Trigonometry

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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)

  • Trigonometry -

    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

  • Trigonometry -

    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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