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Math Help - [SOLVED] trig question

  1. #1
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    [SOLVED] trig question

    Hi, got a problem that I need some help with.

    How can I express cos 4x in cos x?

    Help appreciated.
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  2. #2
    Member great_math's Avatar
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    \cos 4x=\cos(3x+x)=\cos3x\cdot\cos x - \sin 3x\cdot \sin x

    =(4\cos^3x-3\cos x)\cos x-(3\sin x-4\sin^3 x)\cdot \sin x

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

    =4\cos^4x+4\sin^4x-3

    =4\cos^4x+4(1-\cos^2 x)^2-3

    =4\cos^4x+4(1+\cos^4x-2\cos^2x)-3

    =4\cos^4x+4+4\cos^4x-8\cos^2x-3

    =8\cos^4x-8\cos^2x+1
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  3. #3
    Member great_math's Avatar
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    Another way:

    \cos 4x=2\cos^22x-1

    =2(2\cos^2x-1)^2-1

    =2(4\cos^4x+1-4\cos^2x)-1

    =8\cos^4x-8\cos^2x+1
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  4. #4
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    Great, the other one was a bit easier Does the answer satisfy the question in the sense that it requires the answer to be expressed in cosx, whereas the answer you calculated has a + 1 term?
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  5. #5
    Member great_math's Avatar
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    1 is a constant so it does satisfy the question
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  6. #6
    Member great_math's Avatar
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    8\cos^4x-8\cos^2x+1

    This can also be written as (8\cos^4x-8\cos^2x+1)\frac{\cos x}{\cos x}

    =\frac{8\cos^5x-8\cos^3x+\cos x}{\cos x}
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