# Derive the sine sum formula using the geometrical construction given in the figure?

Aug 25, 2016

$\sin \left(\alpha + \beta\right) = \sin \left(\alpha\right) \cos \left(\beta\right) + \cos \left(\alpha\right) \sin \left(\beta\right)$

#### Explanation:

First, construct two additional lines and label the intersections/vertices as follows:

Proceeding, note that from the right triangle $\triangle A C B$, we have
$\sin \left(\alpha + \beta\right) = \frac{A C}{A B} = \frac{A C}{1} = A C$

Then, it remains to calculate $A C$. To do so, we will calculate $A F$ and $F C$ separately, and then add them together.

From the right triangle $\triangle B D E$, we have
$\sin \left(\alpha\right) = \frac{E D}{B E} = \frac{E D}{\cos} \left(\beta\right)$

$\implies E D = \sin \left(\alpha\right) \cos \left(\beta\right)$

$\implies F C = E D = \sin \left(\alpha\right) \cos \left(\beta\right)$

Next, as the sum of the angle of $\triangle B D E$ is $\pi$, we have $\angle B E D = \pi - \frac{\pi}{2} - \alpha = \frac{\pi}{2} - \alpha$.

As $\angle A E B = \frac{\pi}{2}$, this gives $\angle A E D = \frac{\pi}{2} + \left(\frac{\pi}{2} - \alpha\right) = \pi - \alpha$.

Then, as $\angle F E D = \frac{\pi}{2}$ by construction, we have $\angle A E F = \pi - \alpha - \frac{\pi}{2} = \frac{\pi}{2} - \alpha$.

Looking at the right triangle $\triangle A F E$, then, we can calculate $\angle F A E$ as
$\angle F A E = \pi - \frac{\pi}{2} - \left(\frac{\pi}{2} - \alpha\right) = \alpha$.

Calculating the cosine of that angle, we obtain
$\cos \left(\alpha\right) = \frac{A F}{A E} = \frac{A F}{\sin} \left(\beta\right)$

$\implies A F = \cos \left(\alpha\right) \sin \left(\beta\right)$

Putting these together, we get our final result:

$\sin \left(\alpha + \beta\right) = A C$

$= F C + A F$

$= \sin \left(\alpha\right) \cos \left(\beta\right) + \cos \left(\alpha\right) \sin \left(\beta\right)$