Results 1 to 5 of 5

Math Help - Inverse circular relations problems

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    3

    Inverse circular relations problems

    Hey guys, my teacher gave me these two problems which have been keeping me up all night:

    Prove:
    [2tan^-1 (1/3) ]+ [tan^-1(1/7)] = pi/4

    and

    Find all values of θ:
    Tan (π cot θ) = cot (π tan θ)

    My classmates and I have been kinda stumped... Any help is much appreciated

    Sincerely,
    Carlos Jesena
    Last edited by CarlosJesena; April 23rd 2010 at 11:35 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by CarlosJesena View Post
    Hey guys, my teacher gave me these two problems which have been keeping me up all night:

    [2tan^-1 (1/3) ]+ [tan^-1(1/7)] = pi/4
    Hint:
    Spoiler:
    Take the tangent of both sides and remember this



    Find all values of θ:
    Tan (π cot θ) = cot (π tan θ)
    Hint:
    Spoiler:
    \cot=\frac{1}{\tan} and \arctan(x)=\text{arccot}\left(\frac{1}{x}\right)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2010
    Posts
    3
    Quote Originally Posted by Drexel28 View Post
    Hint:
    Spoiler:
    Take the tangent of both sides and remember this





    Hint:
    Spoiler:
    \cot=\frac{1}{\tan} and \arctan(x)=\text{arccot}\left(\frac{1}{x}\right)

    Thanks for the help! For this first problem, I turned it into 2A + B = pi/4 then that becomes tan (2A+B) = 1. Then I expanded the left side and got:

    [tan(2A)+tan(B)] / [1-(tan(2A)tan(B))]

    Am I on the right track? I was thinking of expanding the tan 2A's into double angle formula....
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by CarlosJesena View Post
    Thanks for the help! For this first problem, I turned it into 2A + B = pi/4 then that becomes tan (2A+B) = 1. Then I expanded the left side and got:

    [tan(2A)+tan(B)] / [1-(tan(2A)tan(B))]

    Am I on the right track? I was thinking of expanding the tan 2A's into double angle formula....
    Yes, you're on the right track.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2010
    Posts
    3
    Quote Originally Posted by Drexel28 View Post
    Yes, you're on the right track.

    Thanks so much for the help! I get number one now. Just number two is still a mystery
    Last edited by CarlosJesena; April 24th 2010 at 01:50 AM. Reason: got number 1
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Relations and Functions - Inverse Relations Question
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 13th 2011, 01:20 PM
  2. Circular Problems
    Posted in the Geometry Forum
    Replies: 4
    Last Post: April 8th 2011, 01:51 PM
  3. Inverse circular functions
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 5th 2011, 05:05 PM
  4. more Right Circular Cylinder problems...
    Posted in the Geometry Forum
    Replies: 1
    Last Post: January 12th 2010, 04:33 AM
  5. Inverse Circular Function, help!!!
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: August 23rd 2008, 06:45 PM

Search Tags


/mathhelpforum @mathhelpforum