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courtrigrad
Sep3-04, 07:13 AM
Hello all

In my textbook I encountered the following problem:

Find the six trigonometric values of cos^ -1 (3/7). They must be exact. I gather what they mean is that I find arccos (3/7). I tried applying basic identities, but didn't work. Any help would be appreciated.

Thanks

recon
Sep3-04, 07:42 AM
If cos^-1 (3/7) = x, do they allow you to give x a value of more than 360 degrees?

HallsofIvy
Sep3-04, 08:30 AM
3/7?? I'm going to have to think about that!

courtrigrad
Sep3-04, 09:33 AM
in the answer book it says cos (theta) = 3/7

BobG
Sep3-04, 10:06 AM
cos^{-1}(\frac{3}{7}) is saying "the angle whose cosine is 3/7", so the cosine is already given. You don't need to actually figure out the angle, since the sine, cosine, tangent, etc. all have set relationships between each other.

It would help to have a range for the angle, though, since the sign of the sine, tangent, and cosecant are all going to depend on whether

cos^{-1}(\frac{3}{7}) lies between 0 and \frac{\pi}{2} or between 0 and -\frac{\pi}{2}