Guest clover Posted June 17, 2004 Posted June 17, 2004 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
Bryn Posted June 18, 2004 Posted June 18, 2004 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.
Dave Posted June 19, 2004 Posted June 19, 2004 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.
dryan Posted June 19, 2004 Posted June 19, 2004 "Deriving sin(a+b), cos(a+b)" http://www-istp.gsfc.nasa.gov/stargaze/Strig5.htm
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