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.
$\displaystyle \sqrt{(cos(a+b)- cos(a))^2+ (sin(a+b)- sin(a))^2}= \sqrt{(cos(b)-1)^2+ sin^2(b)}$
Squaring both sides of that gives
$\displaystyle cos^2(a+b)- 2cos(a)cos(a+b)+ cos^2(a)+ sin^2(a+b)- 2sin(a)sin(a+b)+ sin^2(a)$
$\displaystyle = cos^2(b)- 2cos(b)+1 + sin^2(b)$
so $\displaystyle 2- 2cos(a)cos(a+b)- 2sin(a)sin(a+b)= 2- 2cos(b)$
$\displaystyle cos(a)cos(a+b)+ sin(a)sin(a+b)= cos(b)$
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
$\displaystyle cos(y- x)= cos(x)cos(y)+ sin(x)sin(y)$
Replacing x by -x and using the fact that cosine is an even function and sine is an odd function.
$\displaystyle cos(y+x)= cos(x)cos(y)- sin(x)sin(y)$
sin(y+x) can be done now by using the fact that $\displaystyle sin(a)= cos(\pi/2- a)$ and $\displaystyle cos(a)= cos(\pi/2- a)$.
$\displaystyle sin(x+y)= cos(\pi/2-x-y)= cos((\pi/2- x)cos(y)+ sin(\pi/2- x)sin(y)$
$\displaystyle = sin(\pi/2-(\pi/2-x))cos(y)+ cos(\pi/2- (\pi/2-x))sin(y)$
$\displaystyle = sin(x)cos(y)+ cos(x)sin(y)$