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Math Help - Using identities

  1. #1
    Member Furyan's Avatar
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    Using identities

    Hello,

    The question Iím working on is

    f(x) = \4cos^2{x} - \3sin^2{x}

    Show that f(x) = \dfrac{1}{2} +\dfrac{7}{2}\cos2{x}

    I have tried using \sin^2{x} + \cos^2{x} = 1

    And the double angle identity for \cos2{x}

    The best Iíve got is

    f(x) = \3cos2{x} + \cos^2{x}, which it might not be.

    Maybe I should be using the half angle identity, but I'm not sure how.

    I know I have to work at these, but if someone could point me in the right direction with this Iíd be very grateful.

    Thank you
    Last edited by Furyan; December 18th 2011 at 11:50 AM. Reason: Correction
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Using identities

    Hi Furyan!

    Almost there!
    What is the double angle identity for \cos 2x in terms of \cos^2 x?
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  3. #3
    Member Furyan's Avatar
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    Re: Using identities

    Hi ILikeSerena

    Almost there, really? Cool

    It's \cos2{x} = \2cos^2{x} - 1?

    I'll try using it and see what I get.

    Thank you.
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Using identities

    We have given:
    f(x)=4\cos^2(x)-3\sin^2(x)
    which we can write as:
    f(x)=4\cos^2(x)-3[1-\cos^2(x)]
    =4\cos^2(x)-3+3\cos^2(x)
    =7\cos^2(x)-3
    =\frac{7}{2}\left[2\cos^2(x)-\frac{6}{7}]
    =\frac{7}{2}\left[2\cos^2(x)-1+\frac{1}{7}\right]
    =\frac{7}{2}\left[\cos(2x)+\frac{1}{7}\right]
    =\frac{7}{2}\cos(2x)+\frac{1}{2}
    Last edited by Siron; December 18th 2011 at 12:37 PM.
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  5. #5
    Member Furyan's Avatar
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    Re: Using identities

    Hi LikeSerena

    Thank you for letting me know I was on the right track, it really helped. I think I have it.

    \cos2{x} = \2cos^2{x} - 1

    \cos2{x} + 1 = \2cos^2{x}

     \dfrac{1}{2}\(cos2{x}+1) =\cos^2{x}

    Substituting I get

    \3cos2{x}+ \dfrac{1}{2}(\cos2{x}+1)

    \3cos2{x}+ \dfrac{1}{2}\cos2{x}+ \dfrac{1}{2}

    \dfrac{7}{2}\cos2{x}+ \dfrac{1}{2}

    That appears to be it. Yipee, thank you!
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  6. #6
    Super Member ILikeSerena's Avatar
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    Re: Using identities

    Good!
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  7. #7
    Member Furyan's Avatar
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    Re: Using identities

    Hi Siron

    =7\cos^2(x)-3

    This is another approach I had tried, but I had got completely stuck, my factoring is not so good. I am very grateful for your reply and I will enjoy working through your method. Thank you very much for taking the time to reply, I am very grateful. It is enormously helpful.
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  8. #8
    Member Furyan's Avatar
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    Re: Using identities

    Last edited by Furyan; December 18th 2011 at 02:05 PM. Reason: duplicate
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  9. #9
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    Re: Using identities

    You could also try

    cos2x=cos^2x-sin^2x

    4cos^2x-3sin^2x=cos^2x+3\left(cos^2x-sin^2x\right)=cos^2x+3cos2x

    =\frac{2cos^2x}{2}-\frac{cos2x}{2}+\frac{7cos2x}{2}

    =\frac{2cos^2x-cos^2x+sin^2x}{2}+\frac{7cos2x}{2}=\frac{cos^2x+si  n^2x}{2}+\frac{7cos2x}{2}

    =\frac{1}{2}+\frac{7}{2}cos2x
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  10. #10
    Member Furyan's Avatar
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    Re: Using identities

    Thank you ILikeSerena You are great teacher!
    Last edited by Furyan; December 18th 2011 at 02:24 PM. Reason: Correction
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  11. #11
    Member Furyan's Avatar
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    Re: Using identities

    Thank you very much Archie Meade I will try that too.

    Awesome!
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