# Equivalence of Definitions of Cosine of Angle

## Theorem

Let $\theta$ be an angle.

The following definitions of the concept of cosine are equivalent:

### Definition from Triangle

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 coordinate plane.

Let $P = \left({x, y}\right)$ 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 $AP$.

## 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 Interior Angles, $\angle OPA = \theta$.

Thus:

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

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 Interior 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:

 $\displaystyle \frac {\text{Adjacent} } {\text{Hypotenuse} }$ $=$ $\displaystyle \frac {AB} {AC}$ by definition $\displaystyle$ $=$ $\displaystyle \frac {AP} {OP}$ by definition of similarity $\displaystyle$ $=$ $\displaystyle AP$ as $OP$ is the radius of the unit circle $\displaystyle$ $=$ $\displaystyle \cos \theta$ by definition

That is:

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

$\blacksquare$