Posted by Waldo on Monday, March 14, 2011 at 6:30pm.
Cos2T= sin^2T-Cos^2T=1-2cos^2T
3-6cos^2T=7cosT
let u=cosT
6u^2+7u-3=0
(3u-1)(2u+3)=0
u= 1/3 u= -2/3
so what angles are those?
3cos(2Ø)=7cos(Ø)
3(2cos^2 Ø - 1) = 7cosØ
6cos^ Ø - 7cosØ - 3 = 0
(3cosØ+1)(2cosØ-3) = 0
cosØ = -1/3 or cosØ = 3/2, the last is not possible
cosØ = -1/3
Ø must be in II or III
Ø = 180 - 70.529 or Ø = 180 + 70.529
= 109.471° or 250.053°
I assume you know how to change those decimals to degrees and minutes.
I must respectfully disagree with bob
cos 2T = cos^2 T - sin^2 T = 2cos^2 T - 1
bob had it backwards, time for a strong coffee.
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