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Math Help - Trigonometric simplification

  1. #1
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    Trigonometric simplification

    I worked a derivative problem all the way down to this:

    \frac {-csc^2 x}{(cot + 1)^2}

    Answer: \frac {-1}{(sinx + cosx)^2}

    I'm pretty sure there is a way to simplify this, but I don't see how, even after studying trigonometric identities.

    I think csc^2 + cot^2= 1 has something to do with this.

    Or maybe make csc and cot as their corresponding sin and cos values...

    I'm just not sure!!!

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  2. #2
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    Quote Originally Posted by Truthbetold View Post
    I worked a derivative problem all the way down to this:

    \frac {-csc^2 x}{(cot + 1)^2}

    Answer: \frac {-1}{(sinx + cosx)^2}

    I'm pretty sure there is a way to simplify this, but I don't see how, even after studying trigonometric identities.

    I think csc^2 + cot^2= 1 has something to do with this.

    Or maybe make csc and cot as their corresponding sin and cos values...

    I'm just not sure!!!

    By
    -\frac{csc^2(x)}{cot(x) + 1}

    = -\frac{\frac{1}{sin^2(x)}}{\frac{cos(x)}{sin(x)} + 1}

    = -\frac{1}{sin(x) cos(x) + sin^2(x)}

    = -\frac{1}{sin(x) ~ (cos(x) + sin(x) )}

    There is no way this is going to turn into
    -\frac{1}{(sin(x) + cos(x))^2} = -\frac{1}{1 + 2sin(x)cos(x)}

    -Dan
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