Cosine of Sum/Proof 4

Theorem

 * $\cos \left({a + b}\right) = \cos a \cos b - \sin a \sin b$


 * $\sin \left({a + b}\right) = \sin a \cos b + \cos a \sin b$

where $\sin$ and $\cos$ are sine and cosine.

Proof

 * [[File:Tri1.PNG]]

$AB$, $AC$, $AE$, 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 $E$:

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

$\forall x, y \in \R^2: d \left({x, y}\right) = \sqrt {\left({x_1 - y_1}\right)^2 + \left({x_2 - y_2}\right)^2}$

where $x = \left({x_1, y_1}\right), y = \left({x_2, y_2}\right)$.

Thus $DB \cong CE \iff d \left({D, B}\right) = d \left({C, E}\right)$.

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

In the above, repeated use is made of the identity $\cos^2 \theta + \sin^2 \theta \equiv 1$ from Sum of Squares of Sine and Cosine.

Now, using the identity $\cos \left({\frac \pi 2 - a}\right) = \sin a$ from Sine equals Cosine of Complement, we have: