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December 22, 2014

December 22, 2014

Posted by **Chris** on Monday, April 26, 2010 at 9:43pm.

Cos(2 Arccos(5/13))

2. Solve the following equation for

0° ≤ Θ < 360°

Sec(Θ)= tan(Θ) + cos(Θ)

- Trigonometry -
**Reiny**, Monday, April 26, 2010 at 11:26pmCos(2 Arccos(5/13))

let's look at the arccos 5/13

we would have a right-angled triangle with angle Ø so that cos Ø = 5/13

so then we want cos(2Ø)

but cos 2Ø = 2cos^2 Ø - 1

= 2(25/169) - 1 = -119/169

- Trigonometry -
**Reiny**, Monday, April 26, 2010 at 11:31pmSec(Θ)= tan(Θ) + cos(Θ)

1/cosØ = sinØ/cosØ + cosØ

1 = sinØ + cos^2Ø

1 = sinØ + 1 - sin^2Ø

sin^2Ø - sinØ = 0

sinØ(sinØ - 1) = 0

sinØ = 0 ----> Ø = 0,180,360

or sinØ = 1 ---> Ø = 90

so Ø = 0,90,180, and 360°

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