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    Can someone please show me how to do parts a b and c on this question, thanks...........
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    (Original post by melbourne)
    Can someone please show me how to do parts a b and c on this question, thanks...........
    If you sub in the values of x = pi/2 and y = -1 then you get an equation in a and b

    Do the same for the other given values of x and y and you have a second equation in a and b.

    Then solve these simultaneous equations
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    When you find the modulus, anything negative becomes positive. If you imagine the y-axis to be the f(x)-axis, then any negative y becomes it's positive counterpart for the same value of x.
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    (Original post by seankhn)
    When you find the modulus, anything negative becomes positive. If you imagine the y-axis to be the f(x)-axis, then any negative y becomes it's positive counterpart for the same value of x.
    1) I thought the modulus part was only for a)

    2) You have the equation

    y=a+bcosec x

    so

    -1=a+b...... and what about the cosec x??????

    Sorry im being a bit dim today im tired! lol
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    (Original post by melbourne)
    1) I thought the modulus part was only for a)

    2) You have the equation

    y=a+bcosec x

    so

    -1=a+b...... and what about the cosec x??????

    Sorry im being a bit dim today im tired! lol
    cosecx = 1/sinx
    Does that help?
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    (Original post by Widowmaker)
    cosecx = 1/sinx
    Does that help?
    no not really i knew that

    I just dont get why you can get rid of the cosec x and simply have y= a+b and y= a-b (which i still dont understand is a minus)......

    :confused: lol
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    (Original post by melbourne)
    no not really i knew that

    I just dont get why you can get rid of the cosec x and simply have y= a+b and y= a-b (which i still dont understand is a minus)......

    :confused: lol
    lol, i'm only trying to help.

    f(x) = a + bcosecx = a + b/sinx
    Points on graph are; (½π,-1) and (3π/2,-5)

    -1 = a + b/sin[½π]
    -1 = a + b/1
    -1 = a + b (1)

    -5 = a + b/sin[3π/2]
    -5 = a + b/-1
    a - b = -5
    a = -5 + b (2)

    (2) into (1)
    -1 = -5 + b + b
    2b = -1 + 5
    b = 4/2 = 2

    a - b = -5
    a = -5 + b
    a = -5 + 2
    a = -3

    Just sub the values that you know lie on the graph into the equation to form simultaneous equations.
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    (Original post by Widowmaker)
    lol, i'm only trying to help.

    f(x) = a + bcosecx = a + b/sinx
    Points on graph are; (½π,-1) and (3π/2,-5)

    -1 = a + b/sin[½π]
    -1 = a + b/1
    -1 = a + b (1)

    -5 = a + b/sin[3π/2]
    -5 = a + b/-1
    a - b = -5
    a = -5 + b (2)

    (2) into (1)
    -1 = -5 + b + b
    2b = -1 + 5
    b = 4/2 = 2

    a - b = -5
    a = -5 + b
    a = -5 + 2
    a = -3

    Just sub the values that you know lie on the graph into the equation to form simultaneous equations.
    lol sorry widowmaker i didnt mean to sound rude! hehe

    I repped you for the help
 
 
 
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