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Exercise arctan

  1. Jul 23, 2010 #1
    2Arctan(1/3) + arctan(1/7) = π/4

    I try to solve it like this:

    = sin(2arctan(1/3)) + sin(arctan(1/7) = sin(π/4)

    But this isn't correct

    What have I done wrong?
  2. jcsd
  3. Jul 23, 2010 #2


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    If we let A=Arctan(1/3) and B= arctan(1/7), then tanA=1/3 and tanB=1/7 right?

    So now you have 2A+B=π/4, what happens if you take the tan of both sides?
  4. Jul 24, 2010 #3


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    The problem is that [tex]sin(A+B)\neq sinA+sinB[/tex] for all A and B, which is what you've done on the left side of the equation.
    Since you took the sine of both sides, it should be [tex]sin\left(2arctan(1/3)+arctan(1/7)\right)=sin(\pi/4)[/tex]
    and from the mistake above that I showed you, you can't separate each part in the sine to make two sine functions as you've done.

    Try follow on and see where rockfreak is leading you. You need to know how to expand tan double angles and sums.
  5. Jul 25, 2010 #4
    Thanks! Now I found it
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