trigonometry

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express 3 cos x -2 sin x in th eform R cos (x + a) and hence write down the maximum and minimum values of 3 cos x - 2 sin x.

  • trigonometry -

    let 3cosx - 2sinx = Rcos(x+a)

    Rcos(x+a) = R(cosxcosa - sinxsina)
    = Rcosxcosa - Rsinxsina

    so we have the identity
    Rcosxcosa - Rsinxsina = 3cosx-2sinx
    this must be valid for any x
    so let's pick x's that simplify this

    let x = 0
    then
    Rcos0cosa - Rsin0sins = 3cos0 - 2sin0
    Rcosa = 3
    cosa = 3/R

    let x = 90°
    Rcos90cosa - Rsin90sina = 3cos90 - 2sin90
    -Rsina = -2
    sina = 2/R

    but sin^2a + cos^2a = 1
    4/R^2 + 9/R^2 = 1
    R^2 = 13
    R = √13

    also : sina/cosa = (2/R) / (3/R) = 23
    tana = 2/3
    a = arctan (2/3) = 33.69°

    thus 3cosx - 2sinx = √13cos(x + 33.69°)

    check by taking any angle x
    let x = 26°
    LS =1.8196...
    RS = √13 cos(5969) = 1.8196
    ........ how about that !!

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