Results 1 to 7 of 7

Math Help - Proving

  1. #1
    Member maybeline9216's Avatar
    Joined
    Sep 2008
    From
    Singapore~
    Posts
    108

    Proving

    Please help me with this.thanks
    Attached Thumbnails Attached Thumbnails Proving-gypartb.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    It says, prove:
    2csc^2(2A) -csc^2(A) = -2cot(2A)csc(2A)
    where A = theta

    As usual, develop the LHS to see if it matches the RHS.

    LHS =
    = 2csc^2(2A) -csc^2(A)
    = [2 /sin^2(2A)] -[1 /sin^2(A)]
    = [2 / (2sinAcosA)^2] -[1 /sin^2(A)]
    = [1 / 2sin^2(A)cos^2(A)] -[1 /sin^2(A)]

    Combine the two fractions into one fraction only,

    = [1 -2cos^2(A)] / [2sin^2(A)cos^2(A)]
    = [-(2cos^2(A) -1)] / [(2sinAcosA)(sinAcosA)]
    = [-cos(2A)] / [sin(2A)*(1/2)sin(2A)]
    = [-cos(2A) /sin(2A)]*[1 / (1/2)sin(2A)]
    = [-cot(2A)]*[2 /sin(2A)]
    = [-cot(2A)]*[2csc(2A)]
    = -2cot(2A)csc(2A)
    = RHS

    Proven.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member maybeline9216's Avatar
    Joined
    Sep 2008
    From
    Singapore~
    Posts
    108
    WOW!Exactly 10 steps lol.


    [1 -2cos^2(A)] / [2sin^2(A)cos^2(A)]
    = [-(2cos^2(A) -1)] / [(2sinAcosA)(sinAcosA)]
    = [-cos(2A)] / [sin(2A)*(1/2)sin(2A)]

    There is something wrong here..did u just take out -1
    from [1 -2cos^2(A)] ? then its supposed to be [-(2cos^2(A) +1)] isn't it??

    anyway thks alot..

    Umm...wad abt this??is there something wrong here?

    [-cos(2A)] / [sin(2A)*(1/2)sin(2A)]
    = [-cos(2A) /sin(2A)]*[1 / (1/2)sin(2A)]

    where did the 1/(1/2) come from??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2008
    Posts
    10
    Quote Originally Posted by maybeline9216 View Post
    WOW!Exactly 10 steps lol.


    [1 -2cos^2(A)] / [2sin^2(A)cos^2(A)]
    = [-(2cos^2(A) -1)] / [(2sinAcosA)(sinAcosA)]
    = [-cos(2A)] / [sin(2A)*(1/2)sin(2A)]

    There is something wrong here..did u just take out -1
    from [1 -2cos^2(A)] ? then its supposed to be [-(2cos^2(A) +1)] isn't it??

    anyway thks alot..

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2008
    Posts
    10
    Quote Originally Posted by maybeline9216 View Post
    WOW!Exactly 10 steps lol.


    [1 -2cos^2(A)] / [2sin^2(A)cos^2(A)]
    = [-(2cos^2(A) -1)] / [(2sinAcosA)(sinAcosA)]
    = [-cos(2A)] / [sin(2A)*(1/2)sin(2A)]

    There is something wrong here..did u just take out -1
    from [1 -2cos^2(A)] ? then its supposed to be [-(2cos^2(A) +1)] isn't it??

    anyway thks alot..

    \begin{array}{l}<br />
 1 - 2\cos ^2 2A \\ <br />
  =  - 2\cos ^2 2A + 1 \\ <br />
  =  - (2\cos ^2 2A - 1) \\ <br />
 \end{array}

    did u just take out -1
    from [1 -2cos^2(A)] ? then its supposed to be [-(2cos^2(A) +1)]
    Note that [-(2cos^2(A) +1) is exactly the same as [1 -2cos^2(A)]...



    Umm...wad abt this??is there something wrong here?

    [-cos(2A)] / [sin(2A)*(1/2)sin(2A)]
    = [-cos(2A) /sin(2A)]*[1 / (1/2)sin(2A)]

    where did the 1/(1/2) come from??
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    [1 -2cos^2(A)] / [2sin^2(A)cos^2(A)]
    = [-(2cos^2(A) -1)] / [(2sinAcosA)(sinAcosA)]
    = [-cos(2A)] / [sin(2A)*(1/2)sin(2A)]

    There is something wrong here..did u just take out -1
    from [1 -2cos^2(A)] ? then its supposed to be [-(2cos^2(A) +1)] isn't it??


    1 -2cos^2(2A)
    Multiply that by (-1)/(-1) ......(which is really 1 only),
    = (-1)/(-1) *[1 -2cos^2(A)]
    = [-1 +2cos^2(A)] / (-1)
    = -[2cos^2(A) -1]
    = -(2cos^2(A) -1)

    With practice, you'd master those changing of signs eventually.

    -------------------------
    Umm...wad abt this??is there something wrong here?

    [-cos(2A)] / [sin(2A)*(1/2)sin(2A)]
    = [-cos(2A) /sin(2A)]*[1 / (1/2)sin(2A)]

    where did the 1/(1/2) come from??


    [-cos(2A)] is also [(-cos(2A))*(1)]
    The (-cos(2A)) is made to be the numerator over sin(2A), so the (1) remained.
    Then, that (1) is made to be the numerator over the (1/2)sin(2A)

    More practice, or, in time you'd master all these.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member maybeline9216's Avatar
    Joined
    Sep 2008
    From
    Singapore~
    Posts
    108
    i do know how to chg the sign...Erm...its because i was so confused by so many brackets, presenting it in latex form is much easier to see..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving?
    Posted in the Geometry Forum
    Replies: 2
    Last Post: January 24th 2011, 05:41 PM
  2. Proving the SST = SSE + SSR
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: November 9th 2010, 10:10 AM
  3. proving sin^2(x)<=|sin(x)|
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: September 22nd 2010, 01:14 PM
  4. proving
    Posted in the Number Theory Forum
    Replies: 9
    Last Post: October 25th 2009, 09:27 AM
  5. Proving an identity that's proving to be complex
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 21st 2009, 02:30 PM

Search Tags


/mathhelpforum @mathhelpforum