Trigonometry
posted by Chris .
1. Find the exact value of the following (Think identity)
Cos(2 Arccos(5/13))
2. Solve the following equation for
0° ≤ Θ < 360°
Sec(Θ)= tan(Θ) + cos(Θ)

Cos(2 Arccos(5/13))
let's look at the arccos 5/13
we would have a rightangled 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 
Sec(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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