Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By Soroban

Math Help - trig identities

  1. #1
    Newbie
    Joined
    Dec 2012
    From
    tirana
    Posts
    6

    trig identities

    1+cosx/sinx =cotg(x/2)

    it's mashing up my mind
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,751
    Thanks
    651

    Re: trig identities

    Hello, enea54!

    Two identities: . \sin\tfrac{x}{2} \;=\;\sqrt{\tfrac{1-\cos x}{2}} \qquad \cos\tfrac{x}{2} \;=\;\sqrt{\tfrac{1+\cos x}{2}}



    \text{Prove: }\:\frac{1+\cos x}{\sin x} \:=\:\cot\tfrac{x}{2}

    The right side is: . \cot\tfrac{x}{2} \;\;=\;\;\dfrac{\cos\frac{x}{2}}{\sin\frac{x}{2}} \;\;=\;\; \dfrac{\sqrt{\frac{1+\cos x}{2}}}{\sqrt{\frac{1-\cos x}{2}}}

    . . =\;\;\sqrt{\dfrac{\frac{1+\cos x}{2}}{\frac{1-\cos x}{2}}} \;\;=\;\;\sqrt{\dfrac{1+\cos x}{1-\cos x}} \;\;=\;\; \sqrt{\frac{1+\cos x}{1-\cos x}\cdot\color{blue}{\frac{1+\cos x}{1+\cos x}}}

    . . =\;\;\sqrt{\frac{(1+\cos x)^2}{1-\cos^2\!x}} \;\;=\;\; \sqrt{\frac{(1+\cos x)^2}{\sin^2\!x}} \;\;=\;\;\frac{1+\cos x}{\sin x}
    Thanks from enea54
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2012
    From
    tirana
    Posts
    6

    Re: trig identities

    thanks a lot man it was really helpful
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: November 26th 2012, 03:57 PM
  2. Trig identities.
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: November 13th 2011, 10:34 AM
  3. Trig identities
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: September 30th 2009, 05:40 PM
  4. trig identities
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: January 13th 2009, 04:21 PM
  5. Trig identities!!!!!!!!!
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: June 18th 2008, 08:48 PM

Search Tags


/mathhelpforum @mathhelpforum