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Math Help - prov trig indentity

  1. #1
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    prov trig indentity

    prove that  cos4x = 8cos^{4}x - 8cos^{2}x - 1

    I have tried for quite a while, but am completely stuck now.

    Can anyone have ago?

    Thanks.
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  2. #2
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    Quote Originally Posted by Tweety View Post
    prove that  cos4x = 8cos^{4}x - 8cos^{2}x - 1

    I have tried for quite a while, but am completely stuck now.

    Can anyone have ago?

    Thanks.
    You should use the identity

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

    Now make u = 2x

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

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

    Now use it again

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

    Now expand the RHS, you should have your answer.
    Last edited by pickslides; January 12th 2010 at 01:51 PM. Reason: bad latex
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  3. #3
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    Hi Tweety,

    first factorise..

    8Cos^{2}x(Cos^{2}x-1)-1=8Cos^{2}x(-Sin^{2}x)-1

    =-8Cos^{2}xSin^{2}x-1=-8\frac{1}{2}(1+cos2x)\frac{1}{2}(1-cos2x)-1

    =-2(1-Cos^{2}2x)-1=-2[1-\frac{1}{2}(1+Cos4x)]-1

    =-2+1+Cos4x-1

    I guess you have a typo,
    i reckon your original expression should have +1 instead of -1.
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  4. #4
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    Quote Originally Posted by pickslides View Post
    You should use the identity

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

    Now make u = 2x

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

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

    Now use it again

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

    Now expand the RHS, you should have your answer.

    Thank you,

    Could you please explain the last step, what happened to 2x? How did it become just 'x'?
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  5. #5
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    Cos2x=2Cos^{2}x-1

    Cos^{2}2x=(Cos2x)^2
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  6. #6
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    Quote Originally Posted by Tweety View Post
    Thank you,

    Could you please explain the last step, what happened to 2x? How did it become just 'x'?
    Archie Meade has explained the substitution I have used, is it clear?
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  7. #7
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    Quote Originally Posted by pickslides View Post
    Archie Meade has explained the substitution I have used, is it clear?
    Yes thank you!
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  8. #8
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    The method that pickslides used was very neat, Tweety.

    He used the same identity at each step.

    You may see the identity written as

    cos^{2}A=\frac{1}{2}(1+cos2A).

    Rearranging this, we get

    2cos^{2}A=1+cos2A

    so

    cos2A=2cos^{2}A-1

    He simply used this, by letting A=2x to get cos4x
    and the powers of x follow.

    I started with the powers and worked back, eventually using that identity,
    but i also used

    cos^{2}x+sin^{2}x=1

    in writing

    cos^{2}x-1=-sin^{2}x

    The other identity i used is

    sin^2x=\frac{1}{2}(1-cos2x)
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