cos(A+B)=cosAcosB-sinAsinB

bcos(A+B)=bcosAcosB-asin^2B

You can use the triangle that i've drawn in the attachment with this thread.

attachment.doc

i just spent the last week working on this kind of trig (it's a barstard) so lets see if i can remeber.

 

.. nope damn well forgoten it, i'll get me coat.

i was always rubbish at trig.

Only thing I can really think of is that angle ACD is A+B, and then try using some sin rules to get rid of lengths, but could be completely wrong.

"Deriving sin(a+b), cos(a+b)"

http://www-istp.gsfc.nasa.gov/stargaze/Strig5.htm