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Homework Help: Inverse trigonometry - Finding solutions

  1. Aug 18, 2013 #1
    1. The problem statement, all variables and given/known data
    Which of the following is the solution set of the equation

    Ans: A

    2. Relevant equations

    3. The attempt at a solution
    I start by rewriting LHS in terms of ##\arctan##.
    Using ##2\arctan(y)=\arctan\left(\frac{2y}{1-y^2}\right)##, it can be further simplified to
    Hence, both LHS and RHS are identical. It can be concluded that they satisfy all values of x in their domain. The domain is (-1,1)-{0} i.e answer is B. But the given answer is A.

    Any help is appreciated. Thanks!
  2. jcsd
  3. Aug 18, 2013 #2


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    Through such manipulations, there's always a risk of introducing extra solutions. Since all offered answers include continuous ranges, it was almost inevitable that simplification would lead to tautology. There are values where the original equation makes no sense. Clearly x = 0 and x >= 1 are problematic. Also bear in mind that arccos(x) is a defined function that only takes one of the y values that satisfy x = cos(y), likewise arccot.
  4. Aug 19, 2013 #3
    I still don't see how to eliminate the extra solutions. :confused:
  5. Aug 19, 2013 #4


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    Hi Pranav-Arora! :smile:

    arctan is defined to be in (-π/2,π/2)

    arccos (and arcsin) is defined to be in [0,π)

    where they don't agree, you get an error "translating" from one to the other
    eg that isn't true for x = -1/√2 …

    the LHS is 2*3π/4, but the RHS is 2*-π/4 :wink:
  6. Aug 19, 2013 #5
    I see that for the above to be possible, x must be greater than 0. This does give the right answer but how am I supposed to think this way. I mean I always face the problem of this "principal range" and end up with the wrong answer. In this case, can you please explain how should I have approached the problem? Thank you!
  7. Aug 19, 2013 #6


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    Hi Pranav-Arora! :smile:
    When the principal ranges differ, the solution has to be in the overlap.

    If it isn't, it can't be a solution …

    eg in this case, if x < 0, the the RHS is obviously negative, but the LHS can never be negative! :wink:
  8. Aug 19, 2013 #7
    Thanks tiny-tim! :smile:

    I have one more problem similar to this. Instead of creating a new thread, I will post it here as I think its related to this "principal range" problem.

    Here goes the question:
    Solution set of the equation, ##\arccos(x)-\arcsin(x)=\arccos(\sqrt{3}x)##
    A)is a unit set
    B)consists of two elements
    C)B)consists of three elements
    D)is a void set

    The equation is defined when ##-1/\sqrt{3}\leq x \leq 1/\sqrt{3}##. I can rewrite ##\arcsin(x)## as ##\arccos(\sqrt{1-x^2})##. I guess here I have to apply the condition that ##x>0## for this to be true. Correct?

    Now using ##\arccos(a)-\arccos(b)=\arccos(ab+\sqrt{(1-a^2)(1-b^2)})##
    Since x>0, |x|=x. Solving further gives x=0 and 1/2 which suggests that answer is B and this is incorrect. Where did I go wrong this time? :confused:
  9. Aug 19, 2013 #8


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    No, if x < 0, then arcsin < 0 but ##\arccos(\sqrt{1-x^2})## > 0 :wink:
  10. Aug 19, 2013 #9
    Do I write ##\arcsin(x)=-\arccos(\sqrt{1-x^2})## for ##x<0##?
  11. Aug 19, 2013 #10


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    yes, that seems to work :smile:

    oh, and you can combine the two into

  12. Aug 19, 2013 #11
    Great, thanks a lot tiny-tim! :smile:
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