Equivalence of Definitions of Cosine of Angle

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Theorem

Let $\theta$ be an angle.


The following definitions of the concept of cosine are equivalent:

Definition from Triangle

SineCosine.png

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

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

Definition from Circle

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


CosineFirstQuadrant.png


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

Let $AP$ be the perpendicular from $P$ to the $y$-axis.


Then the cosine of $\theta$ is defined as the length of $AP$.

Hence in the first quadrant, the cosine is positive.


Proof

Definition from Triangle implies Definition from Circle

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

Consider the triangle $\triangle OAP$.

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

From Parallelism implies Equal Alternate Angles:

$\angle OPA = \theta$


Thus:

\(\ds \cos \theta\) \(=\) \(\ds \frac {AP} {OP}\)
\(\ds \) \(=\) \(\ds \frac {AP} 1\) as $OP$ is the radius of the unit circle
\(\ds \) \(=\) \(\ds AP\)

That is:

$\cos \theta = AP$

$\Box$


Definition from Circle implies Definition from Triangle

Let $\cos \theta$ be defined as the length of $AP$ in the triangle $\triangle OAP$.

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

From Parallelism implies Equal Alternate Angles:

$\angle OPA = \theta$

We have that:

$\angle CAB = \angle OPA = \theta$
$\angle ABC = \angle OAP$ which is a right angle

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

By definition of similarity:

\(\ds \frac {\text {Adjacent} } {\text {Hypotenuse} }\) \(=\) \(\ds \frac {AB} {AC}\) by definition
\(\ds \) \(=\) \(\ds \frac {AP} {OP}\) Definition of Similar Triangles
\(\ds \) \(=\) \(\ds AP\) as $OP$ is the radius of the unit circle
\(\ds \) \(=\) \(\ds \cos \theta\) by definition

That is:

$\dfrac {\text {Adjacent} } {\text {Hypotenuse} } = \cos \theta$

$\blacksquare$