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Presentation about Tri.

  1. Feb 16, 2005 #1
    Sometimes, I don't know how to present an answer well.
    For [itex] sin (x+y)=sin (5x+z) [/itex]
    For y and z are constant. You may subsitute some value into it.
    If the range of x is restricted to [itex] -\pi < x < \pi [/itex],
    naturally we will use [2pi-(5x+z)] or [pi-(5x+z)]. However, in some situation as I mentioned before, the required x value cannot be found in this way. Perhaps,
    we need to use [6pi-(5x+z)]=x+y instead to find out the value in the required range. How should I present? Generally, no one subs. pi and 2pi , then 5 pi , it's very strange, some marks maybe conserved for this reason.
     
  2. jcsd
  3. Feb 16, 2005 #2
    I'll give you a hint.... (I hope it's not the whole thing or I'll get fired at).

    The general solution for the equation [itex]\sin x = \sin\alpha[/itex] is [itex]x = n\pi + (-1)^n\alpha[/itex] where n is an integer (positive, negative or zero). A proof of this assertion is available in Plane Trigonometry by SL Loney. Try and convince yourself that this is so.

    Also, sometimes you may have (or come up with) two different looking solutions to an equation. Plug in values for the integer parameter to make sure they yield the same values for the angle.
     
    Last edited: Feb 16, 2005
  4. Feb 17, 2005 #3
    I know of it. I have found it myself too. But the problem is, if the answer was restricted to a range. Sometimes, as I mentioned before, the solution needed to use 5pi, then 2pi, then 10pi, it seems that you are just "creating" the method rather than finding it with a superb way. What should I write? [BY FINDING]?
     
  5. Feb 19, 2005 #4
    Well, depending on your texbook, instructor or course conventions, there can be different meanings to the word principal angle and particular solution. For a particular solution, you must first find a general solution and specialize to the case required. As a concrete example, suppose you have to find a solution of the equation [itex]\sin\alpha = \frac{1}{2}[/itex] in [itex][- \frac{\pi}{2},\frac{\pi}{2} ][/itex]. While you can write [itex]\sin\alpha = \sin\frac{\pi}{6}[/itex] and then write [itex]\alpha = \frac{\pi}{6}[/itex], this is conceptually incorrect as the first equation does not lead to the second unless you concern yourself only with a particular value of the angle. In general, it is safer and logically correct to write instead the general solution, which is, [itex]\alpha = n\pi + (-1)^{n}\frac{\pi}{6}[/itex] and specialize by using a value of n (in this case 0) to make sure that [itex]\alpha[/itex] lies in the required range.

    In a problem therefore write the general solution first and then give values ot the integer parameter to determine the values which lie in the required range. This method is good also because sometimes there is more than one solution over a specified range and it's not always obvious. But why miss it? There's no general rule to determine what value of n you have to assign but that is usually obvious from the form of the general solution obtained and the required constraints.

    Hope that helps.....

    Cheers
    Vivek
     
    Last edited: Feb 19, 2005
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