than half a circle, is called an obtuse angle.
is a standard symbol in the literature (although found more
often in geometry texts than trigonometry texts). When it shows
up in a diagram, it designates the labeled angle as a right
angle.
For the
record: Triangles are constructed by picking three points, then
joining each pair by a straight line segment. We assume we can
neglect curvature of the surface. That is, we assume the geometry
is Euclidean. [You will not get the same kind of triangles on
a sphere as on a flat desk or piece of paper. The sphere is
visibly curved, the desk and the flat piece of paper are not
supposed to be curved.] I will occasionally use the fact the
geometry is Euclidean.
The sum
of the three (interior) angles of a triangle is half a circle,
i.e. 180° or p radians or 200 gradients. I will sometimes denote
this by 2
in the rest of this crash review; this is *not* standard notation.
An implicit exercise will be translating 2
to angular measure [degrees, radians, and/or gradients.].
If we pick
a point on a straight line, the two parts of the straight line
formed by taking out the point form an angle of half a circle,
i.e. 180° i.e. p radians i.e. 200 gradients. In general, I will
omit "interior" when talking about angles of a triangle.
We say a
triangle, that contains a right angle, is a right triangle.
If a triangle has two sides of equal length and one side of
different length, we call it an isosceles triangle. If a triangle
has three equal sides, we say it is an equilateral triangle.
Note:
-
For
an isosceles triangle, the two angles opposite the sides
with equal length (i.e., the two angles whose bounding sides
have one of the two sides of equal length, and the side
with different length) are equal.
-
For
an equilateral triangle, all three angles are equal, and
equal 60° i.e. p/3 radians i.e. 200/3 gradients.
The angle
measure in radians is simply arc length on a circle of radius
1. In general, for a circle with radius R and an arc with central
angle X in radians,
There are
two pictures to keep in mind here, the generic right triangle:
and the
unit circle:
Here, A,
H, and O are acronyms for Adjacent [to angle], Hypotenuse, and
Opposite [from angle], respectively.

What
does a negative angle (measure) mean?
First of
all, all physical angles have some size. We cannot visualize
an angle with negative physical size. They do not exist in anything
sufficiently similar to the physical space we live in.
However,
(especially when dealing with the unit circle), it is often
convenient to measure angles in a specific direction: counterclockwise.
[I'm not going to digress into "stratosphere" math now, but
that is dictating the convention here.]
Then, a
negative sign means we are measuring the angle "unconventionally",
i.e. clockwise.
This will
simplify the use of some of the trigonometric identities we
are going to look at.

Why
should I know the generic right triangle?

Given
a right triangle, the trig function values for the two acute
angles [angles smaller than a right angle] can be computed without
knowing the angles. I prefer to remember the formulae this way
[X is an angle]:
In all cases,
division by zero should be interpreted as "undefined".
If H=1,
then O and A are simply cos(X) and sin(X).
A mnemonic
for the formulae for sin(X), cos(X), and tan(X) is the (fictitious)
Indian Chief Soh-cah-toa, who had no problems with this part
of trigonometry. [I got this from Mr. Coole, a long time ago
-- I was in grade school then.]
The way
I wrote the formulae, above, emphasizes the following identities:
These identities
do work hold when both functions involved are defined for the
angle X, regardless of size.
Another
fact, of some use, is the Pythagorean theorem: H²=A²+O². If
we relabel A as a, O as b, and H as c, we get the familiar form
of the Pythagorean theorem: c²=a²+b². This *only* works for
right triangles. The generalization of the Pythagorean theorem to non-right
triangles is called the law of cosines.
A common
strategy is to memorize how to compute tan(X), cot(X), sec(X),
and csc(X) in terms of sin(X) and cos(X), and then reduce everything
to this. This is not necessarily the least painful way to do
a trig problem, but it is often more important to start the
problem, than figure out how to do it elegantly. [EXERCISE:
derive this from the A, H, O formulation for acute angles. The
formulae work for arbitrary angles.]
The generic
right triangle also motivates some terminology [which we inherited
from twelfth century Arabian mathematics].
First of
all, we say two angles are complementary if they
add up to a right angle. That is, for an angle X, we say -X
is complementary to X. For instance,

the angle
in the upper right-hand corner is complementary to, i.e. the
complement of, the marked angle in the lower left-hand corner.
A related
piece of terminology is supplementary angles:
two angles are said to be supplementary if they add up to half
a circle, i.e. 180° i.e. p radians i.e. 200 gradients. That
is, 2
- X is supplementary to the angle X. To get an intuition for
this, draw a unit circle with the horizontal axis.. Draw an
arbitrary radius from the center to somewhere on the unit circle.
The two angles formed between the radius, and the horizontal
axis, are supplementary.
Now, let's
look at the trig function-computation rules again:
What happens
if we swap A and O in these formulae? This is computing the
trig function of the complement of X, i.e. -X.
Go ahead, do it :) It's easy.
What you
should get are the cofunction identities:
-
sin( -X)=cos(X)
-
cos( -X)=sin(X)
-
tan( -X)=cot(X)
-
cot( -X)=tan(X)
-
sec( -X)=csc(X)
-
csc( -X)=sec(X)
We say that
sin and cos [sine and cosine], tan and cot [tangent and cotangent],
and sec and csc [secant and cosecant] are cofunctions, and that
the trig function of the complement of an angle X is equal to
the trig cofunction of the angle. This is explicitly in the
function names: cosine is the cofunction of sine, cotangent
is the cofunction of tangent, and cosecant is the cofunction
of secant.
These identities
do work for arbitrary angles. If one side is undefined, both
sides are undefined.

What
are the reference triangles?
The reference
triangles are right triangles that are "easy to construct".
They provide easily-memorized values for the angles with measure
30°, 45°, and 60°, i.e. p/6, p/4, and p/3 radians, i.e. 100/3,
50, and 200/3 gradients.
There are
two reference triangles. They [in the U.S.A.] are known by their
degree names:
The 45-45-90
(degree) right triangle can be constructed from a square with
sides of length one. The hypotenuse is the line connecting two
opposite vertices of the square; it has length by the
Pythagorean Theorem.
[Exercise: compute this!] The legs of the triangle are sides
of the square. Note: 45° is its own complementary angle.
When we
put A=1=O and H=
into the generic right triangle, we get:
[Do it.]
The 30-60-90
triangle is constructed from an equilateral triangle with sides
of length 2. We put one side on the horizontal axis, and bisect
the angle opposite this side. [Note that an equilateral triangle
has three vertex angles, all of which are 60°.]

We now
have a hypotenuse of length 2, one leg of length 1 [from bisecting
the horizontal side], and one side with length [from
the bisecting line].
We get:
[Again,
do it.]

Why
should I know the unit circle?
The unit
circle provides a picture on which to memorize reference values
of the trig functions.

Think of
it as the set of possible endpoints for a length 1 hypotenuse
H, with one endpoint of H fixed at the origin. We can construct
generic right triangles with hypotenuse 1 in it. Pick a point
on the circumference, draw a line segment from it to the origin,
and then draw a perpendicular line segment down to the x-axis.

Notice that
the coordinates of the vertex, in Cartesian coordinates, is
(cos(X), sin(X)), where X is the central angle. The horizontal
side (on the x-axis) is A, and the vertical side (parallel to
the y-axis) is O. The radius (length 1) is H. The slope of the
hypotenuse H is tan(X).
However,
observe the four quadrants. Our example triangle has its hypotenuse
in the upper-right quadrant [both coordinates positive]. Horizontal
and vertical hypotenuses create line segments rather than triangles.
Also, at least one of the coordinates go negative in the other
three quadrants [upper-left, lower-left, lower-right].
We proceed
by assuming that the coordinates of the vertex, in Cartesian
coordinates, is (cos(X), sin(X)), regardless of where the vertex
is on the unit circle. This immediately leads us to the Pythagorean
identities:
If we interpret
undefined as equal to undefined, these identities hold for arbitrary
angles.
Incidentally,
the computation of the slope of the hypotenuse H via tan(X)
also works for arbitrary angles. [Recall that an undefined slope
corresponds to a vertical line].
The next
point is that the signs of the various trigonometric functions
are controlled by the quadrant the function is evaluated in.
We number the quadrants I through IV [Roman numerals] counterclockwise,
as follows:

Now, the
trigonometric function signs are controlled by the quadrants
as follows:
As should
be clear, the boundaries between the quadrants do not behave
this way. In fact, they are reference values for the trigonometric
functions. [For ease of memorization, I will put all of the
reference values for the first quadrant and its boundaries in
one place, elsewhere in this document.]
-
Boundary
between Quadrants I, IV: X is 0° i.e. 0 radians i.e. 0 gradients;
point (1,0)
-
Boundary
between Quadrants I, II:
i.e. 90° i.e. p/2 radians i.e. 100 gradients
-
Boundary
between Quadrants II, III: 2
i.e. 180° i.e. p radians i.e. 200 gradients
-
Boundary
between Quadrants III, IV: 3
i.e. 270° i.e. 3p/2 radians i.e. 300 gradients
Now, let's
look at another way to reconstruct the cofunction identities
i.e. the identities relating complementary angles.

What happens
if we reflect the unit circle about the line y=x? We are swapping
the x and y coordinates. That is, (letting X be the central
angle), we are swapping sin(X) and cos(X). We are also physically
reflecting the central angle X to the angle -X.
So, we find that (here are the cofunction identities again):
It is no
coincidence that the slope of the line y=x is 1. This gives
a central angle between the line y=x, and the x-axis, of 45°
i.e. p/4 i.e. 50 gradients [think of the 45-45-90 triangle];
note that this is exactly half of a right angle.
Now, let
us consider reflecting the unit circle about the x-axis:

Note that
the x-coordinate [cos(X)] is unaffected, while the y-coordinate
[sin(X)] is negated. The resulting point is the point we get
from rotating through the angle -X from the angle 0. This gives
the following formulae:
[Exercise:
derive the last four [tan, cot, sec, and csc versions] from
the first two [sin and cos versions.]
These formulae
classify the trigonometric functions into even functions and
odd functions. As you may recall from College Algebra, even
functions are those functions whose value is unchanged by negating
the argument, and odd functions are those functions whose value
is negated by negating the argument.
It is no
coincidence that the slope of the x-axis is 0. The central angle
of the x-axis with the x-axis is clearly 0 [degrees, radians,
gradients, it matters not which unit]. 0 is also exactly half
of 0.
Now, let
us consider reflecting the unit circle about the y-axis:

Note that
the y-coordinate [sin(X)] is unaffected, while the x-coordinate
[cos(X)] is negated. The resulting point is the point we get
from rotating through the angle -X from the angle 0. This gives
the following formulae:
[Exercise:
derive the last four (tan, cot, sec, and csc versions) from
the first two (sin and cos versions.)]
It is no
coincidence that the slope of the y-axis is undefined. The central
angle of the y-axis with the x-axis is clearly .
This is also exactly half of 2 .
Now, we
can directly compute the reference values of cos and sin for
45° i.e. p/4 radians i.e. 50 gradients, i.e. ½
, without using the 45-45-90 reference triangle. 45° is the
angle that is its own complementary angle. Thus,
Substitute
the above into the Pythagorean identity
to get
i.e. [solve
for cos(45°); we know we need the positive root because 45°
is in the first quadrant, so I can omit ±]
-
cos(45°)
= 1/ =
/2
We can also
directly compute the reference values of cos and sin for 60°,
i.e. p/3 radians, i.e. 200/3 gradients, i.e. [2/3]
. To do this, we first consider inscribing an equilateral hexagon
in the unit circle:

[Just
like an equilateral triangle, an equilateral hexagon has equal
lengths for all of its sides. In general, an equilateral polygon
(regardless of how many sides it has) has equal lengths for
all of its sides.]
Note that
the inscribed hexagon can be considered as the 'splicing together'
of six equilateral triangles.
[If we draw
the line segments from the vertices to the center of the inscribed
hexagon, since the sides have the same length, the angles as
viewed from the center will have the same size. 360/6° = 60°.
(The prior two sentences use Euclidean geometry). We know that
the triangle we just created is isosceles, (two of its sides
are radii), so the two angles opposite the radii have the same
size, and add up to 120°: they are both 60°. All three angles
are equal. Thus all three sides of the triangle have equal length.
(The very last sentence also uses Euclidean geometry)]

Now, observe
that the horizontal side "above the x-axis" is bisected by the
y-axis. Thus, the length of this side in the first quadrant
of the unit circle is ½ [the triangle is equilateral, so the
side being bisected has the length of the radius, i.e. is length
1]. That is, cos(60°)=½.
[The y-axis
has an angle of 90° i.e. p/2 radians i.e. 100 gradients. The
angle created by drawing line segments from the vertices of
this side, to the center, starts at 60° and ends at 120°. So,
the y-axis is 90° - 60° = 30° into the above angle, which means
the y-axis bisects this angle.]
From the
Pythagorean
identity, we find
-
(½)²+sin²(60°)=1,
i.e.
-
¼+sin²(60°)=1
i.e. [solve
for sin(60°); we know we need the positive root because 60°
is in the first quadrant, so I can omit ±]
-
sin(60°)
= /2
Since 30°
is the complementary angle to 60°, we also have computed sin(30°)
and cos(30°).

How
does "wrapping the real number line around the unit circle"
work?
The unit
circle has a circumference of 2p. "Thus", all trig functions
will have the same value when evaluated 2p radians apart. We
say that all trig functions have a period. [In "stratosphere"
math, this period is arrived at in a very different way.]
The trig
function period identities are [X is an angle, n is a positive
integer]:
These identities
hold even in the undefined case [if one side is undefined, they
both are.]
To formally
demonstrate the n part, I would use natural induction, which
should be buried somewhere in College Algebra. If you don't
recall this term clearly, don't worry about it. However, I'm
only going to explain it for n=1.
We read
these from the unit circle immediately:
We use the
rewrite of sec(X) and csc(X) in cos(X) and sin(X) to derive
these:
Now, to
deal with tan(X) and cot(X), we have to be a little more clever.
For tan(X), write:
i.e. tan(X+[½][circle])=tan(X).
The identity cot(X+[½][circle]) = cot(X) is similar, but works
with the multiplicative reciprocals throughout.

What
are the triangle area formulae?
There are
two basic formulae, and one "impractical" one.
"½
base times height"
(Area)=(½)(length
of base)(length of height)
To use this
formula, pick one side of the triangle as the "base". Note its
length. Then, draw a perpendicular line segment from the vertex
of the triangle not in the "base", to the "base", and note this
line segment's length (this is the height). [This formula does
work for obtuse triangles. Extend the base to where it would
hit the perpendicular line.].
This formula
is easily visualized for a right triangle [a rectangle with
a line segment between two opposite vertices gives two right
triangles, both clearly with half the original area since they
are congruent]. It behaves reasonably for line segments [it
gives zero area, which is correct for a line segment; either
base or height is zero for a line segment.]
EXERCISE:
learn to use this formula by applying it to the reference triangles.
The 45-45-90 triangle with hypotenuse should
have area ½, and the 30-60-90 triangle with hypotenuse length
2 should have area .
This formula
has useful analogies in higher dimensions. The volume of either
a pyramid or a cone, for instance, is given by:
The "obvious
generalization" for n-dimensional Euclidean space (n a positive
integer) is
and for
4-d Euclidean space, English permits a simplification:
The n-dimensional
formula is clearly dimensionally consistent: using a length
unit, both sides have dimension (length)n. The n-dimensional
formula can be directly computed if one is familiar with iterated
integrals (say, from Calculus III or a decent physics course).
"½
of the product of the lengths of two sides, and the sine of
the included angle"
We can see
this, from the immediately prior formula, by taking side 1 to
be the base, side 2 to be the hypotenuse of the right triangle
formed between the [extended, if necessary] base and the height,
and then solving for the height in terms of side 2 and the angle
between sides 1 and 2 [an acute angle of the right triangle
we just constructed].
We get:
EXERCISE:
Learn to use this formula by applying it to the reference triangles.
[The correct areas are the same as before.]
EXERCISE:
Now, learn to use this formula by applying it to the equilateral
triangle with all sides length 2. This triangle is essentially
two copies of the 30-60-90 triangle referred to earlier, so
its area is twice as large -- 2 .
I do not
know how to generalize the above formula to n-dimensional Euclidean
geometry.
Heron's
area formula for a triangle:
This formula
gives zero when the length of one side is the same as the length
of the other two sides, but will malfunction when one side has
length zero. The Heron referred to here is a Greek mathematician
(B.C.), so the formula can (or should be able to, at least)
be derived in straight Euclidean geometry without coordinates.
However, I have not read this, so I cannot explain it.
EXERCISE:
Learn to use this formula by applying it to the reference triangles,
and also to the equilateral triangle with all sides length 2.
The mathematician
Cartan generalized this formula to n-dimensional Euclidean geometry,
using matrix determinants [this is the determinant of a certain
3x3 matrix]. Cartan's generalization is definitely beyond the
scope of this crash review.

What
is the law of sines?
Note that
the angles A, B, and C are strictly between 0 and 2
in angular measure. This means that solving for the sine of
an angle by the law of sines does *not* strictly determine the
angle, normally. [It does if the angle is a right angle; then
sin(angle)=1].
If the solved-for
sin(angle) is strictly between 0 and 1, then some work is required
to determine the actual angles. The inverse sine function on
a calculator, or spreadsheet, is programmed to give an acute
angle [strictly between 0 and
in angular measure]. However, since the sine of an angle X is
equal to the sine of its supplementary angle 2 -X
[sin(X)=sin(2 -X)],
the supplementary angle is *also* a viable choice.
[EXERCISE:
Learn to use the sine law by explicitly writing out the equalities
for the 45-45-90 triangle with hypotenuse , the
30-60-90 triangle with hypotenuse 2, and the equilateral triangle
with all sides length 2.]

What
is the law of cosines?
This is
the generalization of the Pythagorean theorem to
non-right triangles. I'm going to present it "deus ex machina".
However,
this formula does behave correctly in the extreme cases [worked
for C, others are similar]:
-
C=
-
C=0
-
cos(0)=1.
We get:
c²=a²+b²-2ab, i.e.
c²=a²-2ab+b², i.e. [this is a standard reverse-squaring
rule from College Algebra]
c²=(a-b)²
-
The
last equation is physically correct: if C=0, then the
triangle is really a line segment, and the side c physically
has length |a-b|. This is what the algebra states:
-
c²=(a-b)²
i.e.
|c|=|a-b| i.e. [c is guaranteed to be non-negative,
since it represents a physical length]
c=|a-b|
-
C=2
-
cos(2 )=-1.
We get:
c²=a²+b²+2ab, i.e.
c²=a²+2ab+b², i.e. [this is a standard reverse-squaring
rule from College Algebra]
c²=(a+b)²
-
The
last equation is physically correct: if C=2 ,
then the triangle is really a line segment, and the
side c physically has length a+b. This is what the algebra
states:
-
c²=(a+b)²
i.e.
|c|=|a+b| i.e. [a, b, c are all guaranteed to be non-negative,
since they represent physical lengths]
c=a+b
[EXERCISE:
Learn to use the cosine law by explicitly writing out the equalities
for the 45-45-90 triangle with hypotenuse , the 30-60-90 triangle
with hypotenuse 2, and the equilateral triangle with all sides
length 2.]

What
are the angle sum, difference, and/or doubling formulae?
I'm going
to present these "deus ex machina". [The least painful method,
to me, is 2x2 matrices -- inappropriate here.]
The basic
angle sum formulae are [A,B are arbitrary angles here]:
The other
four trigonometric functions can be recovered from these two.
Note: the tan(A+B) formula has a relatively clean version (assuming
neither tan(A) nor tan(B) is undefined):
That is,
Now, using
A-B = A+(-B), we can use the even-odd identities to easily derive
the following:
Also, by
setting A=B, we get the following double-angle formulae:
[EXERCISES:
Learn to use these formulae with the reference values we already
have: 0°, 30°, 45°, 60°, and 90°. Namely: try A=0° [should reduce
to trigonometric function of B or -B, respectively] and B=0°
[trigonometric function of A]. Also, try 2*0°=0°, 2*45°=90°,
2*30°=60°, and 30°+60° = 90°. Ranging into the second quadrant:
try 90°+(another reference angle), 2*60° = 120°, and 2*90° =
180°. The tangent identities should not be usable when one of
the angles is a right angle.]
Note: several
entries in Trigonometry Survival 201 are/will be based on this.

What
are the angle-halving formulae?
The angle-halving
formulae are easily derived from the Pythagorean Identity and
the formula for cos(2A). In general, we need to know which quadrant
the angle A/2 is in to decide on the correct sign.
Observe:
In both
cases, we can isolate the trigonometric function on the RHS
[right-hand side] by dividing by 2, then taking the square root.
This yields:
Rename A
as A/2 to get:
If we know
which quadrant the angle A/2 is in, we can decide on the correct
sign in the above formulae.
We also
have [exercise: derive the following]:
The Pythagorean
identity allows two easy rewrites of the above [observe that
the ± negates the usefulness of | |, here]:
[EXERCISES:
Learn to use the above with reference values. Namely: 0° = [½]0°,
30°
= [½]60°, 45° = [½]90°, 60° = [½]120°, and 90° = [½]180°. For
the tangent identities, both sides should be undefined for the
90° case.]
What
are the trig(onometric) sum and difference formulae?
What are the trig(onometric) product formulae?
These formulae
may be derived from the angle sum and difference formulae. We
have:
All are
trig(onometric) sum or difference formulae. The first four are
also product formulae.
EXERCISE:
Derive these from the angle sum and difference formulae, as
follows:
-
Set
X=A+B and Y=A-B
Substitute
in the angle sum and difference formulae for the affected
functions, and simplify.
Solve
for A and B in terms of X and Y, and then replace A, B with
their expressions in X, Y.
EXERCISE:
Numerically use the sum and difference identities for X=A+B,
Y=A-B where A, B are reference angles. The tangent ones will
break down when 90° is A, B, X, or Y.
EXERCISE:
Numerically use the product identities for A=(X+Y)/2, B=(X-Y)/2
where X, Y are reference angles.
Wait!
I'm not completely sure how to solve for A and B in terms of
X and Y!
Until I
get a College Algebra page going, here's a quick summary on
how to solve systems of linear equations. (This domain can use
several.) [While not all systems of linear equations
are solvable, the one we want to is solvable.]
-
Substitution
-
We
hope that by cleverly adding multiples of pairs of equations,
we can get equations with reasonably isolated variables.
This works fairly well with two variables.
-
How
this works in practice:
-
X=A+B,
Y=A-B.
-
To
solve for A, eliminate B from the resulting sum
and then solve. Since 1+(-1)=0, we simply add both
equations: X+Y=2A. Dividing by 2 yields (X+Y)/2=A.
-
To
solve for B, eliminate A from the resulting sum
and then solve. Since 1+(-1)*1=0, we subtract Y=A-B
from X=A+B: X-Y=2B. Dividing by 2 yields (X-Y)/2=B.
-
Gaussian
Elimination (named after Gauss, the mathematician)
-
Taking
variables from left to right (in our case, A, then B):
-
Pick
a linear equation using the "leading variable".
If this variable's coefficient is not 1, divide
the equation by this coefficient. This equation
is now the "topmost equation".
-
By
adding a suitable multiple of the "topmost equation"
to the other equations, remove A from the resulting
equations.
-
We
are now done with the "topmost equation". Set it
aside.
-
When
the only equations left have a multiple of a single
variable equal to a constant, solve those variables.
Then replace those variables in the equations that
have been set aside. Repeat until all variables
have been explicitly solved.
-
How
this works in practice:
-
X=A+B,
Y=A-B: A is "leading variable". Both X=A+B and Y=A-B
have coefficient 1 for A.
-
However,
the multiple of the "topmost equation" I am subtracting
is 1[=1/1; numerator is from the equation I am subtracting
from, denominator is from the "topmost equation".
I arbitrarily choose my topmost equation to be Y=A-B.
-
We
subtract Y=A-B (i.e. 1*[Y=A-B]) from X=A+B to get
X-Y=2B.
-
We
solve for B: B=(X-Y)/2.
-
We
then subsitute this into the "topmost equation"
X=A+B, getting X=A+(X-Y)/2
-
Isolating
A, we end up at (X+Y)/2=A.
-
Cramer's
Rule: This is theoretically interesting, since it directly
informs you when the system of linear equations does not
have a unique solution. However, it requires the introduction
of even more terminology. I won't cover it in this refresher.

Angle
classifications

Angle
relations
For angles
X, Y:
Cofunction
relations
For an angle
X:
-
sin( -X)=cos(X)
-
cos( -X)=sin(X)
-
tan( -X)=cot(X)
-
cot( -X)=tan(X)
-
sec( -X)=csc(X)
-
csc( -X)=sec(X)
- csc(90°
-X)=sec(X)
csc(p/2
radians-X)=sec(X)
csc(100
gradients-X)=sec(X)

Supplementary
angle relations
For an angle
X:
-
sin(2 -X)=cos(X)
-
cos(2 -X)=sin(X)
-
tan(2 -X)=cot(X)
-
cot(2 -X)=tan(X)
-
sec(2 -X)=csc(X)
-
csc(2 -X)=sec(X)

Basic
trigonometric reference tables
|
sin
|
cos
|
tan
|
cot
|
sec
|
csc
|
0°
|
0
|
1
|
0
|
undefined
|
1
|
undefined
|
30°
|
½
|
/2
|
/3
|
|
(2 )/3
|
2
|
45°
|
/2
|
/2
|
1
|
1
|
|
|
60°
|
/2
|
½
|
|
/3
|
2
|
(2 )/3
|
90°
|
1
|
0
|
undefined
|
0
|
undefined
|
1
|
Exercise:
convert the angles to radians. If you plan to use gradients,
also convert the angles to gradients.
If you find
the above table difficult to memorize for sin and cos, my ex-fianceé
suggests this alternative table [it uses =1 and
=2]:
|
0°
|
30°
|
45°
|
60°
|
90°
|
sin
|
0/2
|
/2
|
/2
|
/2
|
/2
|
cos
|
/2
|
/2
|
/2
|
/2
|
0/2
|
If you find
this alternative table convenient, you may want to check that
the entries that look different, do algebraically reduce to
those in the first table.

Reference
values -- 15° and 75° i.e. p/12 and 5p/12 radians i.e. 100/6
and 500/6 gradients
EXERCISE:
verify that these tables are correct:
Suggestions:
Note that 15° is computable either with the half-angle or the
difference-angle formula from the standard reference table.
Use both of these methods. Also, 75° can be computed by the
sum formula from the standard reference table; use this method.
Once you are confident that these tables are correct, and if
you want more practice with the angle sum, difference, halving,
and doubling formulae, use these additional reference values
in combination with the earlier ones [0°, 30°, 45°, 60°, and
90°, i.e. the standard table].
[EXERCISE:
translate everything into radians. If you plan to use gradients,
also translate everything into gradients.]

Reference
values -- 18°, 36°, 54°, and 72° i.e. p/10, p/5, 3p/10, and
2p/5 radians i.e. 20, 40, 60, and 80 gradients
EXERCISE:
Verify that these tables are correct:
|
sin
|
cos
|
18°
|
[  -1]/4
|
 /4
|
36°
|
 /4
|
[  +1]/4
|
54°
|
[  +1]/4
|
 /4
|
72°
|
 /4
|
[  -1]/4
|
|
tan
|
cot
|
18°
|
[5-  ]  /20
|
 [  +1]/4
|
36°
|
 [  -1]/4
|
[5+  ]  /20
|
54°
|
[5+  ]  /20
|
 [  -1]/4
|
72°
|
 [  +1]/4
|
[5-  ]  /20
|
|
sec
|
csc
|
18°
|
  /5
|
 +1
|
36°
|
 -1
|
  /5
|
54°
|
  /5
|
 -1
|
72°
|
 +1
|
  /5
|
Suggestions:
First, let X be an angle such that 5X works out to be 0, 1,
2, 3, or 4 full circles.
[That is,
we are directly interested in 72°, 144°, 216°, or 288°. I also
included 0°, but that is a standard reference value. We will
recover:
It may provide
some intuition about the following equations to subtract 1 circle
off 216° [-144°] and 288° [-72°].]
Next, use
the angle addition formulae
to rewrite in terms of sin(X) and cos(X)
These are
the equations that describe an inscribed equilateral pentagon.
They *are* solvable using nothing more than the quadratic formula(!),
and what we already are supposed to know.
This would
occur to me from sin(X) being an (algebraically) odd function, combined
with the angle comments above:
sin(0°)=0,
so we can factor sin(X) out of our expanded version of sin(5X)=0.
Then, use the Pythagorean identity to replace cos²(X) with 1-sin²(X).
Expand the results. This should give a quartic in sin(X) [oops],
which is also a quadratic in sin²(X) [great]. Use the quadratic
formula to solve for sin²(X). Then take square roots on the
solutions we get for sin²(X); we need both the positive and
negative roots. One of these root sets is for 72° and -72°,
and the other one is for 144° and -144°.
Geometrically,
the larger positive root goes with 72°, and the smaller positive
root goes with 144°. Solve for cosine of 72° and 144° with the
Pythagorean identity [be
sure to use the quadrants to force the
correct sign]. Then fill in the table as summarized initially.
Also: when
completing the table, tan(18°) and tan(54°) [and their corresponding
cofunction values, cot(72°) and cot(36°)] are technically difficult
to algebraically compute directly from the sine and cosine values.
[The denominator needs two stages to cancel out correctly.]
I tried that three times in a row, and got three different answers,
all of them wrong. The method I used for the table was to compute
cot(18°) and cot(54°) respectively, and then take the multiplicative
inverse algebraically to get tan(18°) and tan(54°).

Instant
numerical trigonometric tables for sin, cos, and tan near 0
This isn't
really "fair", since it relies on the "stratosphere" math approach.
The first mention of the basis for this, traditionally, is in
the middle of a Calculus series [business or conventional].
However,
it is very practical. (Especially if you plan to use trigonometry
in an engineering course.) I'm going to present how the result
is arrived at in just enough detail, that those readers who
actually know the relevant material will be able to see that
I'm doing it right.
The key
point is: for an angle X with radian measure 0<X<1, we
have
[If you
know what a power series is, and the alternating series test
for convergence, you will immediately recognize what I'm doing.
If not -- well, I said this wasn't "fair". Put these equations
and inequalities into a spreadsheet to see why the above is
plausible. If you don't have access to a spreadsheet, then use
a calculator. If you don't have even that, at least do it for
the numbers 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, and 0.9.]
For three
decimal place accuracy (the least we are interested in, for
practical purposes), we need to control the error to less than
5*10-4 i.e. [½]10-3. This error term is:
Now, let's
solve for X:
That is,
for X in radians:
What if
you want the "instant trig table" to more decimal places? Just
work the above bound calculations with 5*10-(number of
decimal places+1) instead of 5*10-4. Note that
at the final stage, I took the decimal my TI-36SX Solar calculator
gave out, and found the three-decimal place numbers that bracketed
it. For a different number of decimal places, bracket the result
with decimals to as many places as you need precision.
Computing
tan(X) and sec(X) from the results of this "instant trig table"
will give reasonably accurate numerical results. Computing cot(X)
and csc(X) will not work reasonably "near 0"; in general, you
cannot get more significant digits out than you put in, and
sin(X) will lose significant digits as X ends up near 0. This
means you are dividing a large number by a small, imprecise
number, which will *not* give reasonable numerical results.
[The zero decimal places between the decimal point and the first
non-zero digit are "lost significant digits". For example, sin(0.021)=0.021
to three decimal places -- but there are only two significant
digits.]

Primer:
Read
An Overview of Trigonometry with Kenny Felder's Waves
Circles And Triangles
Read what Excel can do for you with a copy of the Cataclysm
SRD-XL Utility help file.
Online:
Kenneth
Boyd's Math
page, or the Zaimoni.com
Home page.
Kenny Felder's The
Math and Physics Help Home Page
Email:
Send comments or questions to the author of this page, Kenneth
Boyd
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