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===Proof of Pythagorean theorem===

[[File:Pythagoras01.png|thumb|width=100px|Fig. 1: Geometrical construction made to prove the Pythagorean theorem.]]

Based on geometric considerations, we are going to give proof for
one of the most widely known identities in mathematics - the Pythagorean theorem. It states that the sides of a
right triangle (that is, a triangle
for which one of the inner angles measures <math>90^\circ</math> or
<math>\pi/2</math> radians) of lengths <math>a</math>, <math>b</math> and <math>c</math> satisfy the following identity:

<math>c^2 = a^2 + b^2.\qquad\qquad(1)</math>

Consider the triangle of sides <math>a</math>, <math>b</math> and <math>c</math> marked in bold
of Fig. 1. If we construct the system of triangles shown in
the figure, then it is clear that the area of the bigger square
is <math>(a+b)^2</math>. The area of all the elements contained within must
of course equal <math>(a+b)^2</math>. There are four triangles of area
<math>\frac{1}{2} a b</math> and one square of area <math>c^2</math>. Hence, it must be

<math>(a+b)^2 = 4 \frac{1}{2} a b + c^2.\qquad\qquad(2)</math>

Expanding the parenthesis in Eq. (2) we have

<math>a^2 + b^2 + \cancel{2ab} = \cancel{2 a b} + c^2,\qquad\qquad(3)</math>

or

<math>a^2 + b^2 = c^2,\qquad\qquad(4)</math>

as we wanted to prove.

===Law of cosines===

[[File:Pythagoras02.png|thumb|width=100px|Fig. 2: Non right triangle to which the law of cosines applies.]]

[[File:Pythagoras03.png|thumb|width=100px|Fig. 3: Geometric proof that <math>\sin (\pi - \alpha) = \sin \alpha</math> and
<math>\cos (\pi - \alpha) = - \cos\alpha</math>.]]

The Pythagorean theorem is only valid for right triangles. There is
a more general theorem, the ''law of cosines'', that relates the lengths
of the sides of any given triangle as

<math>c^2 = a^2 + b^2 - 2ab \cos\alpha,\qquad\qquad(5)</math>

where <math>\alpha</math> is the angle formed by the sides of lengths <math>a</math> and <math>b</math>, and
opposite <math>c</math>, as shown in Fig. 2. In that figure, the side of
length <math>a</math> is extended to form a right triangle that will help
prove the identity given by Eq. (5).

It is clear from Fig. 2 that <math>\beta = \pi - \alpha</math>, in radians,
or <math>\beta = 180^\circ - \alpha</math> in degrees. Here
we prefer to work in radians, and so will use the former expression.
Although usually it would suffice to say that

<math>\cos \beta = \cos \left( \pi - \alpha \right) = - \cos \alpha \qquad \text{and}</math>

<math>\sin \beta = \sin \left( \pi - \alpha \right) = \sin \alpha,\qquad\qquad(6)</math>

we are trying to
make as few assumptions as possible. Therefore the geometric proof to
Eq. (6) is given in Fig. 3.
Having constructed a right triangle in Fig. 2
we can now relate the length of <math>c</math> to the other sides of
the bigger triangle through the Pythagorean theorem:

<math>c^2 = \left( a - b\cos\alpha \right)^2 + \left( b\sin\alpha \right)^2.\qquad\qquad(7)</math>

Expanding the powers in parenthesis, one easily gets to

<math>c^2 = a^2 + b^2 \cos^2 \alpha -2ab\cos\alpha + b^2\sin^2\alpha.\qquad\qquad(8)</math>

Given the well known trigonometric relation
<math>\cos^2\alpha + \sin^2\alpha = 1</math> (the relation
<math>\cos^2\alpha + \sin^2\alpha = 1</math>
can easily be proven using the Pythagorean theorem for a triangle for
which <math>c=1</math> and <math>\alpha</math> is the angle between <math>a</math> and <math>c</math>. It follows
immediately that <math>a=\cos\alpha</math> and <math>b=\sin\alpha</math>.),
Eq. (8) reduces to

<math>c^2 = a^2 + b^2 - 2ab\cos\alpha,\qquad\qquad(9)</math>

which is the relation we wanted to prove.

Pythagorean theorem and law of cosines - Revision history

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