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Math Help - i need help proving this identity

  1. #1
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    i need help proving this identity

    sin4x-sin2x/sin2x=cos3x/cosx



    i got left side equals right side..which is 1-4sin^2x
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  2. #2
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    1) Please add parentheses to clarify meaning.

    Is it as you have written it or did you mean (sin(4x)-sin(2x))/sin(2x)?

    2) Please show your work. It's nice that you managed some result, but how did you do that?

    3) When you get left side = right side, you are done.

    Where does that leave us?
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  3. #3
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    Hello, nithu!

    Working from one side only (the recommended method) is difficult.
    I worked on both sides and made them "meet in the middle" (as you did)
    . . then devised a way to connect them.


    Prove: . \frac{\sin4x-\sin2x}{\sin2x} \;=\;\frac{\cos3x}{\cos x}

    The right side is: . \frac{4\cos^3\!x - 3\cos x}{\cos x} \;\;=\;\;4\cos^2\!x - 3 \;\;=\;\;4\left(\frac{1+\cos2x}{2}\right)-3

    . . . . . = \;\;2 + 2\cos2x - 3 \;\;=\;\;2\cos2x - 1


    Multiply by \frac{\sin2x}{\sin2x}

    . . \frac{\sin2x}{\sin2x}\cdot(2\cos2x - 1) \;\;=\;\;\frac{2\sin2x\cos2x - \sin2x}{\sin2x} \;\;=\;\;\frac{\sin4x - \sin2x}{\sin2x}

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  4. #4
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    can u show me the left side and right side work...thx
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  5. #5
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    Hello, nithu!

    For the left-side work, read up from the bottom of my solution.

    You should be able to see how they "meet in the middle."

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