Edited by Paul Ducham

LAW OF SINES DERIVATION

The law of sines is relatively easy to prove using the right triangle properties. We will also use the fact that
sin (180° x) sin x
which is readily obtained using a difference identity. Referring to Figure 2, we proceed as follows: Angles and in Figure 2(a), and also in Figure 2(b), satisfy

Solving each equation for h, we obtain
h = b sin α       and         h = a sin  β
So,

Suppose that an angle of a triangle and its opposite side are known. Then the ratio of Theorem 1 can be calculated. So if one additional part of the triangle, either of the other angles or either of the other sides, is known, then the law of sines can be used to solve the triangle.
Therefore, the law of sines is used to solve triangles, given:
1. Two sides and an angle opposite one of them (SSA), or
2. Two angles and any side (ASA or AAS)
If the given information for a triangle consists of two sides and the included angle (SAS) or three sides (SSS), then the law of sines cannot be applied.
We will apply the law of sines to the easier ASA and AAS cases first, and then will turn to the more challenging SSA case.

SOLVING THE ASA AND AAS CASES

Note that the AAS case can always be converted to the ASA case by first solving for the third angle. For the ASA or AAS case to determine a unique triangle, the sum of the two angles must be between 0° and 180°, because the sum of all three angles in a triangle is 180° and no angle can be zero or negative.

SOLVING THE SSA CASE—INCLUDING THE AMBIGUOUS CASE

We now look at the case where we are given two sides and an angle opposite one of the sides—the SSA case. This case has several possible outcomes, depending on the measures of the two sides and the angle. Table 2 illustrates the various possibilities.

It is unnecessary to memorize Table 2 to solve triangles in the SSA case. Instead, given sides a, b, and angle , we use the law of sines to solve for the angle opposite side b. The number of triangles is equal to the number of solutions β, 0°<β<180°, of the law of sines equation

that satisfy
α β < 180°
In practice, we check each solution of equation (3) to determine whether inequality (4) is satisfied. If it is, we can easily solve for the remaining parts of the triangle. Examples 2–4 will make the procedure clear.

The law of sines is useful in many applications.

LAW OF COSINES DERIVATION

Theorem 1 states the law of cosines.

The law of cosines is used to solve triangles, given:

1. Two sides and the included angle (SAS), or
2. Three sides (SSS)

We will establish the first equation in Theorem 1. The other two equations then can be obtained from this one simply by relabeling the figure. We start by locating a triangle in a rectangular coordinate system. Figure 2 shows three typical triangles.
For an arbitrary triangle located as in Figure 2, the distance formula is used to obtain

a2 = (h = c)2 + k2
= h2 - 2hc + c2 + k2

From Figure 2, we note that
b2 = h2 + k2
Substituting b2 for h2 + k2 in equation (1), we obtain
a2 = b2 + c2 + 2hc
But
cos α =h/b
h = b cos α
By replacing h in equation (2) with b cos , we reach our objective:
a
2 = b2 + c2 - 2bc cos α
[Note: If is acute, then cos is positive; if is obtuse, then cos is negative.]

SOLVING THE SAS CASE

For the SAS case, start by using the law of cosines to find the side opposite the given angle. Then use either the law of cosines or the law of sines to find a second angle. Because of the simpler computations, the law of sines will generally be used to find the second angle. This gives the following strategy for solving the SAS case.

SOLVING THE SSS CASE

Starting with three sides of a triangle, the problem is to find the three angles. Subsequent calculations are simplified if we solve for the obtuse angle first, if present. The law of cosines is used for this purpose. A second angle, which must be acute, can be found using either law, although computations are usually simpler with the law of sines.
The preceding discussion leads to the following strategy for solving the SSS case:

VECTORS

A vector v is a quantity that has both magnitude and direction. We picture a vector as an arrow from an initial point O to a terminal point P with this provision: arrows that have the same length (magnitude) and direction represent the same vector (Fig. 1).
The vector v of Figure 1 is also denoted by . We use boldface letters such as v to denote vectors. But because it is difficult to write boldface by hand, we suggest that you use  as a substitute for v when you want to denote a vector by a single letter.
The magnitude of the vector v =, denoted by |v|, || or || is the length of the line segment OP. Two vectors have the same direction if they are parallel and point in the same direction. Two vectors have opposite directions if they are parallel and point in opposite directions. The zero vector, denoted by 0 or , has magnitude 0 and arbitrary direction. Two vectors are equal if they have the same magnitude and direction. So a vector can be translated from one location to another as long as the magnitude and direction do not change.

Any vector in a rectangular coordinate system can be translated so that its initial point is the origin O. The vector  such that  = is said to be the standardvector for  (Fig. 2). Note that  is the standard vector for infinitely many vectors—all vectors with the same magnitude and direction as .

Given the coordinates of the endpoints of vector how do we find its corresponding standard vector  ? The coordinates of the origin O, the initial point of OP, are always (0, 0). The coordinates of P, the terminal point of  are given by

(xp, yp) = (xb - xa, yb - ya)
where A = (xa, ya) and B = (xb, yb).

EXAMPLE 1 Finding a Standard Vector for a Given Vector
Given the geometric vector  with initial point A = (3, 4) and terminal point B = (7, 1), find the coordinates of the point P such that  =

SOLUTION   The coordinates of P are given by
(xp, yp) = (xb - xa, yb - ya)
= (7 - 3, -1 - 4)
= (4, -5)
Note in Figure 3 that if we start at A, then move to the right four units and down five units, we will be at B. If we start at the origin, then move to the right four units and down five units, we will be at P.

Example 1 suggests that there is a one-to-one correspondence between vectors in a rectangular coordinate system and points in the system. Any vector  is completely specified by the point P = (xp, yp) such that  =  (we are not concerned that  has a different position than ; we are free to translate a vector anywhere we please). Conversely, any point P of the system corresponds to the vector .

A vector can be denoted by an ordered pair of real numbers. To avoid confusion, we use {c, d} to denote the vector  with initial point (0, 0) and terminal point (c, d) (Fig. 4). The real numbers c and d are called the scalar components of the vector {c, d}. Two vectors u = {a, b} and v = {c, d} are equal if their corresponding components are equal, that is, if a = c and b = d. The zero vector is 0 {0, 0}. The magnitude of the vector {a, b} is the length of the line segment from (0, 0) to (a, b) [Fig. 5].

The sum u + v of two vectors u and v is defined by the tail-to-tip rule: Translate v so that its tail (initial point) is at the tip (terminal point) of u. Then, the vector from the tail of u to the tip of v is the sum, denoted u + v, of the vectors u and v (Fig. 6).
If u and v are not parallel, the parallelogram rule gives an alternative description of u + v: The sum of two nonparallel vectors u and v is the diagonal of the parallelogram formed using u and v as adjacent sides (Fig. 7).
The vector u + v is also called the resultant of the two vectors u and v, and u and v are called vector components of u + v.
The scalar product ku of a scalar (real number) k and a vector u is the vector with magnitude |k||u| that has the same direction as u if k is positive and the opposite direction if k is negative. For example, 2u has twice the magnitude of u and the same direction. Similarly, 0.5u has half the magnitude of u and the opposite direction (Fig. 8).
Both the sum u + v and the scalar product ku are easy to calculate if the scalar components of u and v are given: for the sum, just add corresponding components; for the scalar product, multiply each component by the scalar.

Vector addition and scalar multiplication possess algebraic properties similar to the real numbers. These properties enable us to manipulate symbols representing vectors and scalars in much the same way we manipulate symbols that represent real numbers in algebra. The algebraic properties are listed here for convenient reference.

ALGEBRAIC PROPERTIES OF VECTORS
A. Addition Properties. For all vectors u, v, and w,
1. u + v = v + u                     Commutative Property
2. u + (v + w) = (u + v) + w   Associative Property
3. u + 0 = 0 + u = u              Additive Identity
4. u + (-u) = (-u) + u = 0      Additive Inverse
B. Scalar Multiplication Properties.
For all vectors u and v and all scalars m and n:
1. m(nu) = (mn)u                   Associative Property
2. m(u + v) = mu + mv         Distributive Property
3. (m + n)u = mu + nu          Distributive Property
4. 1u = u                               Multiplicative Identity

UNIT VECTORS

Any vector that has magnitude 1 is called a unit vector. If v is an arbitrary nonzero vector and k is a scalar, then the scalar product kv has magnitude |k||v|. Therefore, by choosing k to be 1/ |v|, the scalar product kv will be a unit vector with the same direction as v.

The unit vectors in the directions of the positive x axis and the positive y axis are denoted by i and j, respectively.

Why are the i and j unit vectors so important? Any vector v =   can be expressed as a linear combination of these two vectors; that is, as ai + bj.

EXAMPLE 6 Algebraic Operations on Vectors Expressed in Terms of the i and j Unit Vectors
For u = i - 2j and v = 5i + 2j, compute each of the following:
(A) u + v (B) u - v (C) 2u - 3v

SOLUTIONS
(A) u + v = (i - 2j) (5i + 2j) = i - 2j + 5i + 2j = 6i + 0j = 6i
(B) u - v = (i - 2j) - (5i + 2j) = i - 2j - 5i - 2j = -4i - 4j
(C) 2u + 3v = 2(i - 2j) + 3(5i + 2j) = 2i - 4j + 15i + 6j = 17i + 2j

VELOCITY VECTORS

A vector that represents the direction and speed of an object in motion is called a velocity vector. Problems involving objects in motion often can be analyzed using vector methods. Many of these problems involve the use of a navigational compass, which is marked clockwise in degrees starting at north as indicated in Figure 9.

EXAMPLE 7 Apparent and Actual Velocity
An airplane has a compass heading (the direction the plane is pointing) of 85° and an airspeed (relative to the air) of 140 miles per hour. The wind is blowing from north to south at 66 miles per hour. The velocity of a plane relative to the air is called apparent velocity, and the velocity relative to the ground is called resultant, or actual, velocity. The resultant velocity is the vector sum of the apparent velocity and the wind velocity. Find the resultant velocity; that is, find the actual speed and direction of the airplane relative to the ground. Directions are given to the nearest degree and magnitudes to two significant digits.

FORCE VECTORS

A vector that represents the direction and magnitude of an applied force is called a force vector. If an object is subjected to two forces, then the sum of these two forces, the resultant force, is a single force. If the resultant force replaced the original two forces, it would act on the object in the same way as the two original forces taken together. In physics it is shown that the resultant force vector can be obtained using vector addition to add the two individual force vectors. It seems natural to use the parallelogram rule for adding force vectors, as is illustrated in Example 8.

EXAMPLE 8 Finding the Resultant Force
Two forces of 30 and 70 pounds act on a point in a plane. If the angle between the force vectors is 40°, what are the magnitude and direction (relative to the 70-pound force) of the resultant force? The magnitudes of the forces are to two significant digits and the angles to the nearest degree.

SOLUTION   We start with a diagram (Fig. 12), letting vectors represent the various forces. Because adjacent angles in a parallelogram are supplementary, the measure of angle OCB = 180° - 40° = 140°. We can now find the magnitude of the resultant vector R using the law of cosines (Fig. 13).

Instead of adding vectors, many problems require the resolution of vectors into components. As we indicated earlier, whenever a vector is expressed as the sum or resultant of two vectors, the two vectors are called vector components of the given vector. Example 9 illustrates an application of the process of resolving a vector into vector components.

An object at rest is said to be in static equilibrium. Example 10 illustrates how important physics and engineering problems can be solved using the condition for static equilibrium: For an object to remain in static equilibrium, the sum of all the force vectors acting on the object must be the zero vector.

EXAMPLE 10 Tension in Cables
A cable car, used to ferry people and supplies across a river, weighs 2,500 pounds fully loaded. The car stops when partway across and deflects the cable relative to the horizontal, as indicated in Figure 16. What is the tension in each part of the cable running to each tower? (The tension in a cable is the magnitude of the force it exerts in the direction parallel to the cable.)

SOLUTION
Step 1. Draw a force diagram with all force vectors in standard position at the origin (Fig. 17). The objective is to find |u| and |v|.
Step 2. Write each force vector in terms of the i and j unit vectors:
u = |u|(cos 7°)i + |u|(sin 7°)j
v = |v|(-cos 15°)i + |v|(sin 15°)j
w = -2,500j
Step 3. For the system to be in static equilibrium, the sum of the force vectors must be the zero vector. That is,
u + v + w = 0
Replacing vectors u, v, and w from step 2, we obtain
[ |u|(cos 7°)i + |u|(sin 7°)j ] + [ |v|(-cos 15°)i + |v|(sin 15°)j ] - 2,500j = 0i + 0j
which, upon combining i and j vectors, becomes
[ |u|(cos 7°) + |v|(cos 15°)] i + [ |u|(sin 7°) + |v|(sin 15°) - 2,500] j = 0i + 0j
Because two vectors are equal if and only if their corresponding components are equal, we are led to the following system of two equations in the two variables |u| and |v|:
(cos 7°)|u| + (-cos 15°)|v| = 0
(sin 7°)|u| + (sin 15°)|v| - 2,500 = 0
Solving this system by standard methods, we find that
|u| = 6,400 pounds and |v| = 6,600 pounds
Did you expect that the tension in each part of the cable is more than the weight hanging from the cable?

POLAR COORDINATE SYSTEM

To form a polar coordinate system in a plane (Fig. 1), start with a fixed point O and call it the pole, or origin. From this point draw a half-line, or ray (usually horizontal and to the right), and call this line the polar axis.
If P is an arbitrary point in a plane, then associate polar coordinates (r,θ ) (Fig. 1) with it as follows: Starting with the polar axis as the initial side of an angle, rotate the terminal side until it, or the extension of it through the pole, passes through the point. The θ coordinate in (r, θ) is this angle, in degree or radian measure. The angle θ is positive if the rotation is counterclockwise and negative if the rotation is clockwise. The r coordinate in (r,θ) is the directed distance from the pole to the point P. It is positive if measured from the pole along the terminal side of θ and negative if measured along the terminal side extended through the pole.
Figure 2 illustrates a point P with three different sets of polar coordinates. Study this figure carefully. The pole has polar coordinates (0, θ) for arbitrary θ. For example, (0, 0°), (0, π/3), and (0, -371°) are all coordinates of the pole.
We now see a distinct difference between rectangular and polar coordinates for the given point. For a given point in a rectangular coordinate system, there exists exactly one set of rectangular coordinates. On the other hand, in a polar coordinate system, a point has infinitely many sets of polar coordinates.
Just as graph paper with a rectangular grid is readily available for plotting rectangular coordinates, polar graph paper is available for plotting polar coordinates.

EXAMPLE 1 Plotting Points in a Polar Coordinate System
Plot the following points in a polar coordinate system:
(A) A = (3, 30°), B = (-8, 180°), C = (5, -135°), D = (-10, -45°)
(B) A = (5, π/3), B = (-6, 5π/6), C = (7, π/2), D = (-4, -π/6)

CONVERTING FROM POLAR TO RECTANGULAR FORM, AND VICE VERSA

Often, it is necessary to transform coordinates or equations in rectangular form into polar form, or vice versa. The following polar–rectangular relationships are useful in this regard:

Many calculators can automatically convert rectangular coordinates to polar form, and vice versa. (Read the manual for your particular calculator.)
Example 2 illustrates calculator conversions in both directions.

EXAMPLE 2 Converting from Polar to Rectangular Form, and Vice Versa
(A) Convert the polar coordinates (-4, 1.077) to rectangular coordinates to three decimal places.
(B) Convert the rectangular coordinates (-3.207, -5.719) to polar coordinates with θ in degree measure, -180° < θ ≤ 180° and r ≥ 0.

SOLUTIONS
(A) Use a calculator set in radian mode.
(r, θ) = (-4, 1.077)
x = r cos θ = (-4) cos 1.077 = -1.896
y = r sin θ = (-4) sin 1.077 = -3.522
The rectangular coordinates are (-1.896, -3.522).

Generally, a more important use of the polar–rectangular relationships is in the conversion of equations in rectangular form to polar form, and vice versa.

EXAMPLE 3 Converting an Equation from Rectangular Form to Polar Form
Change x2 + y2 - 4y = 0 to polar form.

SOLUTION    Use r2 = x2 + y2 and y = r sin θ.
x
2 + y2 - 4y = 0
r
2 - 4r sin θ = 0
r(r - 4 sin θ) = 0
r = 0     or     r - 4 sin θ = 0
The graph of r = 0 is the pole. Because the pole is included in the graph of r - 4 sin θ = 0 (let θ = 0), we can discard r = 0 and keep only
r - 4 sin θ = 0
or
r = 4 sin
θ The polar form of x2 + y2 - 4y = 0.

EXAMPLE 4 Converting an Equation from Polar Form to Rectangular Form
Change r=-3 cos θ to rectangular form.

SOLUTION   The transformation of this equation as it stands into rectangular form is fairly difficult. With a little trick, however, it becomes easy. We multiply both sides by r, which simply adds the pole to the graph. But the pole is already part of the graph of r=-3 cos θ (let θ = π/2), so we haven’t actually changed anything.
r = -3 cos θ            Multiply both sides by r.
r2 = -3r cos θ          r2 - x2 + y2and r cos θ = x
x
2 + y2 = -3x
x
2 + y2 + 3x = 0

GRAPHING POLAR EQUATIONS

We now turn to graphing polar equations. The graph of a polar equation, such as r = 3θ or r = 6 cos θ, in a polar coordinate system is the set of all points having coordinates that satisfy the polar equation. Certain curves have simpler representations in polar coordinates, and other curves have simpler representations in rectangular coordinates.
To establish fundamentals in graphing polar equations, we start with a point-by-point graph. We then consider a more rapid way of making rough sketches of certain polar curves. And, finally, we show how polar curves are graphed in a graphing utility.
To plot a polar equation using point-by-point plotting, just as in rectangular coordinates, make a table of values that satisfy the equation, plot these points, then join them with a smooth curve. Example 5 illustrates the process.

If only a rough sketch of a polar equation involving sin θ or cos θ is desired, you can speed up the point-by-point graphing process by taking advantage of the uniform variation of sin θ and cos θ as θ moves around a unit circle. This process is referred to as rapid polar sketching. It is convenient to visualize Figure 6 in the process. With a little practice most of the table work in rapid sketching can be done mentally and a rough sketch can be made directly from the equation.

We now turn to graphing polar equations in a graphing calculator. Example 8 illustrates the process.

EXAMPLE 8 Graphing in a Graphing Calculator
Graph each of the following polar equations in a graphing calculator (parts B and C are from Examples 6 and 7).
(A) r = 3θ, 0 ≤ θ ≤ 3π/2 (Archimedes’ spiral)
(B) r = 4 + 4 cos θ (cardioid)
(C) r = 8 cos 2θ (four-leafed rose)

SOLUTIONS   Set the graphing calculator in polar mode and select polar coordinates and radian measure. Adjust window values to accommodate the whole graph. A squared graph is often desirable in showing the true shape of the curve, and is used here. Many graphing calculators, including the one used here, do not show a polar grid. When using TRACE, many graphing calculators offer a choice between polar coordinates and rectangular coordinates for points on the polar curve. The graphs of the preceding equations are shown in Figure 9.

SOME STANDARD POLAR CURVES

In a rectangular coordinate system the simplest types of equations to graph are found by setting the rectangular variables x and y equal to constants:
x = a   and    y = b
The graphs are straight lines: The graph of x = a is a vertical line, and the graph of y = b is a horizontal line. A glance at Table 1 shows that horizontal and vertical lines do not have simple equations in polar coordinates.
Two of the simplest types of polar equations to graph in a polar coordinate system are found by setting the polar variables r and θ equal to constants:
r = a    and    θ = b
Figure 10 illustrates the graphs of θ = π/4 and r = 5.
Table 1 illustrates a number of standard polar graphs and their equations. Polar graphing is often made easier if you have some idea of the final form.

POLAR COORDINATES AND GRAPHS: APPLICATION

Serious sailboat racers make polar plots of boat speeds at various angles to the wind with various sail combinations at different wind speeds. With many polar plots for different sizes and types of sails at different wind speeds, they are able to accurately choose a sail for the optimum performance for different points of sail relative to any given wind strength. Figure 11 illustrates one such polar plot, where the maximum speed appears to be about 7.5 knots at 105° off the wind (with spinnaker sail set).

RECTANGULAR FORM

A complex number is any number that can be written in the form
x + yi
where x and y are real numbers and i is the imaginary unit. (We use x + yi and x + iy interchangeably; each has its advantages in keeping notation simple.) So associated with each complex number x + yi is a unique ordered pair of real numbers (x, y), and vice versa. For example,
5 + 3i   corresponds to    (5, 3)
Associating these ordered pairs of real numbers with points in a rectangular coordinate system, we obtain a complex plane (Fig. 1). When complex numbers are associated with points in a rectangular coordinate system, we refer to the x axis as the real axis and the y axis as the imaginary axis. The complex number x + yi is said to be in rectangular form.

EXAMPLE 1 Plotting in the Complex Plane
Plot the following complex numbers in a complex plane:
A = 2 + 3i   B = -3 + 5i   C = -4   D = -3i

SOLUTION

POLAR FORM

Each point (x, y) of the plane corresponds to a unique complex number z, namely, in rectangular form, z = x + iy. But the point (x, y) can also be specified by polar coordinates. Therefore, the complex number z can be given a polar form that depends on r and θ. The polar form of z is written z = re. (When convenient, we write reθi in place of re.)
Points         Complex numbers
Rectangular form   (x, y)                  x + iy
Polar form              (r, θ)                    re
The point with rectangular coordinates (1, 1) has polar coordinates (√2,∏/4). (Why?) Therefore, the complex number z = 1 + i has the polar form z = √2ei∏/4. A graphing calculator can convert a complex number in rectangular form to polar form and vice versa (see Fig. 2, where θ is in radians and numbers are displayed to two decimal places).
The polar–rectangular relationships lead to the following connections between the rectangular and polar forms of a complex number.

If z = re, then the number r is called the modulus, or absolute value, of z and is denoted by mod z or |z|. The angle (in radians or degrees) is called the argument of z and is denoted by arg z. Recall that (r, θ) and (r, θ + 2∏) represent the same point in polar coordinates. Therefore, z = re = rei(θ + 2∏). So the argument of a complex number is not unique. But we usually choose the argument θ so that -∏ < θ ≤ ∏ (or -180° < θ ≤ 180°).

MULTIPLICATION AND DIVISION

There is a particular advantage in representing complex numbers in polar form: multiplication and division become very easy. Theorem 1 provides the reason. (The polar form of a complex number obeys the product and quotient rules for exponents: bmbn = bm + n and bm/bn = bm - n.)
Theorem 1 says that to multiply two complex numbers, you multiply their moduli and add their arguments. Similarly, to divide two complex numbers, you divide their moduli and subtract their arguments.
We will establish the multiplication property and leave the quotient property.

POWERS—DE MOIVRE’S THEOREM

Abraham De Moivre (1667–1754), of French birth, spent most of his life in London tutoring, writing, and publishing mathematics. He belonged to many prestigious professional societies in England, Germany, and France. He was a close friend of Isaac Newton. The theorem that bears his name gives a formula for raising any complex number to the power n where n is a natural number.

THEOREM 2 De Moivre’s Theorem

If z = re and n is a natural number, then zn = rnei(nθ).

De Moivre’s theorem follows from the formula for the product of complex numbers in polar form. If n = 2 and z = re,then

z2 = rere = r2ei(2θ)

In other words, to square a complex number, you square the modulus and double the argument. Similarly, to cube a complex number you cube the modulus and triple the argument. De Moivre’s theorem says that to raise a complex number to the power n, you raise the modulus to the power n and multiply the argument by n.

COMPLEX NUMBERS AND DE MOIVRE’S THEOREM: ROOTS

Let n > 1 be an integer. A complex number w is an nth root of z if wn = z. For example, 2 and -2 are square roots (second roots) of 4 because 22 = 4 and (-2)2 = 4. Similarly, 3i and -3i are square roots of -9 because (3i)2 = -9 and (3i)29. The nth root theorem gives a formula for all of the nth roots of any nonzero complex number.

THEOREM 3 nth Root Theorem
Let n > 1 be an integer and let z = re be a nonzero complex number. Then z has n distinct nth roots given by
r1/nei(θ/n+k360°/n)    k = 0, 1, . . . , n - 1

The nth root theorem implies that every nonzero complex number z has two square roots, three cube roots, four fourth roots, and so on. Furthermore, all n of the nth roots of z have the same modulus, so they all lie on the same circle centered at the origin, and they are equally spaced around that circle.

COMPLEX NUMBERS AND DE MOIVRE’S THEOREM: HISTORICAL NOTE

There is hardly an area in mathematics that does not have some imprint of the famous Swiss mathematician Leonhard Euler (1707–1783), who spent most of his productive life at the New St. Petersburg Academy in Russia and the Prussian Academy in Berlin. One of the most prolific writers in the history of the subject, he is credited with making the following familiar notations standard:
f (x) function notation
e natural logarithmic base
i imaginary unit, √-1
For our immediate interest, he is also responsible for the extraordinary relationship eiθ = cos θ + i sin θ

If we let θ = ∏, we obtain an equation that relates five of the most important numbers in the history of mathematics:
ei∏ + 1 = 0