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Pertanyaan

jika sin (A - π /4) - 5cos (A- π /4) = 0 maka tan A =..

1 Jawaban

  • sin (A - π /4) - 5cos (A- π /4) = 0

    sin(A - π /4) = -cos(A)*sin(π /4) + sin(A)*cos(π /4)
                        = -cos(A)*√2/2 + sin(A)*√2/2
                        = -(√2/2)cos(A) + (√2/2)sin(A)

    cos(A - π /4) = cos(A)*cos(π /4) + sin(A)*sin(π /4)
                        = cos(A)*√2/2 + sin(A)*√2/2
                        = (√2/2)cos(A) + (√2/2)sin(A)

    (√2/2)cos(A) = (√2.cos(A))/2
                         = (2^(1/2).cos(A))/2
                         = cos(A)/2^(1/2)
                         = cos(A)/√2

    (√2/2)sin(A) = (√2.sin(A))/2
                         = (2^(1/2).sin(A))/2
                         = sin(A)/2^(1/2)
                         = sin(A)/√2

    = cos(A)/√2 + sin(A)/√2
    = (cos(A) + sin(A))/√2

    = 5.((√2/2)cos(A) + (√2/2)sin(A))
    = 5.((cos(A)/√2)+(sin(A)/√2)
    = 5.((cos(A)+(sin(A))/√2)

    = sin (A - π /4) - 5cos (A- π /4)
    = -(√2/2)cos(A) + (√2/2)sin(A) -5((√2/2)cos(A) + (√2/2)sin(A))
    = -((cos(A)+sin(A))/√2) - 5.((cos(A)+(sin(A))/√2)
    = -((cos(A)+sin(A)) - 5.(cos(A)+(sin(A)))/√2
    = (-4sin(A)-6cos(A))/√2
    = -2(2sin(A)-3cos(A))/√2 
    = -√2(2sin(A)-3cos(A))

    -√2(2sin(A)-3cos(A)) = 0
    -√2(2sin(A)-3cos(A))/cos(A) = 0/cos(A)
    (-2√2sin(A))/cos(A))-3√2 = 0
    (-√2(2tan(A) + 3) = 0
    (-√2(2tan(A) + 3)/-√2 = 0/-√2
    2tan(A) + 3 = 0
          2tan(A) = -3
            tan(A) = -3/2

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