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

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Answer 1 · 2016-08-25T20:56:37

$sin(alpha+beta)=sin(alpha)cos(beta)+cos(alpha)sin(beta)$

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

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Proceeding, note that from the right triangle $triangleACB$ , we have
$sin(alpha+beta) = (AC)/(AB) = (AC)/1=AC$

Then, it remains to calculate $AC$ . To do so, we will calculate $AF$ and $FC$ separately, and then add them together.

From the right triangle $triangleBDE$ , we have
$sin(alpha) = (ED)/(BE) = (ED)/cos(beta)$

$=> ED = sin(alpha)cos(beta)$

$=> FC = ED = sin(alpha)cos(beta)$

Next, as the sum of the angle of $triangleBDE$ is $pi$ , we have $angleBED = pi-pi/2-alpha = pi/2-alpha$ .

As $angleAEB=pi/2$ , this gives $angleAED = pi/2+(pi/2-alpha)=pi-alpha$ .

Then, as $angleFED = pi/2$ by construction, we have $angleAEF = pi-alpha-pi/2 = pi/2-alpha$ .

Looking at the right triangle $triangleAFE$ , then, we can calculate $angleFAE$ as
$angleFAE = pi-pi/2-(pi/2-alpha) = alpha$ .

Calculating the cosine of that angle, we obtain
$cos(alpha) = (AF)/(AE) = (AF)/sin(beta)$

$=> AF = cos(alpha)sin(beta)$

Putting these together, we get our final result:

$sin(alpha+beta) = AC$

$=FC+AF$

$=sin(alpha)cos(beta)+cos(alpha)sin(beta)$