Sine of Sum/Proof 7

Theorem

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

Proof

Let two triangles $\triangle ABC$ and $\triangle ADC$ be inscribed in a circle on opposite sides of diameter $AC$.

By Thales' Theorem, they are both right triangles and $\angle ADC$ and $\angle ABC$ are right angles.

Let the diameter $AC = 1$.

Let $\angle DAC = \alpha$ and $\angle CAB = \beta$.

From the construction above, we have the following:

$\cos \alpha = AD$

$\cos \beta = AB$

$\sin \alpha = DC$

$\sin \beta = BC$

By Length of Chord of Circle:

$DB \mathop = 2 r \map \sin {\alpha + \beta}$

Since $2r \mathop = 1$:

$DB \mathop = \map \sin {\alpha + \beta}$

By Quadrilateral is Cyclic iff Opposite Angles sum to Two Right Angles:

$\Box ABCD$ is a cyclic quadrilateral.

By Ptolemy's Theorem:

$DB \times AC = AB \times CD + BC \times AD$

Substituting:

$\map \sin {\alpha + \beta} \times 1 = \cos \beta \sin \alpha + \sin \beta \cos \alpha$

$\map \sin {\alpha + \beta} = \sin \alpha \cos \beta + \sin \beta \cos \alpha$

By Equivalence of Definitions of Sine of Angle, the definition of sine from the circle, from the triangle and as a real function are equivalent.

It follows that all real numbers $x$ and $y$ correspond to values of $\alpha$ and $\beta$ for which the proof above applies, with one exception.

The exception occurs when both $\alpha$ and $\beta$ are equal to $\dfrac {\pi} 2$.

But then the result is simply:

$ \sin {\pi} = \sin \dfrac {\pi} 2 \cos \dfrac {\pi} 2 + \sin \dfrac {\pi} 2 \cos \dfrac {\pi} 2$

$ 0 = 0 \cdot 1 + 1 \cdot 0$

The result follows.

$\blacksquare$