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Math Help - trigonometric equations

  1. #1
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    trigonometric equations

    The diagram shows part of the curve with equation y=f(x) , where f(x)=1+2sin(px+q) , p and q being positive constants and q ≤ 90. The curve cuts the y-axis at the point A and the x-axis at the points C and D. The point B is a maximum point on the curve.


    Given that the coordinates of A and C are (0, 2) and (45, 0) respectively:
    (a) Calculate the value of q.
    (b) Show that p=4.
    (c) Find the coordinates of B and D.

    I can do question a, but I am really stuck on b and c.
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  2. #2
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    Quote Originally Posted by Tweety View Post
    The diagram shows part of the curve with equation y=f(x) , where f(x)=1+2sin(px+q) , p and q being positive constants and q ≤ 90. The curve cuts the y-axis at the point A and the x-axis at the points C and D. The point B is a maximum point on the curve.


    Given that the coordinates of A and C are (0, 2) and (45, 0) respectively:
    (a) Calculate the value of q.
    (b) Show that p=4.
    (c) Find the coordinates of B and D.

    I can do question a, but I am really stuck on b and c.
    f(0) = 2 = 1 + 2\sin(0+q)

    \sin(q) = \frac{1}{2}

    q = 30

    f(45) = 0 = 1 + 2\sin(45p + 30)

    \sin(45p + 30) = -\frac{1}{2}

    45p + 30 = 210

    45p = 180

    p = 4

    B is at a maximum ...

    1+2\sin(4x+30) = 3

    4x+30 = 90

    x = 15

    B is at (15,3)

    I'll leave it for you to find the coordinates of D
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  3. #3
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    The graph looks like a slightly-delays cosine curve, one that isn't quite up to its maximum value at the y-axis. The curve appears to be shifted upward somewhat.

    You are given that y = 1 + 2sin(px + q), so this is really a sine curve that was shifted backwards somewhat. The curve starts back before the y-axis, already having nearly reached its maximum by the time it arrives (horizontally) at the y-axis. The upward shift is seen, from the above, to be 1. The amplitude is given as being 2.

    You are given that the y-intercept of the curve is at the point A, having coordinates (0, 2). Then:

    2 = 1 + 2sin(p*0 + q)

    2 = 1 + 2sin(q)

    1 = 2sin(q)

    1/2 = sin(q)

    What then must be the measure of q?

    The point B appears to be the maximum point of the curve. The points C and D are the first and second x-intercepts, respectively.

    You are given that C has coordinated (45, 0), with "45" being in degrees. Then:

    0 = 1 + 2sin(p*45 + q)

    -1/2 = sin(p*45 + q)

    What then must be the value of 45p + q? You have already found (from above) the value of q, so what then must be the value of p?

    If you get stuck, please reply showing how far you have gotten.

    Thank you!
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  4. #4
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    for part 'b' I dont understand where 210 comes from ?

     sin(45p + 30 ) = -\frac{1}{2}

     45p + 30 = -30
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  5. #5
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    Quote Originally Posted by Tweety View Post
    for part 'b' I dont understand where 210 comes from ?

     sin(45p + 30 ) = -\frac{1}{2}

     45p + 30 = -30
    p and q being positive constants ...
    your equation above makes p < 0
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  6. #6
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    Quote Originally Posted by skeeter View Post
    your equation above makes p < 0
    Does that not mean that you could add any number to make it positive? And would you not have to add it to both sides?
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  7. #7
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    Quote Originally Posted by Tweety View Post
    Does that not mean that you could add any number to make it positive? And would you not have to add it to both sides?
    adding any number to both sides of the equation will not change the value of p
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