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Math Help - simplifying trig expression (I think)

  1. #1
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    simplifying trig expression (I think)

    Not sure if this is a Calc or a trig question. But I wanted to take the derivative of this function:

    y = -csc\,x - sin\,x

    Through simple steps I got:

    -\frac{d}{dx}(csc\,x)\,-\,\frac{d}{dx}sin\,x

    csc\,x\,cot\,x\,-\,cos\,x

    However the answer in the book is...

    cos\,x\,cot^2\,x

    So presumably my calculus is wrong, or I didn't simplify enough. I have tried various transformations but never quite end up with that. Can anybody show me how to simplify to that?
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  2. #2
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    Re: simplifying trig expression (I think)

    Hello, infraRed!

    \text{Find the derivative of: }\:y \:=\: -\csc x - \sin x

    \text{I got: }\:y' \:=\:\csc x\cot x - \cos x

    \text{However, the answer in the book is: }\,\cos x \cot^2\!x

    \text{So presumably my calculus is wrong, or I didn't simplify enough.}

    Your calculus is correct!


    \csc x\cot x - \cos x \;=\;\csc x\cdot\tfrac{\cos x}{\sin x} - \cos x

    . . . . . . . . . . . . . . . =\;\cos x\left(\tfrac{\csc x}{\sin x} - 1\right)

    . . . . . . . . . . . . . . . =\;\cos x\left(\frac{\csc x}{\frac{1}{\csc x}} - 1\right)

    . . . . . . . . . . . . . . . =\;\cos x\underbrace{\left(\csc^2\!x - 1\right)}_{\text{This is }\cot^2x}

    . . . . . . . . . . . . . . . =\;\cos x\cot^2\!x
    Thanks from infraRed
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