# trig question

• Nov 20th 2008, 08:02 PM
brian311
trig question
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,
• Nov 20th 2008, 09:02 PM
David24
Quote:

Originally Posted by brian311
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,

Hey mate,

why not let 3u = t and solve for cos(t) etc, once you solve t, u is simply t/3

Hope this helps,

David
• Nov 21st 2008, 03:17 AM
HallsofIvy
Quote:

Originally Posted by brian311
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)= \$\displaystyle cos^(u)- sin^2(u)\$ and sin(2u)= sin(u+ u)= 2 sin(u)cos(u) so cos(3u)= \$\displaystyle (cos^2(u)- sin^2(u))cos(u)- (2 sin(u)cos(u))sin(u)\$
\$\displaystyle = cos^3(u)- sin^2(u)cos(u)- 2sin^(u)cos(u)= cos^3(u)- 3sin^2(u)cos(u)\$
\$\displaystyle = cos^3(u)- 3(1- cos^2(u))cos(u)= cos^3(u)- 3cos(u)+ 3cos^3(u)\$

\$\displaystyle = 4 cos^3(u)- 3 cos(u)\$