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Math Help - Write (Cos(5x))^2 using half angle formula

  1. #1
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    Write (Cos(5x))^2 using half angle formula

    I need to know how to write some thing like (Cos(5x))^2 using a half angle formula, Such that I have something that looks like

    (Cos(5x))^2=_______+_______cos(___x)

    I know with a half angle formula that
    Cos(x/2)=+/- sqrt((1+cos(x))/2)
    but that ^2 is throwing me off, I don't know what to do with it per se.

    Would something like = +/- sqrt((1+cos(5x)^2)/2)
    Then using a power reducing formula work?
    cos(x)^2=(1+cos2x)/(2)

    \sqrt{\frac{1+\cos\!\left(\frac{2\cdot 5x}{2}\right)}{2}}" alt="\sqrt{\frac{1+\cos\!\left(\frac{2\cdot 5x}{2}\right)}{2}}" />
    (i do not know why font is coming up)

    I'm not sure what to do with 5x if I'm supposed to end up with 2x
    would it be cos(2(5x)) or am I going down the wrong path?

    Any suggestions?
    Thank you for your time.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    \sqrt{\dfrac{1+\cos\left(\frac{2\cdot 5x}{2}\right)}{2}}


    Why did you put \cos(\frac{2\cdot 5x}{2})

    If you take

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

    This becomes:

    \cos(5x)= \sqrt{\dfrac{1+\cos(10x)}{2}}

    But you don't need to take the square root...
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  3. #3
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    I ended up with
    (Cos(5x))^2 = 1/2+1/2(cos(10x))

    but I must admit that I'm not sure how I came up with this answer I just made a few assumptions and thus guess works. Can someone help me step this out using trig identities?
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  4. #4
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    e^(i*pi)'s Avatar
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    I suspect you're using the identity for \cos(2 \theta) because it looks that way.

    \cos(2 \theta) = 2\cos^2 \theta - 1 (If you don't know this then learn it, and learn it quickly! You'll need it)

    Rearrange that using algera to get \cos^2 \theta as the subject:

    \dfrac{1}{2}\,\left(\cos(2 \theta) +1\right) = \dfrac{1}{2} + \dfrac{1}{2}\cos(2 \theta)

    In your question use the above and simply subsitute \theta = 5x
    Last edited by e^(i*pi); November 1st 2010 at 12:56 PM. Reason: tag
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  5. #5
    MHF Contributor Unknown008's Avatar
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    Yep, that's it.

    You can remember all the three forms of cos2theta in one go.

    \begin{array}{ccl}<br />
\cos2\theta &=& \cos^2\theta - \sin^2\theta \\<br />
&=& 2 \cos^2\theta - 1\\<br />
&=& 1 - 2\sin^2\theta \end{array}

    If you can remember at least the first one, you can use Pythagoras' Theorem to substitute the sin^2 and cos^2 ratios.
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  6. #6
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    Ok, I think I get it. If I take 5x out of the equation and plug it back in later.

    The power reduction formula
    cos^2 A = (1 + cos2A)/2
    seems a bit of a better shortcut to cos2x=2cosx^2-1

    *sigh*

    Anyway. Thanks for all the help! I'm getting much better at throwing this stuff around!
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