1. Find the exact value of the following (Think identity)

Question

Cos(2 Arccos(5/13))

2. Solve the following equation for
0° ≤ Θ < 360°

Sec(Θ)= tan(Θ) + cos(Θ)

Answer 1

Cos(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

Answer 2

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°