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Trigonometry

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Let ABC be a triangle. Prove that (cos(A/2))^x, (cos(B/2))^x, and (cos(C/2))^x are the lengths of a triangle for any x greater than or equal to 0.

From what I have found in my books, it is impossible to solve for side lengths of a triangle using AAA b/c there is no formula to do so. It is possible to find similar triangles, though. From talking to the professor, he said that the side lengths are for another triangle. When I run "sample problems" the lengths that I get are less than or equal to 1. I understand by closure that the (A/2) is just an ordinary angle and that they are all positive numbers. I also believe that one of the original angles would have to be 90 degrees. I have also thought that the unit circle might be the key to the problem but I could not get that to work either. I'm sure that this problem isn't as hard as I think it is, but I cannot figure it out for the life of me.

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Solution Summary

The expert uses triangle ABC to show that a form is able to find the side lengths.

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Let ABC be a triangle. Prove that (cos(A/2))^x, (cos(B/2))^x, and (cos(C/2))^x are the lengths of a triangle for any x greater than or equal to 0.

Three edges make a triangle if the sum of any two edges is greater than the third.
If form the sides of a triangle for any x , we need to show that
(a)
(b)
(c)
Let us show the very first result and the other ...

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