Equivalence of Definitions of Cotangent of Angle

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\theta$ be an angle.


The following definitions of the concept of Cotangent of Angle are equivalent:

Definition from Triangle

SineCosine.png

In the above right triangle, we are concerned about the angle $\theta$.

The cotangent of $\angle \theta$ is defined as being $\dfrac {\text{Adjacent}} {\text{Opposite}}$.

Definition from Circle

Consider a unit circle $C$ whose center is at the origin of a cartesian coordinate plane.


CotangentFirstQuadrant.png


Let $P$ be the point on $C$ in the first quadrant such that $\theta$ is the angle made by $OP$ with the $x$-axis.

Let a tangent line be drawn to touch $C$ at $A = \left({0, 1}\right)$.

Let $OP$ be produced to meet this tangent line at $B$.


Then the cotangent of $\theta$ is defined as the length of $AB$.


Proof

Definition from Triangle implies Definition from Circle

Let $\cot \theta$ be defined as $\dfrac {\text{Adjacent}} {\text{Opposite}}$ in a right triangle.

Consider the triangle $\triangle OAB$.

By construction, $\angle OAB$ is a right angle.


From Parallelism implies Equal Alternate Angles:

$\angle OBA = \theta$

Thus:

\(\displaystyle \cot \theta\) \(=\) \(\displaystyle \frac {AB} {OA}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {AB} 1\) as $OA$ is the radius of the unit circle
\(\displaystyle \) \(=\) \(\displaystyle AB\)

That is:

$\cot \theta = AB$

$\Box$


Definition from Circle implies Definition from Triangle

Let $\cot \theta$ be defined as the length of $AB$ in the triangle $\triangle OAB$.

Compare $\triangle OAB$ with $\triangle ABC$ in the diagram above.

From Parallelism implies Equal Alternate Angles:

$\angle OBA = \theta$

We have that:

$\angle CAB = \angle OBA = \theta$
$\angle ABC = \angle OAB$ which is a right angle

Therefore by Triangles with Two Equal Angles are Similar it follows that $\triangle OAB$ and $\triangle ABC$ are similar.

By definition of similarity:

\(\displaystyle \frac {\text{Adjacent} } {\text{Opposite} }\) \(=\) \(\displaystyle \frac {AB} {BC}\) by definition
\(\displaystyle \) \(=\) \(\displaystyle \frac {AB} {OA}\) Definition of Similar Triangles
\(\displaystyle \) \(=\) \(\displaystyle AB\) $OA$ is Radius of the Unit Circle
\(\displaystyle \) \(=\) \(\displaystyle \cot \theta\) by definition

That is:

$\dfrac {\text{Adjacent} } {\text{Opposite} } = \cot \theta$

$\blacksquare$