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January 25, 2015

January 25, 2015

Posted by **Waldo** on Monday, March 14, 2011 at 6:30pm.

- Trig -
**bobpursley**, Monday, March 14, 2011 at 6:39pmCos2T= 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?

- Trig -
**Reiny**, Monday, March 14, 2011 at 6:39pm3cos(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.

- Trig -
**Reiny**, Monday, March 14, 2011 at 6:48pmI 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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