Guest clover Posted June 17, 2004 Share 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 Link to comment Share on other sites More sharing options...
Bryn Posted June 18, 2004 Share 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. Link to comment Share on other sites More sharing options...
bloodhound Posted June 18, 2004 Share Posted June 18, 2004 i was always rubbish at trig. Link to comment Share on other sites More sharing options...
Dave Posted June 19, 2004 Share 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. Link to comment Share on other sites More sharing options...
dryan Posted June 19, 2004 Share Posted June 19, 2004 "Deriving sin(a+b), cos(a+b)" http://www-istp.gsfc.nasa.gov/stargaze/Strig5.htm Link to comment Share on other sites More sharing options...
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