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Math Help - Various Trig problems involving trigonometric identities

  1. #1
    zll
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    Various Trig problems involving trigonometric identities

    Couldn't think of a title, as i'll be using this thread more than once today for related, but different problems.

    for starters:
    <br />
\frac{1}{2}\sin{2x}[\tan^2{x}-\cos{2x}+2\cos^2{2x}]<br />

    For this, I ended up with -2sinxcosx, it's supposedly wrong.

    Here's my work, for reference:

    <br />
\frac{1}{2}\sin{2x}[\tan^2{x}-\cos{2x}+2\cos^2{2x}]<br />

    <br />
\frac{1}{2}\sin{2x}[\tan^2{x}-2\cos^2{x}-1+2\cos^2{2x}]<br />

    <br />
\frac{1}{2}\sin{2x}[\tan^2{x}-1]<br />

    <br />
\sin{x}\cos{x}(\tan^2{x}-1)<br />

    <br />
\sin{x}\cos{x}(\frac{1-\cos{2x}}{1+\cos{2x}}-1)<br />

    <br />
\frac{\sin{x}\cos{x}-\sin{x}\cos{x}\cos{2x}}{1+\cos{2x}}-\sin{x}\cos{x}<br />

    <br />
\frac{-\sin{x}\cos{x}(1+\cos{2x})}{1+\cos{2x}}<br />

    <br />
-2\sin{x}\cos{x}<br />

    Thanks in advance/
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  2. #2
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    earboth's Avatar
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    Only one question:

    Quote Originally Posted by zll View Post
    ...

    <br />
\frac{-\sin{x}\cos{x}(1+\cos{2x})}{1+\cos{2x}}<br />

    <br />
-2\sin{x}\cos{x} <br />
<-- where does the factor 2 comes from?

    Thanks in advance/
    ...
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  3. #3
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    What is the question? Simplify as far as possible, or is there a specific expression you are supposed to find?
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  4. #4
    zll
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    in the previous picture, i accidentally left out the -sinxcosx next to the fraction which would combine with the -sinxcosx that would be left after the 1+cos2x's cancelled out.

    I've since figured this out (the question was to simplify as far as possible, by the way. sorry about that.). It's tanx.
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