Algebraic equivalence of the Law of Sines, Pythagorean Theorem, and the Law of Cosines
I.
Lemma:
In right triangle ABC, let the sides a, b, and c represent the sides opposite angles A, B, and C,
respectively. Then if C is a right angle, we have:
One cannot define
in such a triangle, since the definition would lead to a degenerate triangle –
essentially two concurrent and congruent line segments. So to define this measure we will look at
as extended to angles that are not acute.
In fact it can be shown that
is a continuous function for all real numbers. A rigorous
proof would require an epsilon-delta argument that I will not reproduce here but can be found on
various internet sites and in many textbooks. However I will highlight a few key facts that lead to
this conclusion:
1) In such triangle,
and
. Also
and
. As A gets closer to 90, B gets closer to 0, and vice-versa, so also
and
. Also
and
.
2) Using the addition formula for sine (see II. below),
and consequently:
,
3)
4)
since both A + B = 90 and C = 90
5)
by definition
6) Since the limit and the value are equal, the sine function is continuous at C = 90.
7) Further evidence can justify why this value is 1. Consider the law of sines
. Multiply all three fractions by
triangle,
and you will have three area formulas for this
. But since C is a right angle,
.
. Therefore
II.
Lemma:
This identity is just a restatement of Ptolemy's theorem as outlined at:
http://www.cut-the-knot.org/proofs/sine_cosine.shtml
III.
The Pythagorean Theorem - A trigonometric proof using the Law of Sines
1) In a right triangle ABC, consider the special case of the Law of Sines where angle C has a
sine value of 1:
2)
3) Multiplying all three fractions by
, we have:
4) The sum of the squares of the first two terms equals twice the square of the last one:
5) Now adding
6) Since
to both sides we have:
, we can use co-function identities. On the left, let
. Also on the right let
and
, leaving:
and
7) Factoring the left side along with using the addition formula for sine and simplifying the
right side:
8) Now
, so:
9) Multiplying both sides by
, factoring and dividing we have:
10) Addendum – a much simpler proof of the Pythagorean is possible simply from the addition
formula if one considers the argument in 6) – 8) and the lemmas in I and II. For acute angles
A and B:
IV.
The Law of Cosines – a derivation from the Pythagorean Theorem
There are three cases for triangle ABC: acute, right, and obtuse. In each let the sides a, b,
and c be the lengths of the sides opposite the respective angles.
1) The acute case. Consider the diagram below:
A
Here
and therefore
. Also
. Using triangle ADB, we have
Since
B
D
, then
C
2) The right triangle case. Consider the diagram below:
A
B
The Pythagorean Theorem has already been established in III,
but here we can make an argument similar to Lemma I that
. This reduces the Law of Cosines to:
C
3) The obtuse triangle case. Consider the diagram below:
A
Let
Since
B
C
D
be
and
be . Here
. Also
Using triangle ADB, we have
, and
and
.
then
V.
The Law of Sines derived from the Law of Cosines – using Mollweide’s Formula
1) Mollweide’s Formulas:
The formulas are generally used for checking solutions to triangles, since they contain all
three angles and all three sides in the same equation. Below are two forms of the formula:
2) The Law of Cosines
3) Consider the equation
divide both sides by . We then have:
. Subtract
from both sides and also
4) Now substituting expressions from Mollweide’s formulas, we have:
5) Now using half-angle identities to simplify the left side:
6) Repeat steps 3 through 5 but instead, subtract
by . This will lead to:
from both sides and also divide both sides
7) Using the other forms of the law of cosines from 2), along with the same procedure outlined
in 3), 4), and 5) we have:
8) Now set any two identical fractions that are identical from the Law of Sines equal:
Because sine is an odd function, the denominators are opposites, leaving:
This produces an identity
altitude from angle C to side c.
Using the other terms
or
other altitudes in the triangle.
that can be clearly seen by drawing the
in this manner will produce identities that stem from the
9) Since all of the forms of the law of cosines are equivalent, it then holds that these identities all
are equivalent, leaving:
And therefore the Law of Sines is derived from the Law of Cosines.
VI.
As long as one angle of a given triangle is a right angle, along with I, II, III, IV, and V we
have that the Law of Sines, the Pythagorean Theorem, and the Law of Cosines are all
equivalent statements.