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    hi can someone explain question 5)b when you have a squred ?solomon A
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    (Original post by otrivine)
    https://www.box.com/shared/8quza0w4k...66/735382824/1

    hi can someone explain question 5)b when you have a squred ?solomon A
     6 \cos^2 x + \sin (2x) = 6 \cos^2 x + 2 \sin x \cos x = 2 \cos x ( 3 \cos x + \sin x )

    Now use the result of part (a) for  3 \cos x + \sin x
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    (Original post by raheem94)
     6 \cos^2 x + \sin (2x) = 6 \cos^2 x + 2 \sin x \cos x = 2 \cos x ( 3 \cos x + \sin x )

    Now use the result of part (a) for  3 \cos x + \sin x
    got it thanks
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    (Original post by raheem94)
     6 \cos^2 x + \sin (2x) = 6 \cos^2 x + 2 \sin x \cos x = 2 \cos x ( 3 \cos x + \sin x )

    Now use the result of part (a) for  3 \cos x + \sin x
    and one thing raheem http://www.edexcel.com/migrationdocu...2007_Paper.pdf
    question 1)a) how do u tackle the question
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    (Original post by otrivine)
    and one thing raheem http://www.edexcel.com/migrationdocu...2007_Paper.pdf
    question 1)a) how do u tackle the question
     \sin (3 \theta ) = \sin ( 2 \theta + \theta ) = \sin (2 \theta ) \cos ( \theta ) + \cos (2 \theta) \sin ( \theta )

    Write  \sin ( 2 \theta ) = 2 \sin \theta \cos \theta \ and \ \cos ( 2 \theta ) = 1 - 2 \sin^2 ( \theta )
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    (Original post by raheem94)
     \sin (3 \theta ) = \sin ( 2 \theta + \theta ) = \sin (2 \theta ) \cos ( \theta ) + \cos (2 \theta) \sin ( \theta )

    Write  \sin ( 2 \theta ) = 2 \sin \theta \cos \theta \ and \ \cos ( 2 \theta ) = 1 - 2 \sin^2 ( \theta )
    thanks and last one http://www.edexcel.com/migrationdocu...e_20090115.pdf
    6)a)ii i did the first part now for the second part do i use the equation and rearrange so i sub the sin3X?
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    (Original post by otrivine)
    thanks and last one http://www.edexcel.com/migrationdocu...e_20090115.pdf
    6)a)ii i did the first part now for the second part do i use the equation and rearrange so i sub the sin3X?
    Multiply both sides of the equation given in part (i) by  -2 , and you should be able to see what to do next.
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    (Original post by raheem94)
    Multiply both sides of the equation given in part (i) by  -2 , and you should be able to see what to do next.
    i am not getting ur point?
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    (Original post by otrivine)
    i am not getting ur point?
     8 \sin^3 \theta -6 \sin \theta +1 = 0 \implies -2( -4 \sin^3 \theta +3 \sin \theta ) +1 = 0 \\ \implies -2 \sin ( 3 \theta ) +1 =0

    Does it make sense?
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    (Original post by raheem94)
     8 \sin^3 \theta -6 \sin \theta +1 = 0 \implies -2( -4 \sin^3 \theta +3 \sin \theta ) +1 = 0 \\ \implies -2 \sin ( 3 \theta ) +1 =0

    Does it make sense?
    yep yep thanks
 
 
 
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