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Cos, tan, csc problem

  1. Dec 10, 2004 #1

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    doing a problem but cant figure out the answer:


    Given the following, evaluate csc(theta):

    cos(theta) = 1/2

    tan(theta) = negative

    a> 2

    b> -2(square root of)3/3

    c> 2(square root of)3/3

    d> -2

    any help?
     
  2. jcsd
  3. Dec 10, 2004 #2
    i do not understand the part with the a b c and d in it but i can tel lyou that if you look at your unit circle (CAST) rule then you can fiure out the point where the cos is positive and the tan is negative

    if cos theta = 0.5 and tan theta < 0 then theta can onlky lie in the first or fourth quadrant i.e. between 0 and 90 and 270 and 360
    or if you care in radians 0 to pi/2 and 3pi/2 and 2pi
     
  4. Dec 10, 2004 #3
    if the cos(theta) = 1/2, then (theta) = pi/3, or 5pi/3, on the interval [0, 2pi],
    since it is given that tan(theta) = sin(theta)/cos(theta) < 0,
    sin(theta)/(1/2) = 2sin(theta) < 0 or sin(theta) < 0,
    therefore (theta) has to be (5pi/3)
    now, csc(theta) = 1/sin(theta) = 1/(-3/(2sqrt[3]) = -2sqrt[3]/3
     
    Last edited: Dec 10, 2004
  5. Dec 10, 2004 #4

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    thx for the input, now here's another:

    what is the phase shift for 3sin (3x+(pi/2))

    a> -pi/6

    b> 3

    c> pi/2

    d> -3pi/2

    *the a,b,c,& d are multiple choices on the assignment I'm doing.
     
  6. Dec 10, 2004 #5

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    nevermind, i got it.
     
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