How do you derive the trig addition formulas?

2. Originally Posted by manyarrows
How do you derive the trig addition formulas?
That depends on exactly how you define the trig functions themselves. One common method is this: Give a non-negative number t, start at the point (1, 0) and measure counterclockwise around the circumference a distance t. The final point will have coordinates (cos(t), sin(t)). In other words, cos(t) is defined as the x-coordinate of that point and sin(t) is defined as the y-coordinate. Given that, consider the straight line distance between (cos(a+b), sin(a+b)) and (cos(a),sin(a)) and the straight line distance between (0, 0) and (cos(b), sin(b)). The arc between the first pair of points has length b and the arc between the second part also has length b so the chords (straight lines between them) must also have the same length.

$\sqrt{(cos(a+b)- cos(a))^2+ (sin(a+b)- sin(a))^2}= \sqrt{(cos(b)-1)^2+ sin^2(b)}$
Squaring both sides of that gives
$cos^2(a+b)- 2cos(a)cos(a+b)+ cos^2(a)+ sin^2(a+b)- 2sin(a)sin(a+b)+ sin^2(a)$
$= cos^2(b)- 2cos(b)+1 + sin^2(b)$
so $2- 2cos(a)cos(a+b)- 2sin(a)sin(a+b)= 2- 2cos(b)$
$cos(a)cos(a+b)+ sin(a)sin(a+b)= cos(b)$
Now, that isn't quite what we want but if we take x= a, y= a+b, we can "reverse" that: now b= y- a= y- x so that becomes
$cos(y- x)= cos(x)cos(y)+ sin(x)sin(y)$

Replacing x by -x and using the fact that cosine is an even function and sine is an odd function.
$cos(y+x)= cos(x)cos(y)- sin(x)sin(y)$

sin(y+x) can be done now by using the fact that $sin(a)= cos(\pi/2- a)$ and $cos(a)= cos(\pi/2- a)$.
$sin(x+y)= cos(\pi/2-x-y)= cos((\pi/2- x)cos(y)+ sin(\pi/2- x)sin(y)$
$= sin(\pi/2-(\pi/2-x))cos(y)+ cos(\pi/2- (\pi/2-x))sin(y)$
$= sin(x)cos(y)+ cos(x)sin(y)$