Yeah dave. It is the basically same thing! Because you will need to have -45 degree or -(pi/4) + 60 degrees or (pi/3) to make +15 degrees. Therefore then do all the works.

[math]\cos{\tfrac{\pi}{12}}=\cos{(\tfrac{\pi}{3}-\tfrac{\pi}{4})}[/math]

 

[math]\cos{\tfrac{-\pi}{12}}=\cos{(\tfrac{\pi}{4}-\tfrac{\pi}{3})}[/math]

 

 

Take the double angle form on either of them.

 

[math]\cos{\tfrac{\pi}{3}-\tfrac{\pi}{4}}=\cos{\tfrac{\pi}{3}} \cdot \cos{\tfrac{\pi}{4}}+\sin{\tfrac{\pi}{3}} \cdot \sin{\tfrac{\pi}{3}}[/math]

 

[math]\cos{\tfrac{\pi}{4}-\tfrac{\pi}{3}}=\cos{\tfrac{\pi}{4}} \cdot \cos{\tfrac{\pi}{3}}+\sin{\tfrac{\pi}{3}} \cdot \sin{\tfrac{\pi}{4}}[/math]

 

You'll note that either way, you end up with the same (because it doesn't matter the order in which you multiply, a*b=b*a)

 

general case...

 

(where a-b=c)

 

[math]\cos{c}=\cos(a-b)[/math]

[math]\cos{-c}=\cos(b-a)[/math]

 

[math]\cos(a-b)=\cos{a} \cdot \cos{b} + \sin{a} \cdot \sin{b}[/math]

[math]\cos(b-a)=\cos{b} \cdot \cos{a} + \sin{b} \cdot \sin{a}[/math]

Yeah I can notice that. I thought it won't work, because I know that 1 - 1/2 is .5, and 1/2 - 1 is -.5. But as the additive and subractive identites does multipication, I guess it does not really matter.