If cos u = 4/5 and tan v = – 12/5 while (3pi)/2 < u < 2pi and pi/2 < v < pi , find : sin(u-v)

Question

If cos u = 4/5 and tan v = – 12/5 while (3pi)/2 < u < 2pi and pi/2 < v < pi , find : sin(u-v)

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MichaelMet 4 years 2021-08-18T20:13:34+00:00 1 Answers 22 views 0

Answers ( )

  1. mocmien2
    0
    2021-08-18T20:15:12+00:00
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    Expand sin(u – v) using the angle sum formula for sine:

    sin(u – v) = sin(u) cos(v) – cos(u) sin(v)

    Recall the Pythagorean identity: for all x,

    cos²(x) + sin²(x) = 1

    Dividing both sides by cos²(x) gives

    1 + tan²(x) = sec²(x)

    (since tan(x) = sin(x)/cos(x) and sec(x) = 1/cos(x))

    With 3π/2 < u < 2π, we expect sin(u) < 0, and with π/2 < v < π, we expect cos(v) < 0 and sin(v) > 0.

    Solve for sin(u) :

    sin²(u) = 1 – (4/5)²

    sin(u) = – √(1 – 16/25)

    sin(u) = – √(9/25)

    sin(u) = -3/5

    Solve for sec(v) :

    sec²(v) = 1 + (-12/5)²

    sec(v) = – √(169/25)

    sec(v) = -13/5

    Take the reciprocal of both sides to solve for cos(v) :

    1/sec(v) = 1 / (-13/5)

    cos(v) = -5/13

    Solve for sin(v) :

    sin²(v) = 1 – cos²(v)

    sin(v) = √(1 – (-5/13)²)

    sin(v) = √(144/169)

    sin(v) = 12/13

    Now put everything together:

    sin(u – v) = (-3/5) (-5/13) – (4/5) (12/13) = -33/65

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