Double Angle Formulas/Cosine/Proof 4
Jump to navigation
Jump to search
Theorem
- $\cos 2 \theta = \cos^2 \theta - \sin^2 \theta$
Proof
Consider an isosceles triangle $\triangle ABC$ with base $BC$, and apex $\angle BAC = 2 \alpha$.
Draw an angle bisector to $\angle BAC$ and name it $AH$.
- $\angle BAH = \angle CAH = \alpha$
From Angle Bisector and Altitude Coincide iff Triangle is Isosceles:
- $AH \perp BC$
From Law of Cosines:
| \(\text {(1)}: \quad\) | \(\ds CB^2\) | \(=\) | \(\ds AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos 2 \alpha\) |
From Pythagoras's Theorem:
| \(\ds AC ^ 2\) | \(=\) | \(\ds CH^2 + AH^2\) | in triangle $\triangle AHC$ | |||||||||||
| \(\text {(2.1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds CH^2\) | \(=\) | \(\ds AC^2 - AH^2\) | ||||||||||
| \(\ds \) | \(\) | \(\ds \) | ||||||||||||
| \(\ds AB ^ 2\) | \(=\) | \(\ds BH^2 + AH^2\) | in triangle $\triangle AHB$ | |||||||||||
| \(\text {(2.2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds BH^2\) | \(=\) | \(\ds BC^2 - AH^2\) |
By definition of sine:
| \(\text {(3.1)}: \quad\) | \(\ds CH\) | \(=\) | \(\ds AC \sin \alpha\) | |||||||||||
| \(\text {(3.2)}: \quad\) | \(\ds BH\) | \(=\) | \(\ds AB \sin \alpha\) |
By definition of cosine:
- $AH = AB \cos \alpha = AC \cos \alpha$
So:
| \(\text {(4)}: \quad\) | \(\ds AH^2\) | \(=\) | \(\ds AB \cdot AC \cdot \cos^2 \alpha\) | |||||||||||
| \(\ds \) | \(\) | \(\ds \) | ||||||||||||
| \(\ds CH^2\) | \(=\) | \(\ds AC^2 - AH^2\) | $(2.1)$ | |||||||||||
| \(\text {(5.1)}: \quad\) | \(\ds \) | \(=\) | \(\ds AC^2 - AB \cdot AC \cdot \cos^2 \alpha\) | assigning $(4)$ | ||||||||||
| \(\ds \) | \(\) | \(\ds \) | ||||||||||||
| \(\ds BH^2\) | \(=\) | \(\ds AB^2 - AH^2\) | $(2.2)$ | |||||||||||
| \(\text {(5.2)}: \quad\) | \(\ds \) | \(=\) | \(\ds AB^2 - AB \cdot AC \cdot \cos^2 \alpha\) | assigning $(4)$ |
Now:
| \(\ds CB^2\) | \(=\) | \(\ds (CH + BH)^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds CH^2 + BH^2 + 2 \cdot CH \cdot BH\) | Square of Sum | |||||||||||
| \(\ds \) | \(=\) | \(\ds AC^2 - AB \cdot AC \cdot \cos^2 \alpha + AB^2 - AB \cdot AC \cdot \cos^2 \alpha + 2 \cdot CH \cdot BH\) | assigning $(5.1)$,$(5.2)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos^2 \alpha + 2 \cdot CH \cdot BH\) | simplifying | |||||||||||
| \(\ds \) | \(=\) | \(\ds AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos^2 \alpha + 2 \cdot AB \cdot AC \cdot \sin^2 \alpha\) | assigning $(3.1)$,$(3.2)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds AC^2 + AB^2 - 2 \cdot AB \cdot AC \paren {\cos^2 \alpha - \sin^2 \alpha}\) | simplifying | |||||||||||
| \(\ds \) | \(=\) | \(\ds AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos 2 \alpha\) | equating to $(1)$ |
Hence we get the equation:
| \(\ds AC^2 + AB^2 - 2 \cdot AB \cdot AC \paren {\cos^2 \alpha - \sin^2 \alpha}\) | \(=\) | \(\ds AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos 2 \alpha\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \cos^2 \alpha - \sin^2 \alpha\) | \(=\) | \(\ds \cos 2 \alpha\) | simplifying |
$\blacksquare$