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Why it does not work with a negative (trig problem)

  1. Jan 2, 2014 #1

    i'm looking for an explanation of a rule I used to solve a problem. I came to the correct answer in the end, but I do not understand why my first attempt went wrong.

    1. The problem statement, all variables and given/known data

    Find the general solution to cot4θ = tan5θ

    2. Relevant equations

    tanA = tanB => A = nπ + B

    tanα = a/b = cotβ

    Where α and β are complementary angles.

    3. The attempt at a solution

    cot4θ = tan5θ

    tan(π/2 - 4θ) = tan5θ

    π/2 - 4θ = nπ + 5θ

    9θ = π/2 - nπ

    θ = π/18 (2n - 1)

    The above is not given as a solution in my book and the following is said to be correct.

    5θ = nπ + π/2 - 4θ

    9θ = nπ + π/2

    θ = π/18 (2n + 1)

    Why is this?

  2. jcsd
  3. Jan 2, 2014 #2
    Hi BOAS!

    Don't you think both the answers are same? (Check for different values of n) :)
  4. Jan 2, 2014 #3
    I think for different values of n it is possible to get the same answer from both equations, but is that the same as 'being the same'?

    IE - Is the first one not written in my book because it is the same as the second one?
  5. Jan 2, 2014 #4


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    Staff Emeritus
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    Both answers are the same.

    θ = (π/18) multiplied by any odd integer .

    Write your solution as
    θ = (π/18)(2m-1)

    The book solution is
    θ = (π/18)(2n+1)

    To get the same solution for both n = m-1 .
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