Cosine of Sum/Proof 4

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Theorem

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


Proof



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 Congruence, $\triangle DAB \cong \triangle CAE$.

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
\(\ds \leadsto \ \ \) \(\ds \map \cos {a + b}\) \(=\) \(\ds \cos a \cos b - \sin a \sin b\)

$\blacksquare$