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Homework Help: Simultaneous eq using trig.

  1. Oct 31, 2009 #1
    1. The problem statement, all variables and given/known data

    The simultaneous equations in x,y
    (cos a)x - (sin a) y = 2
    (sin a)x + (cos a)y = 1

    are solvable
    a) for all values of a in the range 0 < a < pi
    b) except for one value of a in the range 0 < a < pi
    c) except for two values of a in the range 0 < a < pi
    d) except for three values of a in the range 0 < a < pi
    (NB the range is supposed to include 0, I don't have that character on my keyboard)

    2. Relevant equations
    possibly trig identities though I haven't found a way to use these! otherwise just normal methods for solving simultaneous eq.

    3. The attempt at a solution
    I got as far as:

    cos a = (1 - x sin a)/y (sub into first eq)

    x(1 - x sin a)/y - y sin a = 2 (multiply by y)

    (y^2 sin a) + 2y - x - (x^2 sin a) = 0

    sin a = (x-2y) / (y^2 - x^2)
    which I think tells me that sin a cannot equal zero, as this would involve dividing by zero on the RHS? Although come to think of it I'm not sure it tells me that anyway... and it doesn't tell me anything about other values of a as far as I can tell.
    Any ideas? Thanks.
  2. jcsd
  3. Oct 31, 2009 #2
    I think that you're paying far too much attention to the fact that you have trigonometric coefficients. This is a system of linear equations in the variables x and y. Why are you solving for the coefficients? Solve for x and y. Then look and see where the solutions you get are defined.
  4. Oct 31, 2009 #3
    Ok.. I now get
    y = cos a - 2 sin a
    x = 2sec a + sin a - [2 (sin^2 a )/cos a]

    which I think are definable as long as cos a doesn't equal zero, which is when a = pi.

    So the answer is b)...?
    I don't have any answers for these questions.. kinda annoying cos there's quite a few I'm stuck on. (and have no idea if the ones I have got an answer for are even right)
  5. Oct 31, 2009 #4
    When does cos a = 0?
  6. Oct 31, 2009 #5
    Stupid mistake.. when a = pi/2 or 3pi/2. So its c.. sorry!
  7. Oct 31, 2009 #6
    Let's try another way. What's the determinate of the system? For what values of a is it zero? There's a mistake being made in the algebra somewhere and I'm not in the mood to sort it out.
  8. Nov 1, 2009 #7
    I've gone over the algebra and pretty sure its right.. not saying how I've interpreted it is though.

    when x is zero, a = arctan (-2)
    when y is zero, a = arctan (2)
    y and x cannot both be zero

    Sorry if I'm being really dense but this doesn't seem to mean anything except that if a does not equal arctan (+/- 2) the graph of these would not cross the y/x-axis.
    I feel like I'm just approaching this completely wrong way- its not supposed to take forever so there must be something I'm missing.
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