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Trigonometric Function

  1. Feb 1, 2006 #1
    I have stuck on this problem for long time

    [tex] sin^2 \alpha = \frac{\alpha}{2} [/tex]

    I never meet this kind of problem before, and I have no idea about this. Could someone tell me how to solve this kind of problem?

    Thanks in advance.

    (Ans: [tex] \alpha = 1.39 rad [/tex])
  2. jcsd
  3. Feb 1, 2006 #2


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    This kind of problem can be done with Newton's method.
    [tex]\sin ^ 2 \alpha = \frac{\alpha}{2}[/tex]
    [tex]\Leftrightarrow 2 \sin ^ 2 \alpha = \alpha[/tex]
    [tex]\Leftrightarrow - 2 \sin ^ 2 \alpha + \alpha = 0[/tex]
    [tex]\Leftrightarrow \cos (2 \alpha) - 1 + \alpha = 0[/tex]
    Now let [tex]f(x) = \cos (2x) + x - 1[/tex]
    Since f(0) = 0, that means x = 0 is 1 solution to the question.
    Now let's choose x0.
    [itex]f(1) = \cos 2[/itex]. Since [tex]2 \in ]\frac{\pi}{2} ; \ \pi[[/tex], we have: [itex]f(1) = \cos 2 < 0[/itex]
    [itex]f(2) = \cos 4 + 1 > 0[/itex].
    So we have 1 more solution on the interval ]1; 2[.
    Pick any x0 on the interval ]1; 2[.
    Then apply the Newton's method:
    [tex]x_{n + 1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}[/tex]. And let n increase without bond. The solution will be [tex]x = \lim_{n \rightarrow \infty} x_n[/tex].
    If you get x = 0 (just try another x0) (we already know this solution, we want to find another on the interval ]1; 2[).
    You may want to try some x0 that's closed to 2.
    Just try it and see what you get.
    So there are 2 solutions to the problem.
    Can you go from here?
    By the way, I believe you cannot get any solution that reads:
    [itex]x \approx 1.39 \mbox{ rad}[/itex]. That may be a typo though.
    Last edited: Feb 1, 2006
  4. Feb 1, 2006 #3
    hmmm.... though a little bit vague, I think I can solve it.

  5. Feb 1, 2006 #4


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    You do understand that [itex]\alpha= 0[/itex] is an obvious solution and are asking for non-zero solutions?

    The point is that there is no "algebraic method" to solve an equation in which the unknown occurs both in and outside of a trascendental function- such as [itex] sin^2 \alpha= \frac{\alpha}{2}[/itex].

    As VietDao29 said, a good way to solve such equation is "Newton's method". A solution to f(x)= 0 is given by xn+1= xn- \frac{f(x_0}{f'(x_0}[/tex]

    In this problem [itex]f(x)= sin^2(x)- \frac{x}{2}. f'(x)= 2 sin(x)cos(x)-\frac{1}{2}[/itex] so [itex]x_{n+1}= x_n- \frac{sin^(x)- \frac{x}{2}}{2sin(x)cos(x)-\frac{1}{2}}[/itex]

    Pick a starting value for x (perhaps [itex]x= \frac{\pi}{2}[/itex] would be keep you away from 0) and do the arithmetic.
  6. Feb 2, 2006 #5


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    A good way to start this type of problem is by plotting the functions [itex] \sin^2( \alpha )[/itex] and [itex] \frac { \alpha } 2 [/itex].

    Roots are at the points of intersection.
  7. Feb 2, 2006 #6
    For small angle, sin^2(alpha) ~= (alpha^2). So a place to look might be when alpha = 1/2

    My calculator converges on .5545
  8. Feb 2, 2006 #7


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    Then, there's probably something wrong with your calculator. :tongue2:
    You can always put x back to the original equation, and find out that it is not a solution... :tongue:
    By the way, the problem reads:
    [tex]\sin ^ 2 \alpha = \frac{\alpha}{2}[/tex].
  9. Feb 2, 2006 #8
    No, that ans is flat out wrong. Plug it in, you will get .9676672 = .695, I dont think so. The two solutions are .5545 and I think something like 1.83ish.
  10. Feb 2, 2006 #9


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    I get 1.849076836 and 0.554572091
  11. Feb 2, 2006 #10


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    Whooops, I almost missed one solution x = 0.554572091... My bad. :cry: :blushing: :frown:
    I must be careful next time... :grumpy:
    But don't forget the obvious solution x = 0.
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