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Math Help - Trignometric Identities

  1. #1
    Newbie Sanjana Das's Avatar
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    Trignometric Identities

     if sinA+cosA=x, Prove that Sin^6A +cos^6A = 4-3(x^2-1)^2 /4
    Last edited by Sanjana Das; September 22nd 2009 at 02:25 AM.
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  2. #2
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    Quote Originally Posted by Sanjana Das View Post
     if sinA+cosA, Prove that Sin^6A +cos^6A = 4-3(x^2-1)^2 /4
    something is wrong here... on the left side you have sin( A ) and on the right side x ???

    Anyway... if A = x...

    sin^6( x ) + cos^6( x ) = ( sin^2( x ) + cos^2( x ) )^3 - 3sin^4( x )*cos^2( x ) - 3sin^2( x )*cos^4( x ) =

    = 1^3 - 3*( sin^2( 2x )/4 )*sin^2( x ) - 3( sin^2( 2x )/4 )*cos^2( x ) =

    = 1 - (3/4)*sin^2( 2x )*( sin^2( x ) + cos^2( x ) ) =

    = 1 - (3/4)*sin^2( 2x )
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  3. #3
    Newbie Sanjana Das's Avatar
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    Quote Originally Posted by josipive View Post
    something is wrong here... on the left side you have sin( A ) and on the right side x ???

    Anyway... if A = x...

    sin^6( x ) + cos^6( x ) = ( sin^2( x ) + cos^2( x ) )^3 - 3sin^4( x )*cos^2( x ) - 3sin^2( x )*cos^4( x ) =

    = 1^3 - 3*( sin^2( 2x )/4 )*sin^2( x ) - 3( sin^2( 2x )/4 )*cos^2( x ) =

    = 1 - (3/4)*sin^2( 2x )*( sin^2( x ) + cos^2( x ) ) =

    = 1 - (3/4)*sin^2( 2x )
    no no the answer is not tht plsss check again ...it is sinA+cosA=x
    and also A=thetha
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  4. #4
    Newbie Sanjana Das's Avatar
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    Quote Originally Posted by sanjana das View Post
    no no the answer is not tht plsss check again ...it is sina+cosa=x
    and also a=thetha
    plsssssss someone solve this question plssssssssssssssss
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  5. #5
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    ok.... than

    you see this part that i solved:

    sin^6( A ) + cos^6( A ) = ... = 1 - (3/4)*sin^2( 2A )

    this is equal to:

    1 - ( 3/4 )*sin^2( 2A ) = 4 - 3*( x^2 - 1 )^2 / 4

    (3/4)*sin^2( 2A ) = ( 3/4 )*( x^2 - 1 )^2 - 3

    sin^2( 2A ) = ( x^2 - 1 )^2 - 4
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