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tan(θ + ϕ); cos(θ) = − 1/3

θ in Quadrant III, sin(ϕ) = 1/4 ϕ in Quadrant II

I am really struggling with this idk why I keep getting the wrong answers

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3 answers
  1. me either. why don't you show your work? I've done several already; your turn. I'm sure we can pinpoint where you go astray.

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  2. i was given this equation Tan(a+b)=sin(a+b)/cos(a+b)

    So I did (square root 8/3)(square root 15/4)+sin(1/4)cos(-1/3) divide that by (-1/3)(square root 15 over 4) - sin( square root 8 over 3)sin(1/4)

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  3. what are all those sin() stuff still hanging around? The final expression should just be a bunch of fractions.

    tan(θ + ϕ); cos(θ) = − 1/3
    θ in Quadrant III, sin(ϕ) = 1/4 ϕ in Quadrant II

    You really need to review the signs of the trig functions in the various quadrants.

    In QIII,
    sinθ = -√8/3
    cosθ = -1/3

    In QII,
    sinϕ = 1/4
    cosϕ = -√15/4

    tan(θ+ϕ) = sin(θ+ϕ)/cos(θ+ϕ)
    = (sinθcosϕ+cosθsinϕ)/(cosθcosϕ-sinθsinϕ)
    = ((-√8/3)(-√15/4)+(-1/3)(1/4))/((-1/3)(-√15/4)-(-√8/3)(1/4))
    = (32√2-9√15)/7

    Or, using tan(θ+ϕ) = (tanθ+tanϕ)/(1-tanθ*tanϕ)

    tanθ = sinθ/cosθ = √8
    tanϕ = sinϕ/cosϕ = -1/√15

    tan(θ+ϕ) = (√8 - 1/√15)/(1+√8/√15) = (32√2-9√15)/7

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