# simplifying trig expression (I think)

• Feb 16th 2013, 05:11 PM
infraRed
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?
• Feb 16th 2013, 06:35 PM
Soroban
Re: simplifying trig expression (I think)
Hello, infraRed!

Quote:

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

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

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

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

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