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Math Help - double angle/half angle formulas

  1. #1
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    double angle/half angle formulas

    I have two problems for homework tonight that I can't get.
    they are over double and half angles.
    if someone can get any of these, and show how they did it or give a short explanation how, that would be awesome.

    1. sin2x[4cos^4(x) - 4cos^2(x)]^2
    I started by changing sin2x to 2sinxcosx & squared the parenthetical part
    then I multiplied it all out but I don't see any way to simplify.


    2. sin(C + D) = 7/8, (C+D) is in quadrant I, and sin C = 2/8, C is in quadrant I
    find sin D
    this one I'm rather lost. I know the formula for sin(C + D), so I can make it
    (sinC)(cosD) + (cosC)(sinD) = 7/8
    change the sin C to 2/8... but can't get it all figured out.


    please help!
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  2. #2
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    Auckland, New Zealand
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    For (Q1),

    \sin(2x) \times (4 \cos^4x - 4 \cos^2x)^2

    = 2 \sin{x} \cos{x} \times [4 \cos^2x ( \cos^2x - 1)]^2

    We know 1 - \cos^2x = \sin^2x:

    So  2 \sin{x} \cos{x} \times [4 \cos^2x ( \cos^2x - 1)]^2

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

    = 2 \sin{x} \cos{x} \times (16 \cos^4x \, \sin^4x)

    = 32 \sin^5x \, \cos^5x
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  3. #3
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    Auckland, New Zealand
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    For (Q2),

    \sin(C + D) = \frac{7}{8}

    We know \sin{C} = \frac{2}{8} = \frac{1}{4}

    C = \arcsin{\left( \frac{1}{4} \right)} = 14.478_{}^{o} (Quadrant I)
    <br />
\sin{(C + D)} = \sin{(14.478_{}^{o} + D)} = \frac{7}{8}
    <br />
{14.478_{}^{o} + D} = \arcsin{\left(\frac{7}{8} \right)} = 61.045_{}^{o} (Quadrant I)

    D = 61.045_{}^{o} - 14.478_{}^{o} = 46.567_{}^{o}

    Now you can find \sin{D}
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  4. #4
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    thank you! I did not think to use arcsin for question 2.
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