Results 1 to 5 of 5

Math Help - prove trigonometric identities...

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    31

    prove trigonometric identities...

    sin^6x + cos^6x = 1-3sin^2x cos^2x how can i prove that identity...tks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    (\sin^2{x})^3 + (\cos^2{x})^3 =

    (\sin^2{x} + \cos^2{x})(\sin^4{x} - \sin^2{x}\cos^2{x} + \cos^4{x}) =

    \sin^4{x} - \sin^2{x}\cos^2{x} + \cos^4{x} =

    \sin^4{x} + 2\sin^2{x}\cos^2{x} + \cos^4{x} - 3\sin^2{x}\cos^2{x} =

    (\sin^2{x} + \cos^2{x})^2 - 3\sin^2{x}\cos^2{x} =

    1 - 3\sin^2{x}\cos^2{x}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2008
    Posts
    31
    \sin^4{x} + 2\sin^2{x}\cos^2{x} + \cos^4{x} - 3\sin^2{x}\cos^2{x} =

    can u explain me what did you do in that step? i didnt get that one...

    why didnt you use \sin^4{x} + 1\sin^2{x}\cos^2{x} + \cos^4{x} - 2\sin^2{x}\cos^2{x} =
    i know that we want that 3 but can we add terms just like that?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    \sin^4{x} + 2\sin^2{x}\cos^2{x} + \cos^4{x} - 3\sin^2{x}\cos^2{x} =

    can u explain me what did you do in that step? i didnt get that one...

    I needed the middle term + 2\sin^2{x}\cos^2{x} so that I could factor the \sin^4{x} + \cos^4{x} ... to compensate adding it, I also had to also subtract it, resulting in the -3\sin^2{x}\cos^2{x} term.

    why didnt you use \sin^4{x} + 1\sin^2{x}\cos^2{x} + \cos^4{x} - 2\sin^2{x}\cos^2{x} =

    why do that? it won't factor.

    i know that we want that 3 but can we add terms just like that?

    sure can ... isn't a^2 - ab + b^2 = a^2  + 2ab  +  b^2 - 3ab ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2008
    Posts
    31
    Quote Originally Posted by skeeter View Post
    \sin^4{x} + 2\sin^2{x}\cos^2{x} + \cos^4{x} - 3\sin^2{x}\cos^2{x} =

    can u explain me what did you do in that step? i didnt get that one...

    I needed the middle term + 2\sin^2{x}\cos^2{x} so that I could factor the \sin^4{x} + \cos^4{x} ... to compensate adding it, I also had to also subtract it, resulting in the -3\sin^2{x}\cos^2{x} term.

    why didnt you use \sin^4{x} + 1\sin^2{x}\cos^2{x} + \cos^4{x} - 2\sin^2{x}\cos^2{x} =

    why do that? it won't factor.

    i know that we want that 3 but can we add terms just like that?

    sure can ... isn't a^2 - ab + b^2 = a^2 + 2ab + b^2 - 3ab ?

    thanks for your time
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove using Trigonometric Identities
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: October 24th 2010, 03:24 PM
  2. trigonometric identities
    Posted in the Trigonometry Forum
    Replies: 8
    Last Post: May 1st 2008, 06:10 PM
  3. Trigonometric Identities
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: April 6th 2008, 05:16 PM
  4. Prove: Trigonometric Identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 29th 2007, 05:10 PM
  5. Trigonometric Identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 21st 2006, 02:14 PM

Search Tags


/mathhelpforum @mathhelpforum