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March 30, 2017

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if cot 2θ = 5/12 with 0≤2θ≤π find cos θ, sin θ, tan θ

  • Trigonometry - ,

    if cot 2Ø = 5/12
    then tan 2Ø = 12/5

    recognizing the familar 5-12-13 right-angled triangle, we would say

    sin 2Ø/cos 2Ø = 12/5 = (12/13) / (5/13)

    so sin 2Ø = 12/13 ----> 2sinØcosØ = 12/13
    and cos 2Ø = 2cos^ Ø -1 = 5/13
    2cos^2 Ø = 18/13
    cos^2 Ø = 9/13
    cosØ = ± 3/√13

    if cosØ = 3/√13
    in 2sinØcoØ = 12/13
    sinØ(3/√13) = 6/13
    sinØ = (6/√13)(√13/3) = 2√13/13 or 2/√13
    tanØ = 2/3

    if cosØ = -3/√13
    (only the signs will change, same method)
    sinØ = -2/√13
    tanØ = 2/3

  • Trigonometry - ,

    but, of course, cos2θ will not be -3/√13, since cot2θ>0 and 2θ<π

  • Steve - Trigonometry - ,

    Steve, you are right, should have looked at that original restriction more carefully.
    ( must have automatically assumed the usual
    0 ≤ Ø ≤ 2π )

    So you will find my other solution comical, where I assumed that domain and got solutions in all 4 quadrants :

    ...


    If cot(2Ø) = 5/12 , then
    tan(2Ø) = 12/5 and 2Ø is in either I or III

    tan 2Ø = 2tanØ/(1 - tan^2 Ø)
    let tanØ = u
    then 12/5 = 2u/(1 - u^2) , (for easier typing)

    10u = 12 - 12u^2
    12u^2 + 10u - 12 = 0
    6u^2 + 5u - 6 = 0
    (2u + 3)(3u - 2) = 0
    u = -3/2 or u = 2/3
    tanØ = -3/2 or tan u = 2/3

    case 1:
    tanØ = -3/2--> Ø in II or IV
    make a sketch of a right right-angled triangle
    hypotenuse = √(3^2+2^2) = √13
    IN II: sinØ = 2/√13 , cosØ = -3/√13, tanØ = -3/2
    IN IV: sinØ = -2/√13 , cosØ = 3/√13, tanØ = -3/2

    case 2:
    tanØ = 2/3---> Ø in I or III
    make a new sketch, hypotenuse is still √13
    IN I: sinØ = 2/√13, cosØ = 3/√13 , tanØ = 2/3
    IN III: sinØ = -2/√13 , cosØ = -3/√13, tanØ = 2/3

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