Math (Trigonometry)

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What is the maximum value of

[9sinθ+2sin(θ+π/3)]^2?

  • Math (Trigonometry) -

    let y = [9sinθ+2sin(θ+π/3)]^2

    dy/dØ = (9sinθ+2sin(θ+π/3)) (9cosØ + 2cos(Ø+π/3)
    = 0 for a max/min

    so (9sinθ+2sin(θ+π/3)) =0 OR (9cosØ + 2cos(Ø+π/3)) = 0

    (Where are you getting these questions from, they are tediously long....)

    Case1:
    9sinØ + 2sin(Ø+π/3) = 0
    9sinØ + 2(sinØcosπ/3 + cosØsinπ/3) = 0
    9sinØ + 2( (1/2)sinØ + (√3/2)cosØ ) = 0
    9sinØ + sinØ + √3cosØ = 0
    10sinØ = -√3cosØ
    sinØ/cosØ = -√3/2
    tanØ = -√3/2
    Ø could be in II or IV
    in II sinØ = √3/√103 and cosØ = -10/√103

    but remember I showed that
    9sinØ + 2sin(Ø+π/3)
    = 10sinØ + √3cosØ

    then y = [9sinØ + 2sin(Ø+π/3)]^2
    = (10sinØ + √3cosØ)^2
    = (10√3/√103 + √3(-10/√103))^2
    = 0 , how about that ?

    in IV sinØ = -√3/√103 and cosØ = 10/√103
    and (10sinØ + √3cosØ)^2
    = 0 as well

    HALF WAY DONE

    Case 2:
    9cosØ + 2cos(Ø+π/3) = 0
    9cosØ + 2(cosØcos π/3 - sinØsin π/3) = 0
    9cosØ + 2( (1/2)cosØ - (√3/2)sinØ ) = 0
    9cosØ + cosØ - √3sinØ = 0
    10cosØ = √3sinØ
    10/√3 = sinØ/cosØ
    tanØ = 10/√3
    Ø could be in I or III

    in I , sinØ = 10/√103 , cosØ = √3/√103

    again, recall that
    9sinθ+2sin(θ+π/3) = 10sinØ +√3cosØ

    y = (10(10/√103) + √3(√3/√103)^2
    = (100/√103 + 3/√103)^2
    = (103/√103)^2
    = 10000/103 = appr 97.09

    in III , sinØ = -√3/√103 , cosØ = -10/√103

    y = same as above, except
    = (-103/√103)^2 = 10000/103 = appr 97.09

    so the max is 97.09 , and the minimum is 0

    Wolfram appears to confirm my answer
    http://www.wolframalpha.com/input/?i=plot+%5B9sinθ%2B2sin%28θ%2Bπ%2F3%29%5D%5E2

  • Math (Trigonometry) -

    Thank you very much! You're good! :)

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