# Double Angle Formulas/Cosine

## Theorem

$\cos 2 \theta = \cos^2 \theta - \sin^2 \theta$

where $\cos$ and $\sin$ denote cosine and sine respectively.

### Corollary 1

$\cos 2 \theta = 2 \cos^2 \theta - 1$

### Corollary 2

$\cos 2 \theta = 1 - 2 \sin^2 \theta$

### Corollary 3

$\cos 2 \theta = \dfrac {1 - \tan^2 \theta} {1 + \tan^2 \theta}$

## Proof 1

 $\displaystyle \cos 2 \theta + i \sin 2 \theta$ $=$ $\displaystyle \paren {\cos \theta + i \sin \theta}^2$ De Moivre's Formula $\displaystyle$ $=$ $\displaystyle \cos^2 \theta + i^2 \sin^2 \theta + 2 i \cos \theta \sin \theta$ $\displaystyle$ $=$ $\displaystyle \cos^2 \theta - \sin^2 \theta + 2 i \cos \theta \sin \theta$ $\displaystyle \leadsto \ \$ $\displaystyle \cos 2 \theta$ $=$ $\displaystyle \cos^2 \theta - \sin^2 \theta$ equating real parts

$\blacksquare$

## Proof 2

 $\displaystyle \cos 2 \theta$ $=$ $\displaystyle \map \cos {\theta + \theta}$ $\displaystyle$ $=$ $\displaystyle \cos \theta \cos \theta - \sin \theta \sin \theta$ Cosine of Sum $\displaystyle$ $=$ $\displaystyle \cos^2 \theta - \sin^2 \theta$

$\blacksquare$

## Proof 3

Starting from the right, we have:

 $\displaystyle \cos^2 \theta - \sin^2\theta$ $=$ $\displaystyle \left({\frac 1 2 \left({e^{i \theta} + e^{-i \theta} }\right)}\right)^2 - \left({\frac 1 {2 i} \left({e^{i \theta} - e^{-i \theta} }\right)}\right)^2$ Cosine Exponential Formulation, Sine Exponential Formulation $\displaystyle$ $=$ $\displaystyle \frac 1 4 \left({e^{i \theta} + e^{-i \theta} }\right)^2 + \frac 1 4 \left({e^{i \theta} - e^{-i \theta} }\right)^2$ $i$ is the imaginary unit $\displaystyle$ $=$ $\displaystyle \frac 1 4 \left({e^{2 i \theta} + 2 + e^{-2 i \theta} + e^{2 i \theta} - 2 + e^{-2 i \theta} }\right)$ Square of Sum, Square of Difference $\displaystyle$ $=$ $\displaystyle \frac 1 2 \left({e^{2 i \theta} + e^{-2 i \theta} }\right)$ Simplifying $\displaystyle$ $=$ $\displaystyle \cos 2 \theta$ Cosine Exponential Formulation

$\blacksquare$

## Proof 4

Consider an isosceles triangle $\triangle ABC$ with base $BC$, and head angle $\angle BAC = 2 \alpha$.

Draw an angle bisector to $\angle BAC$ and name it $AH$.

$\angle BAH = \angle CAH = \alpha$
$AH \perp BC$

From Law of Cosines:

 $\text {(1)}: \quad$ $\displaystyle CB^2$ $=$ $\displaystyle AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos 2 \alpha$

From Pythagoras's Theorem:

 $\displaystyle AC ^ 2$ $=$ $\displaystyle CH^2 + AH^2$ in triangle $\triangle AHC$ $\text {(2.1)}: \quad$ $\displaystyle \leadsto \ \$ $\displaystyle CH^2$ $=$ $\displaystyle AC^2 - AH^2$ $\displaystyle$  $\displaystyle$ $\displaystyle AB ^ 2$ $=$ $\displaystyle BH^2 + AH^2$ in triangle $\triangle AHB$ $\text {(2.2)}: \quad$ $\displaystyle \leadsto \ \$ $\displaystyle BH^2$ $=$ $\displaystyle BC^2 - AH^2$

By definition of sine:

 $\text {(3.1)}: \quad$ $\displaystyle CH$ $=$ $\displaystyle AC \sin \alpha$ $\text {(3.2)}: \quad$ $\displaystyle BH$ $=$ $\displaystyle AB \sin \alpha$

By definition of cosine:

$AH = AB \cos \alpha = AC \cos \alpha$

So:

 $\text {(4)}: \quad$ $\displaystyle AH^2$ $=$ $\displaystyle AB \cdot AC \cdot \cos^2 \alpha$ $\displaystyle$  $\displaystyle$ $\displaystyle CH^2$ $=$ $\displaystyle AC^2 - AH^2$ $(2.1)$ $\text {(5.1)}: \quad$ $\displaystyle$ $=$ $\displaystyle AC^2 - AB \cdot AC \cdot \cos^2 \alpha$ assigning $(4)$ $\displaystyle$  $\displaystyle$ $\displaystyle BH^2$ $=$ $\displaystyle AB^2 - AH^2$ $(2.2)$ $\text {(5.2)}: \quad$ $\displaystyle$ $=$ $\displaystyle AB^2 - AB \cdot AC \cdot \cos^2 \alpha$ assigning $(4)$

Now:

 $\displaystyle CB^2$ $=$ $\displaystyle (CH + BH)^2$ $\displaystyle$ $=$ $\displaystyle CH^2 + BH^2 + 2 \cdot CH \cdot BH$ Square of Sum $\displaystyle$ $=$ $\displaystyle AC^2 - AB \cdot AC \cdot \cos^2 \alpha + AB^2 - AB \cdot AC \cdot \cos^2 \alpha + 2 \cdot CH \cdot BH$ assigning $(5.1)$,$(5.2)$ $\displaystyle$ $=$ $\displaystyle AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos^2 \alpha + 2 \cdot CH \cdot BH$ simplifying $\displaystyle$ $=$ $\displaystyle AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos^2 \alpha + 2 \cdot AB \cdot AC \cdot \sin^2 \alpha$ assigning $(3.1)$,$(3.2)$ $\displaystyle$ $=$ $\displaystyle AC^2 + AB^2 - 2 \cdot AB \cdot AC \paren {\cos^2 \alpha - \sin^2 \alpha}$ simplifying $\displaystyle$ $=$ $\displaystyle AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos 2 \alpha$ equating to $(1)$

Hence we get the equation:

 $\displaystyle AC^2 + AB^2 - 2 \cdot AB \cdot AC \paren {\cos^2 \alpha - \sin^2 \alpha}$ $=$ $\displaystyle AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos 2 \alpha$ $\displaystyle \leadsto \ \$ $\displaystyle \cos^2 \alpha - \sin^2 \alpha$ $=$ $\displaystyle \cos 2 \alpha$ simplifying

$\blacksquare$