Thread: trig question
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Old November 20th, 2008, 11:02 PM
David24 David24 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,
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
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