Thread: cubic equation help
View Single Post
  #6  
Old October 16th, 2008, 09:25 AM
ThePerfectHacker's Avatar
ThePerfectHacker ThePerfectHacker is offline
Global Moderator

  
Join Date: Nov 2005
Location: New York City
Posts: 11,186
Country:
Thanks: 482
Thanked 3,758 Times in 3,070 Posts
ThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond reputeThePerfectHacker has a reputation beyond repute
Default
Quote:
Originally Posted by Opalg View Post
(ThePerfectHacker had \theta = \tfrac{4\pi}{18}, but I think \theta = \tfrac{5\pi}{18} must be correct.)
Yes I made a mistake. But we are going to keep it secret.

The mistake was in the simplification. The solutions that I got to the depressed cubic are correct I just made mistakes in the identities.
Here is the corrected version:

1)2 + 2\cos \frac{2\pi}{9} = 2 - 2\cos \frac{7\pi}{9} = 4\sin^2 \frac{7\pi}{18}

2)2 + 2\cos \frac{8\pi}{9} = 2 - 2\cos \frac{\pi}{9} = 4\sin^2 \frac{\pi}{18}

3)2 + 2\cos \frac{14\pi}{9} = 2 - 2\cos \frac{5\pi}{9} = 4\sin^2 \frac{5\pi}{18}
__________________

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.


"Democracy has proved only that the best way to gain power
over people is to assure the people that they are ruling
themselves. Once they believe that, they make wonderfully
submissive slaves." - Joseph Sobran


To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
Reply With Quote