## Thursday, March 30, 2006

### History of Trigonometric Terms

Trigonometry means the study of trigons (triangles). The term was first used by Bartholomew Pitiscus in 1595. If anyone is not familiar with the definitions for sine, cosine, or tangent, go to the "Right Triangle Definitions" here.

Trigonometry itself emerged from astronomy. The Babylonians, for example, showed evidence of using trigonometric functions and it is from them that we have the concept of a circle consisting of 360 degrees.

The first known person to publish a table of trigonometric functions was the Greek mathematician Hipparchus around 140 B.C. Hipparchus used these functions to calculate the size of chords in a circle from a given angle. Today, Hipparchus is known as the father of trigonometry.

All of Hipparchus's major works have been lost. Most of our information about Hipparchus comes from Claudius Ptolemy. Ptolemy's work, the Almagest, stands one of history's most influential books. Almagest was not its original name. Almagest comes the Arab translation "al-majisti" which means the greatest. It presents an earth-centered universe (from Aristotle) where planetary orbits can be understood through trigonometic functions.

The first use of the sine function comes around 500 AD in the Hindu work Aryabhata. It includes a table of half chords which are called jya. This half chord table is referenced by Brahmagupta in 628 and Bhaskara in 1150.

The Arab mathematicians used the word "jiba" and "jaib" to refer to the half chord table from the Aryabhata. The term "jaib" is Arabic for "fold". So, when the Europeans began to translate the Arab texts, they used the term "sinus" (which is Latin for "fold") as the name of the half chord table. Fibonacci, for example, used the term "sinus rectus arcus". The term "cosinus" was introduced by Edmund Gunter in 1620.

The concept of tangents was well known at the time of Thales and was associated with the measurement of shadows. The first known table of tangents (referred to as the table of shadows) was presented by an Arab mathematician in 860. The term tangent itself was first used by Thomas Finke in 1583. The term cotangent was first used by Gunter in 1620.

The secant and cosecant were not used in tables until the 15th century. Copernicus for example used the concept of the secant which he called hypotenusa.

References

## Wednesday, March 29, 2006

### Equiangular Triangles

In today's blog, I am reviewing the background to the concepts of sin and cosin. The major assumption behind sin and cosin is that the ratios of the sides of right triangles can be calculated based solely on the measurement of an angle. In other words, the ratio between sides ( opposite side over hypotenuse for sin and adjacent side over hypotenuse for cosin) is constant for all similar right triangles.

Two right triangles are similar if they share the same angles but not may not share the same sides. The important idea that is presented in Euclid's Elements is that the ratio of two sides is equal to the ratio of two corresponding sides for any equiangular triangle. This property of similar triangles is enough to show that sin and cosin depend solely on the measurement of the angle. I will talk more about this in a future blog.

I will only go over enough which are necessary to establish the corresponding sides property of similar triangles.

Lemma 1: If two triangles have the same height, then the ratio of their areas is equal to a ratio of their bases.

Proof:

(1) Let a1 be the area of triangle ABC with height h and base b1

(2) Let a2 be the area of triangle DEF with height h and base b2

(3) Now a1/a2 = [(1/2)b1h]/[(1/2)b2h] = b1/b2 [See Lemma 2, here]

QED

Lemma 2: If a parallel line cuts through a triangle, it divides the sides of the triangle proportionally.

Proof:

(1) Let DE be a parallel line that cuts through the triangle ABC

(2) We can see that the areas of triangle DEB and triangle DEC are equal since:

(a) They share the same base DE

(b) They have the same height [Based on Lemma 2, here]

(3) Since triangle DEB and triangle DEC have the same area, we know that:

(area DEB)/(area ADE) = (area DEC)/(area ADE)

(4) Now triangle DEB and triangle ADE have the same height.

(5) So, we also know that from Lemma 1 above:
(area DEB)/(area ADE) = DB/AD

(6) Likewise, triangle AED and triangle DEC have the same height so Lemma 1 gives us:
(area DEC)/(area ADE) = EC/AE

(7) Putting this all together (steps #3, #5, and #6) gives us:
DB/AD = EC/AE

QED

Postulate 1: Parallel Postulate

If two lines intersect the same line and the sum of their intersectings angles is less than 180 degrees, these lines will eventually intersect.

For more details on the Parallel Postulate, see here. This is the most famous postulate of all Euclid's Elements.

Lemma 3: If two triangles are equiangular, then the sides about equal angles are proportional where the corresponding sides are opposite the equal angles.

Proof:

(1) Let ABC and DCE be equiangular triangles.

(2) Let us assume that BC and CE are colinear.

(3) Since ∠ ABC ≅ ∠ DCE, we know that FB is parallel to DC [See here for definition of parallel lines]

(4) Since ∠ BCA ≅ ∠ CED, we know that AC is parallel to FE [See here for definition of parallel lines]

(5) By the Parallel Postulate above, we know that if we extend line AB and line DE, they will intersect at a point F.

(6) Now, from (3) and (4), ACDF is a parallelogram [See here for definition of a parallelogram]

(7) Therefore FA ≅ DC and AC ≅ FD [See Lemma 1 here]

(8) Since AC is a parallel line that cuts through triangle FBE, Lemma 2 above gives us:
BA/AF = BC/CE

which means that

BA * CE = BC * AF

and further that:

BA/BC = AF/CE

(9) Since AF ≅ DC [from #7], we have
BA/BC = DC/CE

(10) Since DC is a parallel line that also cuts through triangle FBE, we get:
BC/CE = FD/DE

(11) Now combining (#7) and (#9), we get:
BC/CE = FD/DE = AC/DE

which means that:

BC * DE = AC * CE

and further that:

BC/AC = CE/DE

(12) Combining (#11) and (#9) gives us:

BA/BC = DC/CE [from #9]
BC/AC = CE/DE [from #11]

So that we have:

BA*CE = BC*DC
BC*DE = AC*CE

And we have:

BA*CE*BC*DE = BC*DC*AC*CE

So that we can divide out CE and BC to get:

BA*DE = AC*DC

And finally that:

BA/AC = DC/DE

(13) So we are done since we have:
BA/BC = DC/CE [Step #9]
BC/AC = CE/DE [Step #11]
BA/AC = DC/DE [Step #12]

QED

References

## Tuesday, March 28, 2006

### Area of Triangles

In today's blog, I present some very elementary proofs regarding area. This is needed by the proofs on similar triangles which I use as background for sin and cosin.

I present these definitions and proofs more for a sense of completeness.

Definition 1: Rectangle

A rectangle is a parallelogram where all angles are 90 degrees.

Definition 2: Area of a rectangle

The area of a rectangle is width * height.

Definition 3: Right Triangle

A right triangle is triangle where one of its angles is 90 degrees.

Lemma 1: The area of a parallelogram is base * height

Proof:

This follows directly from Lemma 2, here since we can construct a rectangle in the same parallel based on the base of the parallelogram.

By Lemma 2, the parallelogram will be congruent to this rectangle so the area of the parallelogram will be the same.

QED

Lemma 2: The area of any triangle is (1/2)height * base

Proof:

(1) Let ABC be a triangle

(2) Let CE be a line parallel to AB

(3) Let AF be a line parallel to BC

(4) let D be the point where CE and AF intersect.

(5) From (2) and (3), we see that ABCD is a parallelogram. [See here for definition of a parallelogram]

(6) Then triangle ABC ≅ triangle CDA by S-A-S since [See here for definition of S-A-S]:

(a) AB ≅ CD and BC ≅ DA since opposite sides of a parallelogram are congruent [See here for proof]

(b) ∠ ABC ≅ ∠ CDA since opposite angles of a parallelogram are congruent [See here for proof]

(7) Now since the area of the parallelogram is itself is base*height (see Lemma 1 above), the area of each triangle is (1/2)base*height.

QED

References