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Trigonometry question solving help

  1. Oct 16, 2007 #1
    1. The problem statement, all variables and given/known data

    Use the algebraic method to write √2 sin x - √2 cos x in the form
    kcos(x + ϕ) where k>0, 0≤ ϕ ≤ 2π

    2. Relevant equations

    auxiliary form??

    3. The attempt at a solution

    my working out:

    1st:

    A= √2 B= -√2

    K = √a^2 + b^2

    = √(√2)^2 + (-√2)^2

    = √2+2 = √4 = 2 therefore k = 2

    2nd: find ϕ

    let √2 sin - √2 cos x = k cos (x+ϕ )

    = k [ cosx cosϕ - sinx sinϕ ]

    therefore √2 sinx - √2 cosx = kcosx cosϕ -ksinx sinϕ

    equate: sinx => √2 = - ksinϕ

    sinϕ = - √2/k

    therfore sinϕ = - √2/2 (k = 2)

    cosx => -√2 = kcosϕ

    cos ϕ = -√2/K

    cos ϕ= -√2/√2

    sinϕ = -√2/√2 < 0 therefore ϕ in 3rd quadrant

    cosϕ = -√2/√2 < 0 therefore ϕ in 3rd quadrant

    tan x = -1

    x tan ^-1 (1) = 45

    = 225. π/180

    ϕ = 5π/4

    therefore √2sinx--√2cosx = k cos(x+ϕ)

    = √2sinx - √2cosx = 2cos (x + 5π/4)

    therefore 2cos (x+ 5π/4)

    is this correct??
    and is there any easier way to show this or solve this ?? if so can u show me the full method

    Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 16, 2007 #2

    mjsd

    User Avatar
    Homework Helper

    welcome!

    looks good to me
     
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