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Math Help - Proving indentity..

  1. #1
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    Proving indentity..

    sin^6 A + cos^6 A= 1 - 3sin^2 A.cos^2 A
    Last edited by snigdha; February 21st 2010 at 07:07 AM. Reason: typing error!
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  2. #2
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    Hello, snigdha!

    Prove: . \sin^6\!A + \cos^6\!A\;=\; 1 - 3\sin^2\!A\cos^2\!A
    The left side is a sum of cubes.


    \text{Factor: }\;\underbrace{(\sin^2\!A + \cos^2\!A)}_{\text{This is 1}}\:(\sin^4\!A - \sin^2\!A\cos^2\!A + \cos^4\!A)

    . . . . . . =\;\;\sin^4\!A \;-\; \sin^2\!A\cos^2\!A \;+\; \cos^4\!A  \;\;=\;\;\sin^4\!A + \cos^4\!A - \sin^2\!A\cos^2\!A



    Add and subtract 2\sin^2\!A\cos^2\!A

    . . . \underbrace{\sin^4\!A {\color{red}\;+\; 2\sin^2\!A\cos^2\!A} + \cos^4\!A} \;-\; \underbrace{\sin^2\!A\cos^2\!A {\color{red}\;-\; 2\sin^2\!A\cos^2\!A}}

    . . = \qquad\underbrace{(\sin^2\!A + \cos^2\!A)^2}_{\text{This is 1}} \qquad\quad -\qquad\quad 3\sin^2\!A\cos^2\!A

    . . . . . . . . . . . . . = \;\;1 - 3\sin^2\!A\cos^2\!A

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, snigdha!

    The left side is a sum of cubes.


    \sin^4\!A - \sin^2\!A\cos^2\!A + \cos^4\!A)" alt="\text{Factor: }\;\underbrace{(\sin^2\!A + \cos^2\!A)}_{\text{This is 1}}\\sin^4\!A - \sin^2\!A\cos^2\!A + \cos^4\!A)" />

    . . . . . . =\;\;\sin^4\!A \;-\; \sin^2\!A\cos^2\!A \;+\; \cos^4\!A  \;\;=\;\;\sin^4\!A + \cos^4\!A - \sin^2\!A\cos^2\!A



    Add and subtract 2\sin^2\!A\cos^2\!A

    . . . \underbrace{\sin^4\!A {\color{red}\;+\; 2\sin^2\!A\cos^2\!A} + \cos^4\!A} \;-\; \underbrace{\sin^2\!A\cos^2\!A {\color{red}\;-\; 2\sin^2\!A\cos^2\!A}}

    . . = \qquad\underbrace{(\sin^2\!A + \cos^2\!A)^2}_{\text{This is 1}} \qquad\quad -\qquad\quad 3\sin^2\!A\cos^2\!A

    . . . . . . . . . . . . . = \;\;1 - 3\sin^2\!A\cos^2\!A

    i got a better clue...
    we know that a^3+b^3 = (a+b)^3 - 3ab(a+b)

    so by this i proceeded with the problem...& got it proved...
    thanks anyway...!!
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