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Homework Help: 2 Trigonometric Problems i'm stuck with. Help!

  1. Nov 25, 2009 #1
    Hi. I would appreciate if anybody could help me with the following:

    a)Solve 3cosec x-5sin x = 2 for 0<x<360degrees

    b) Prove that cosecx - sinx = cosx cotx

    a) f(x) is x + ln(3x-4). Show how to convert f(x) into the iterative formula

    b) If x0=1, write values of x1, x2, 3 until a root is achieved correct to 3 d.p.

    c) Justify your previous answer by showing a sign change of f(x)

    d) Use simpsons rule with 4 intervals to calculate the definite integral between 2 and 4 of
  2. jcsd
  3. Nov 25, 2009 #2
    in 1 (a) and (b)
    just give a hard try .... u will get the answer
  4. Nov 25, 2009 #3
    Yes, I think 1(b) is:
    = cos(x)[cos(x)/sin(x)]
    = cos²(x) / sin(x)
    = [1 - sin²(x)] / sin(x)
    = [1/sin(x)] - [sin²(x)/sin(x)]
    = csc(x) - sin(x)

    but I don't have a clue about 1(a)
    I can only get to:
    [3(1/sinx)]-5sinx-2=0 and then don't know how to carry on
  5. Nov 25, 2009 #4


    Staff: Mentor

    For 1a, multiply both sides of the equation you got by sin(x), which results in an equation that is quadratic in form. Be aware that multiplying by sin(x) might introduce extraneous solutions x = 0, x = pi that aren't solutions of the original equation.
  6. Nov 25, 2009 #5


    Staff: Mentor

    For 2, it's not clear what you are trying to do. Are you trying to find a root of the equation x + ln(3x -4) = 0?

    If that's it, you might have a typo because ln(3x -4) is undefined at x = 1.
  7. Nov 25, 2009 #6
    ... this is not a quadratic is it??
  8. Nov 26, 2009 #7


    Staff: Mentor

    1a. No it isn't. BTW, the usual abbreviation for cosecant is csc.

    You started with 3csc(x) -5sinx = 2, then rewrote this as
    1/sin(x) -5sin(x) - 2 = 0, then you reconverted 1/sin(x) back to csc(x) - not a good move.

    Continue from this equation, 1/sin(x) -5sin(x) - 2 = 0, and reread what I said in post #4.

    For 2, I still need some answers from you.
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