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 plane.
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:
- $\dfrac {\text {Adjacent} } {\text {Hypotenuse} } = \cos \theta$
$\blacksquare$