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Project Maths SEC 2011 Sample Q5

  • 02-06-2012 10:51AM
    #1
    FaoiSin
    Registered Users, Registered Users 2 Posts: 789 ✭✭✭


    Hey I was wondering would anybody be able to explain to me how to get the point P in this question. It's one of those Trig Graphs.


Comments

  • #2
    BL1993
    Closed Accounts Posts: 341 ✭✭BL1993


    2y=1 therfore y=.5

    let f(x)= .5

    sin2x= .5 (ref angle is pi/6)

    therfore, 2x= pi/6 or 2x= pi - pi/6= 5pi/6 (as sin is positive in the 1st and 2nd quadrants)

    general solutions:

    2x= pi/6 +2npi or 2x= 5pi/6 +2npi (where n is a natural number)
    -->x=pi/12 +npi or x= 5pi/12 +npi

    then sub in values for n starting at zero until you get the greatest value for x which is between pi and 2pi


  • #3
    FaoiSin
    Registered Users, Registered Users 2 Posts: 789 ✭✭✭FaoiSin


    BL1993 wrote: »
    2y=1 therfore y=.5

    let f(x)= .5

    sin2x= .5 (ref angle is pi/6)

    therfore, 2x= pi/6 or 2x= pi - pi/6= 5pi/6 (as sin is positive in the 1st and 2nd quadrants)

    general solutions:

    2x= pi/6 +2npi or 2x= 5pi/6 +2npi (where n is a natural number)
    -->x=pi/12 +npi or x= 5pi/12 +npi

    then sub in values for n starting at zero until you get the greatest value for x which is between pi and 2pi

    Ah I see. Are there many questions they can ask on these graphs?


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