Results 1 to 5 of 5
Like Tree1Thanks
  • 1 Post By Siron

Math Help - half angle formula

  1. #1
    Member
    Joined
    Mar 2012
    From
    malaysia
    Posts
    94
    Thanks
    15

    half angle formula

    Express cosA in terms of t when tan(A/2)=t

    i checked the solution but i don't understand it.

    solution:

    using the half-angle formula for tan, tan^2(A/2)=(1-cosA)/(1+cosA)

    therefore cosA=(1-tan^2(A/2))/(1+tan^2(A/2))=(1-t^2)/(1+t^2)

    what i don't understand is how they get cosA in terms of tan^2(A/2).
    how i would solve it is to multiply both sides of tan^2(A/2)=(1-cosA)/(1+cosA) with (1+cosA), but that won't get me cosA=(1-tan^2(A/2))/(1+tan^2(A/2))
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2011
    From
    Crna Gora
    Posts
    420
    Thanks
    64

    Re: half angle formula

    Quote Originally Posted by muddywaters View Post
    Express cosA in terms of t when tan(A/2)=t

    i checked the solution but i don't understand it.

    solution:

    using the half-angle formula for tan, tan^2(A/2)=(1-cosA)/(1+cosA)

    therefore cosA=(1-tan^2(A/2))/(1+tan^2(A/2))=(1-t^2)/(1+t^2)

    what i don't understand is how they get cosA in terms of tan^2(A/2).
    how i would solve it is to multiply both sides of tan^2(A/2)=(1-cosA)/(1+cosA) with (1+cosA), but that won't get me cosA=(1-tan^2(A/2))/(1+tan^2(A/2))
    Note that :

    \cos^2 \left(\frac{A}{2}\right) = \frac{1}{1+t^2} ~\text{and}~\sin^2 \left(\frac{A}{2}\right) = \frac{t^2}{1+t^2}

    hence :

    \cos A = \cos^2 \left(\frac{A}{2}\right) - \sin^2 \left(\frac{A}{2}\right)=\frac{1-t^2}{1+t^2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2012
    From
    malaysia
    Posts
    94
    Thanks
    15

    Re: half angle formula

    sorry, i don't get it >.<

    the question is this:
    half angle formula-question.jpg

    what i dont understand is this:
    question 2.bmp
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: half angle formula

    \tan^2\left(\frac{a}{2}\right)=\frac{1-\cos(a)}{1+\cos(a)}
    \Leftrightarrow \tan^2\left(\frac{a}{2}\right)[1+\cos(a)]=1-\cos(a)
    \Leftrightarrow \tan^2\left(\frac{a}{2}\right)+\tan^2\left(\frac{a  }{2}\right)\cos(a)=1-\cos(a)
    \Leftrightarrow \tan^2\left(\frac{a}{2}\right)\cos(a)+\cos(a)=1-\tan^2\left(\frac{a}{2}\right)
    \Leftrightarrow \cos(a)\left[\tan^2\left(\frac{a}{2}\right)+1\right]=1-\tan^2\left(\frac{a}{2}\right)
    \Leftrightarrow \cos(a)=\frac{1-\tan^2\left(\frac{a}{2}\right)}{1+\tan\left(\frac{  a}{2}\right)}
    \Rightarrow \cos(a)=\frac{1-t^2}{1+t^2}
    Thanks from muddywaters
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2012
    From
    malaysia
    Posts
    94
    Thanks
    15

    Re: half angle formula

    wow awesome. i just didn't see that! thanks so much! super fast reply there's no one i could ask, thanks ^^
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Half-Angle Formula Help
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: November 4th 2009, 04:18 PM
  2. half angle formula
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 27th 2009, 04:39 PM
  3. sum/difference/half angle formula
    Posted in the Pre-Calculus Forum
    Replies: 9
    Last Post: April 13th 2009, 05:51 PM
  4. Half angle formula problem
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 2nd 2009, 07:24 AM
  5. Using half angle formula
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 27th 2006, 03:50 AM

Search Tags


/mathhelpforum @mathhelpforum