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
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)
2 answers
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°
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°