Results 1 to 4 of 4

Thread: [SOLVED] Double angle identities

  1. #1
    Member
    Joined
    Dec 2008
    Posts
    105

    [SOLVED] Double angle identities

    Stuck on all of these and can't see to get my answers anywhere close to the answers

    a) Express $\displaystyle \frac{\sin2x}{1 - 2\cos2x}$ in terms of $\displaystyle \tan(x)$

    b) Prove the identity $\displaystyle \frac{1 + \sin2A - \cos2A}{1 + \sin2A + \cos2A} \equiv \tan(A) $

    and finally

    c) Given that $\displaystyle \cos2A = \tan^{2}x$ show that $\displaystyle \cos2x = \tan^{2}A$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    Quote Originally Posted by Beard View Post
    Stuck on all of these and can't see to get my answers anywhere close to the answers

    a) Express $\displaystyle \frac{\sin2x}{1 - 2\cos2x}$ in terms of $\displaystyle \tan(x)$
    $\displaystyle sin(2x) =2sin(x)cos(x)$

    $\displaystyle cos(2x) = cos^2(x) -sin^2(x) $

    $\displaystyle \frac{\sin2x}{1 - 2\cos2x}$

    $\displaystyle =\frac{2sin(x)cos(x)}{1-2(cos^2(x) -sin^2(x))}$

    Divide numerator and denominator by $\displaystyle cos^2(x)$


    $\displaystyle \frac{\frac{2sin(x)}{cos(x)}} {\frac{1}{cos^2(x)}-\frac{2cos^2(x)}{cos^2(x)}+\frac{2sin^2(x)}{cos^2( x)}} $

    I wont tell you that $\displaystyle tan(x)= \frac{sin(x)}{cos(x)}$


    and that $\displaystyle tan^2(x)+1 = sec^2(x)$

    Neither will I tell you that

    your answer is


    $\displaystyle \frac{2tan(x)} {1+tan^2(x)-2+2tan^2(x)}$

    $\displaystyle = ~\frac{2tan(x)}{-1+3tan^2(x)}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    Quote Originally Posted by Beard View Post
    Stuck on all of these and can't see to get my answers anywhere close to the answers


    b) Prove the identity $\displaystyle \frac{1 + \sin2A - \cos2A}{1 + \sin2A + \cos2A} \equiv \tan(A) $
    Using the formula






    We will get
    numerator = $\displaystyle 1 +2sin(A)cos(A) - cos^2(A) + sin^2(A)$

    Now we use
    $\displaystyle 1-cos^2(x) = sin^2(x) $

    So numerator $\displaystyle = 2sin^2(A) +2sin(A)cos(A) = 2sin(A)[sin(A) +cos(A)] $

    Similarly Denominator will become

    $\displaystyle 1 +2sin(A)cos(A) +cos^2(A)-sin^2(A)$

    $\displaystyle = 2cos^2(A) +2sin(A)cos(A) $

    $\displaystyle = 2cos(A) (sin(A) +cos(A))$
    So our term is now

    $\displaystyle =\frac{2sin(A)(sin(A) +cos(A))}{2cos(A)(sin(A) +cos(A))}$


    Now cancel the terms common in numerator and denominator
    And that's the end of the Second Show
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    Quote Originally Posted by Beard View Post
    c) Given that $\displaystyle \cos2A = \tan^{2}x$ show that
    $\displaystyle \cos2x = \tan^{2}A$
    $\displaystyle cos(2A) = cos^2(A) - sin^2(A) = 2cos^2(x) - 1$

    Hence equation now is
    $\displaystyle 2cos^2(A) - 1 = tan^2(x) $


    $\displaystyle 2cos^2(A)= sec^2(x) $


    $\displaystyle cos^2(A) = sec^2(x)/2$

    Hence $\displaystyle sec^2(A) = \frac{2}{sec^2(x) } = 2 cos^2(x) $

    Thus
    $\displaystyle sec^2(A)=2 cos^2(x) $

    Substract one from both sides

    And
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Double Angle Identities
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Oct 29th 2009, 11:51 PM
  2. Double angle identities
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Oct 4th 2009, 01:36 AM
  3. Half angle or double angle Identities
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Jul 21st 2009, 05:20 PM
  4. Double and Half angle identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Feb 28th 2009, 06:23 PM
  5. [SOLVED] Double Angle Identities
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: Jun 26th 2008, 10:33 PM

Search Tags


/mathhelpforum @mathhelpforum