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Math Help - Using half angle formula

  1. #1
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    Using half angle formula

    Using the half angle formula

    <br /> <br />
cos^2(\theta) = \frac{1}{2} * (1+cos(2\theta))<br />

    How would I go about showing that

    <br /> <br />
cos^2(2\theta) = \frac{1}{2} * cos(4\theta)+\frac{1}{2}<br />
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  2. #2
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    Quote Originally Posted by macca101
    Using the half angle formula

    <br /> <br />
cos^2(\theta) = \frac{1}{2} * (1+cos(2\theta))<br />

    How would I go about showing that

    <br /> <br />
cos^2(2\theta) = \frac{1}{2} * cos(4\theta)+\frac{1}{2}<br />
    Just double the angle in your formula. I suspect you are trying to "over think" the problem here...

    -Dan
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  3. #3
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    Quote Originally Posted by macca101
    Using the half angle formula

    <br /> <br />
cos^2(\theta) = \frac{1}{2} * (1+cos(2\theta))<br />

    How would I go about showing that

    <br /> <br />
cos^2(2\theta) = \frac{1}{2} * cos(4\theta)+\frac{1}{2}<br />
    You know that,
    \cos^2(x)=\frac{1}{2}\cdot (1+\cos 2x)
    Then in the problem,
    \cos^2(2\theta)
    set x=2\theta
    Then,
    \frac{1}{2}\cdot(1+\cos [2(2\theta)])=\frac{1}{2}\cdot (1+\cos 4\theta)
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  4. #4
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    Quote Originally Posted by ThePerfectHacker
    You know that,
    \cos^2(x)=\frac{1}{2}\cdot (1+\cos 2x)
    Then in the problem,
    \cos^2(2\theta)
    set x=2\theta
    Then,
    \frac{1}{2}\cdot(1+\cos [2(2\theta)])=\frac{1}{2}\cdot (1+\cos 4\theta)
    Ok Thanks for that.

    I thought I was stuck then boy I'm I struggling now. The question goes on ...

    Use the answer above and another half angle formula

    I Think


    <br />
sin^2(\theta) = \frac{1}{2} * (1-cos(2\theta))<br />

    To show that

    <br />
sin^4(\theta) = \frac{3}{8}- \frac{1}{2} * cos(2\theta)+\frac{1}{8}* cos(4\theta)<br />

    Apologies if I should have started a new thread for this. Thought it would be OK to continue as it relates to the original post.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by macca101
    Ok Thanks for that.

    I thought I was stuck then boy I'm I struggling now. The question goes on ...

    Use the answer above and another half angle formula

    I Think


    <br />
sin^2(\theta) = \frac{1}{2} * (1-cos(2\theta))<br />

    To show that

    <br />
sin^4(\theta) = \frac{3}{8}- \frac{1}{2} * cos(2\theta)+\frac{1}{8}* cos(4\theta)<br />

    Apologies if I should have started a new thread for this. Thought it would be OK to continue as it relates to the original post.
    sin^4(\theta) = sin^2(\theta) \, sin^2(\theta)=(1/4)(1-cos(2\theta))^2
    =(1/4)(1-2cos(2\theta)+cos^2(2\theta))

    And we know that
    cos^2(2\theta)=(1/2)(1+cos(4\theta))

    So you should just be able to plug that into your sin^4 formula.

    -Dan
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  6. #6
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    Thanks It was the

    <br />
sin^4(\theta) = sin^2(\theta) \, sin^2(\theta)<br />

    Thing that was eluding the empty shell I carry on my shoulders
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by macca101
    Thanks It was the

    <br />
sin^4(\theta) = sin^2(\theta) \, sin^2(\theta)<br />

    Thing that was eluding the empty shell I carry on my shoulders
    Don't feel so bad. Soon you'll be randomly adding and subtracting 1 from equations just like the rest of us. (The tricks we use are seem so inane sometimes!)

    -Dan
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