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Math Help - trig: missing steps?

  1. #1
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    trig: missing steps?

    Hi. One of my math books seems to indicate that

    \sqrt {2 - 2\,sin\,t\,sin\,5t - 2\,cos\,t\,cos\,5t

    is equal to

    \sqrt {2 - 2\,cos\,4t}

    The transformation is I imagine quite obvious to everyone else, but if somebody could bear with me and show the hidden steps, it would help my learning.
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  2. #2
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    Re: trig: missing steps?

    Hello, infraRed!

    We are expected to know this identity:

    . . \cos(A\pm B) \:=\:\cos A\cos B \mp\sin A\sin B


    My math book indicates that: . \sqrt {2 - 2\sin t\sin 5t - 2\cos t\cos5t} \;=\; \sqrt {2 - 2\cos4t}

    The left side has:

    2 - 2\sin t\sin5t - 2\cos t\cos5t

    . . =\;2 - 2(\cos 5t\cos t + \sin5t\sin t)

    . . =\;2 - 2\cos(5t-t)

    . . =\;2 - 2\cos4t
    Thanks from infraRed
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