tex]...

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[tex] \frac{2 \sin( \alpha ) - \sin(2 \alpha ) }{ \cos(2 \alpha ) - 1 } - \cot( \alpha ) [/tex]

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Artem112

Verified answer

Основные формулы:

$\sin^2x+\cos^2x=1$

$2\sin x=2\sin x\cos x$

$\cos2x=2\cos^2x-1$

$\mathrm{ctg}\,x=\dfrac{\cos x}{\sin x}$

Получаем:

$\dfrac{2 \sin \alpha - \sin2 \alpha }{ \cos2 \alpha - 1 } - \mathrm{ctg}\, \alpha=\dfrac{2 \sin \alpha -2 \sin \alpha\cos\alpha }{ 2\cos^2 \alpha-1 - 1 } - \mathrm{ctg}\, \alpha=$

$=\dfrac{2 \sin \alpha(1 -\cos\alpha) }{ 2\cos^2 \alpha-2} - \mathrm{ctg}\, \alpha=-\dfrac{2 \sin \alpha(\cos\alpha-1) }{ 2(\cos^2 \alpha-1)} - \mathrm{ctg}\, \alpha=$

$=-\dfrac{2 \sin \alpha(\cos\alpha-1) }{ 2(\cos \alpha-1)(\cos \alpha+1)} - \mathrm{ctg}\, \alpha=-\dfrac{ \sin \alpha }{\cos \alpha+1} -\dfrac{\cos\alpha }{\sin\alpha } =$

$=-\left(\dfrac{ \sin \alpha }{\cos \alpha+1} +\dfrac{\cos\alpha }{\sin\alpha } \right)=-\dfrac{ \sin \alpha\sin\alpha +\cos\alpha (\cos\alpha +1) }{\sin\alpha (\cos \alpha+1)} =$

$=-\dfrac{ \sin^2 \alpha +\cos^2\alpha +\cos\alpha }{\sin\alpha (\cos \alpha+1)} =-\dfrac{ 1+\cos\alpha }{\sin\alpha (\cos \alpha+1)} =-\dfrac{ 1 }{\sin\alpha}$

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