Thread: trig question
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Old November 21st, 2008, 05:17 AM
HallsofIvy HallsofIvy is offline
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Quote:
Originally Posted by brian311 View Post
Hello,

If I have a problem calling for cos(u) but elsewhere in the same problem I've calculated 3u as follows:

3u=(cos^-1(B/A)) --where B and A are symbolic constants

Is there a way of adjusting the cos(u) to accept the 3u instead so that my cos and cos^-1 will cancel out?

Thanks,
cos(A+ B)= cos(A)cos(B)- sin(A)sin(B) and sin(A+ B)= sin(A)cos(B)+ cos(A)sin(B).

In particular, cos(3u)= cos(2u+ u)= cos(2u)cos(u)- sin(2u)sin(u)

By the same argument cos(2u)= cos(u+ u)= cos^(u)- sin^2(u) and sin(2u)= sin(u+ u)= 2 sin(u)cos(u) so cos(3u)= (cos^2(u)- sin^2(u))cos(u)- (2 sin(u)cos(u))sin(u)
= cos^3(u)- sin^2(u)cos(u)- 2sin^(u)cos(u)= cos^3(u)- 3sin^2(u)cos(u)
= cos^3(u)- 3(1- cos^2(u))cos(u)= cos^3(u)- 3cos(u)+ 3cos^3(u)

= 4 cos^3(u)- 3 cos(u)
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