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Trig equation

  1. Sep 11, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex]\ arcsin \frac{x}{x-1} + 2 arctan \frac{1}{x+1} = \frac{π}{2}[/itex]


    2. Relevant equations


    3. The attempt at a solution

    if [itex]\ arcsin\frac{x}{(x-1)} = α [/itex]

    then [itex] sin α = \frac{x}{(x-1)}[/itex]

    if [itex]2 arctan\frac{1}{x+1} = β[/itex]

    then [itex]tan(\frac{β}{2}) = \frac{1}{(x+1)}[/itex]

    [itex]tan(β) = \frac{2\frac{1}{x+1}}{1-(\frac{1}{x+1})^2}[/itex]

    [itex]tan(β) = \frac{2(x+1)}{x(x+2)}[/itex]

    [itex]α + β = \frac{π}{2}[/itex]

    [itex]sin(α + β) = 1[/itex]

    [itex]sin(α + β) = sinαcosβ + cosαsinβ[/itex]

    [itex]cosβ = \frac{x(x+2)}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}[/itex] by pythagoras' theorem

    [itex]sinβ = \frac{(2(x+1))}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}[/itex] by pythagoras' theorem

    [itex]cosα = \frac{\sqrt{(x-1)^2 - x^2}}{(x-1)}[/itex] by pythagoras' theorem

    [itex](\frac{x}{(x-1)})(\frac{x(x+2)}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}) + (\frac{\sqrt{(x-1)^2 - x^2}}{(x-1)})(\frac{(2(x+1))}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}) = 1[/itex]

    [itex]\frac{x^2(x+2) + (\sqrt{(x-1)^2 - x^2)}(2(x+1))}{(x-1)(\sqrt{(x(x+2))^2 + (2(x+1))^2})} = 1[/itex]

    [itex]x^2(x+2) + (\sqrt{(x-1)^2 - x^2})(2(x+1)) = (x-1)(\sqrt{(x(x+2))^2 + (2(x+1))^2})[/itex]

    At this point I figure I half made an error because I can't see a fruitful direction in which to progress and it feels a bit messy compared to the other problems in the exercise.
     
  2. jcsd
  3. Sep 11, 2013 #2

    eumyang

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    Homework Helper

    I haven't figured the answer myself, but I have a couple of comments:

    The denominators here will simplify into an irreducible quadratic.

    TBH, I don't know if this will be helpful, but the numerator can be simplified a bit (although you won't lose the square root).
     
  4. Sep 11, 2013 #3

    verty

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    Homework Helper

    You were on the right track at this point:

    ##α + β = \frac{π}{2}##

    but the next step was wrong. Can you think of something better to do here?

    Hint: Don't make it any more complicated than it already is.

    Use sin and cos
     
  5. Sep 12, 2013 #4
    Thanks for your comments. After staring at this for a good proportion of the day I noticed that α and β are complementary angles which enabled me to proceed in the following fashion:

    [itex]sinα = \frac{x}{x-1} [/itex]

    [itex]cosβ = \frac{x}{x-1} [/itex] complementary angle

    [itex]cosα = \frac{-2x+1}{x-1} [/itex] pythagoras' theorem

    [itex]sinβ = \frac{-2x+1}{x-1} [/itex] complementary angle

    [itex]sin(α+β) = 1[/itex]

    [itex]sin(α+β) = (\frac{x}{x-1})^2 + (\frac{-2x+1}{x-1})^2 = 1[/itex]

    [itex]\frac{5x^2-4x+1}{x-1} = 1[/itex]

    [itex]5x^2-4x+1 = x-1[/itex]

    [itex]x^2-x = 0[/itex]

    [itex]x(x-1) = 0[/itex]

    [itex]x = 0[/itex] or [itex]x = 1[/itex]

    but [itex] asin \frac{1}{1-1}[/itex] makes no sense

    so [itex]x = 0[/itex]

    However the process I have deployed here seems no less complicated than the process I was following earlier. Is there a quicker/simpler solution?
     
  6. Sep 12, 2013 #5

    eumyang

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    Homework Helper

    Really? I think the above is so much simpler than your first attempt, radicals and all. I want to kick myself for not seeing it earlier.
     
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