Results 1 to 5 of 5

Math Help - 2 hard trig problems

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    18

    2 hard trig problems

    I have tried to solve these through every way i know but i get stuck halfway.

    tan^{2}10^{\circ}+tan^{2}50^{\circ}+tan^{2}70^{\ci  rc}=9

    tan^{2}36^{\circ}\times tan^{2}72^{\circ}=5
    Last edited by ldx2; December 15th 2009 at 11:57 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Hello ldx2
    Quote Originally Posted by ldx2 View Post
    I have tried to solve these through every way i know but i get stuck halfway.

    tan^{2}10^{\circ}+tan^{2}50^{\circ}+tan^{2}70^{\ci  rc}=9

    tan^{2}36^{\circ}\times tan^{2}72^{\circ}=5
    I haven't got a solution for part 1 yet, but here's an answer to part 2, based on the fact that \sin18^o = \frac{\sqrt5-1}{4} (which we proved in an earlier post).

    First note that
    \sin18^o = \frac{\sqrt5-1}{4}
    \Rightarrow \sin^218^o = \frac{6-2\sqrt5}{16}=\frac{3-\sqrt5}{8}
    Then, noting that \tan72^o = \cot18^o = \frac{1}{\tan18^o}:
    \tan36^o\tan72^o = \frac{\tan36^o}{\tan18^o}=\frac{2}{1-\tan^218^o}
    =\frac{2}{1-\dfrac{\sin^218^o}{\cos^218^o}}

    =\frac{2(1-\sin^218^o)}{1-2\sin^218^o}

    = \frac{2\left(1-\dfrac{3-\sqrt5}{8}\right)}{1-\dfrac{3-\sqrt5}{4}}

    = ... etc

    = \sqrt5 (Can you supply the missing lines?)
    Hence \tan^236^o\tan^272^o=5

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2009
    Posts
    18
    I understood all but this part :
    \frac{\tan36^o}{\tan18^o}=\frac{2}{1-\tan^218^o}

    Can you explain what identity you used ??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Hello ldx2
    Quote Originally Posted by ldx2 View Post
    I understood all but this part :
    \frac{\tan36^o}{\tan18^o}=\frac{2}{1-\tan^218^o}

    Can you explain what identity you used ??
    \tan2A = \frac{2\tan A}{1-\tan^2A}. So \tan36^o = ?

    Grandad
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2009
    Posts
    18
    Yeah i get it now, thank you very much for your help. Any luck with the first problem ?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 3 Bad Hard Problems
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 21st 2009, 01:46 PM
  2. 3 hard integral problems
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 28th 2009, 07:46 AM
  3. Really hard pre cal problems (4)~(5)
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: February 21st 2009, 07:05 PM
  4. 4 hard problems please help
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: December 9th 2008, 10:43 PM
  5. Hard Problems
    Posted in the Geometry Forum
    Replies: 1
    Last Post: August 22nd 2008, 10:38 PM

Search Tags


/mathhelpforum @mathhelpforum