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I am given this expression to simplify:

$\frac{\cos(2x)}{\cot(x)-1}-\frac{\sin(2x)}{2}$

and I know the correct answer is $\sin^2(x)$

I was able to reduce the second fraction to a bit nicer $\frac{\sin(2x)}{2}=\frac{2\sin(x)\cos(x)}{2\tan(x)\cot(x)}=\frac{2\sin(x)\cos(x)}{\frac{2\cos(x)\sin(x)}{\cos(x)\sin(x)}}=\frac{2\sin(x)\cos(x)}{1}\cdot\frac{\cos(x)\sin(x)}{2\cos(x)\sin(x)}=\cos(x)\sin(x)$

which changes the original into this

$\frac{\cos(2x)}{\cot(x)-1}-\cos(x)\sin(x)$

which is the prettiest I've got. No matter what I do with the first fraction I get something too complicated. Does anyone have any idea? Thx

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  • $\begingroup$ Unnecessary complicating, $\sin (2x) = 2\sin x \cos x $, then just cut out the 2's $\endgroup$ – windircurse Jun 15 '16 at 15:03
  • $\begingroup$ You could've just written $\frac{\sin (2x)}{2}=\frac{2\sin(x)\cos (x)}{2}=\sin(x)\cos (x)$. There were a lot of unnecessary steps. $\endgroup$ – user236182 Jun 15 '16 at 15:03
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$$\frac{\cos(2x)}{\cot(x)-1}-\frac{\sin(2x)}{2}$$

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

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

$$=\sin(x)(\cos(x)+\sin(x))-\sin(x)\cos(x)=\sin^2(x)$$

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  • $\begingroup$ Oh! $a^2-b^2 = (a+b)(a-b)$ this is the one I was failing to see the whole time. Thx $\endgroup$ – Smejki Jun 15 '16 at 15:54
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$$\frac { \cos 2x }{ \cot x-1 } -\frac { \sin 2x }{ 2 } =\frac { \cos ^{ 2 }{ x } -\sin ^{ 2 }{ x } }{ \frac { \cos { x } -\sin { x } }{ \sin { x } } } -\sin { x } \cos { x } =$$ $$=\frac { \sin { x } \left( \cos { x } +\sin { x } \right) \left( \cos { x } -\sin { x } \right) }{ \cos { x } -\sin { x } } -\sin { x\cos { x } = } $$ $$=\sin { x } \cos { x } +\sin ^{ 2 }{ x } -\cos { x } \sin { x } =\sin ^{ 2 }{ x } $$

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