1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    rock.freak667

    User Avatar
    Homework Helper

    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

    Mentallic

    User Avatar
    Homework Helper

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Exercise arctan
  1. Arctan help (Replies: 1)

  2. Logarithm exercise (Replies: 4)

  3. Arctan of a product (Replies: 2)

Loading...