# Cosine of Sum/Proof 4

## Theorem

$\map \cos {a + b} = \cos a \cos b - \sin a \sin b$

## Proof $AB$, $AC$, $AE$, and $AD$ are radii of the circle centered at $A$.

Let $\angle BAC = a$ and $\angle DAC = \angle BAE = b$.

By Euclid's First Postulate, we can construct line segments $BD$ and $CE$.

By Euclid's second common notion, $\angle DAB = \angle CAE$.

Thus by Triangle Side-Angle-Side Equality, $\triangle DAB \cong \triangle CAE$.

Therefore, $DB = CE$.

We now assign Cartesian coordinates to the points $B$, $C$, $D$, and $E$:

 $\ds B$ $=$ $\ds \tuple {1, 0}$ $\ds C$ $=$ $\ds \tuple {\cos a, \sin a}$ $\ds D$ $=$ $\ds \tuple {\map \cos {a + b}, \map \sin {a + b} }$ $\ds E$ $=$ $\ds \tuple {\cos b, -\sin b}$ Cosine Function is Even and Sine Function is Odd

We use the definition of the distance function on the Euclidean space $\struct {\R^2, d}$ as defined by the Euclidean metric:

$\forall x, y \in \R^2: \map d {x, y} = \sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$

where $x = \tuple {x_1, y_1}, y = \tuple {x_2, y_2}$.

Thus:

$DB \cong CE \iff \map d {D, B} = \map d {C, E}$

So, plugging in the coordinates of $B, C, D, E$, we get:

 $\ds \paren {\map \cos {a + b} } - 1)^2 + \map {\sin^2} {a + b}$ $=$ $\ds \paren {\cos a - \cos b}^2 + \paren {\sin a + \sin b}^2$ $\ds \leadsto \ \$ $\ds \cos^2 \left({a + b}\right) + \sin^2 \left({a + b}\right)$  $\ds$ multiplying out left hand side $\ds {} - 2 \, \map \cos {a + b} + 1$ $=$ $\ds \paren {\cos a - \cos b}^2 + \paren {\sin a + \sin b}^2$ $\ds \leadsto \ \$ $\ds 1 - 2 \, \map \cos {a + b} + 1$ $=$ $\ds \paren {\cos a - \cos b}^2 + \paren {\sin a + \sin b}^2$ Sum of Squares of Sine and Cosine $\ds \leadsto \ \$ $\ds 2 - 2 \, \map \cos {a + b}$ $=$ $\ds \cos^2 a - 2 \cos a \cos b + \cos^2 b$ multiplying out right hand side $\ds$  $\, \ds + \,$ $\ds \sin^2 a + 2 \sin a \sin b + \sin^2 b$ $\ds \leadsto \ \$ $\ds 2 - 2 \, \map \cos {a + b}$ $=$ $\ds 2 - 2 \cos a \cos b + 2 \sin a \sin b$ Sum of Squares of Sine and Cosine $\ds \leadsto \ \$ $\ds \map \cos {a + b}$ $=$ $\ds \cos a \cos b - \sin a \sin b$

$\blacksquare$