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Thread: Double argument property?

  1. #1
    Member wiseguy's Avatar
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    Double argument property?

    I'm not quite sure if I'm answering the question...?

    "Use the double argument property, $\displaystyle cos 2x = 1 - 2 sin2x$ to express cos 60 in terms of sin 30."

    My response:

    $\displaystyle x=30

    cos(2*30)=1-2sin^2(30)

    cos(60)=1-2sin^2(30)

    (cos(60)-1)/-2)=sin^2(30)

    0.25=sin^2(30)

    0.5=sin(30)$
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  2. #2
    Member wiseguy's Avatar
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    The next question asks the exact same thing, except in reverse:

    "Use the double argument property: $\displaystyle cos 2x = 1 - 2sin2 x$ to express cos 60 in terms of sin 30."

    Would I just take my work and swap everything over the equal sign? Assuming I approached this problem correctly...
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  3. #3
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    Quote Originally Posted by wiseguy View Post
    I'm not quite sure if I'm answering the question...?

    "Use the double argument property, $\displaystyle cos 2x = 1 - 2 sin2x$
    This is incorrect- probably a typo. You mean $\displaystyle cos(2x)= 1- 2 sin^2(x)$

    My response:

    $\displaystyle x=30

    cos(2*30)=1-2sin^2(30)

    cos(60)=1-2sin^2(30)p$
    Very good! And you are finished! You have "expressed cos(60) in terms of sin(30)".

    $\displaystyle (cos(60)-1)/-2)=sin^2(30)

    0.25=sin^2(30)

    0.5=sin(30)$
    You were not asked to evaluate cos(60) or sin(30).
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