Question

248. Упростите выражение
С решением
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Answer 1

$\displaystyle\bf\\1)\\\\Sin(\alpha -180^\circ)\cdot Cos(-\alpha )\cdot tg(-\alpha )\cdot Ctg(360^\circ+\alpha )=\\\\=-Sin\alpha \cdot Cos\alpha\cdot(-tg\alpha )\cdot Ctg\alpha =Sin\alpha Cos\alpha \cdot 1=Sin\alpha Cos\alpha =\\\\=0,5\cdot 2Sin\alpha Cos\alpha =0,5Sin2\alpha \\\\2)\\\\\frac{Cos(\alpha -90^\circ)\cdot tg(\alpha -360^\circ)}{Cos\Big(\dfrac{3\pi }{2}+\alpha \Big) \cdot Ctg(360^\circ-\alpha )} =\frac{Sin\alpha \cdot tg\alpha }{Sin\alpha \cdot (-Ctg\alpha )} =$

$\displaystyle\bf\\=-\frac{tg\alpha }{Ctg\alpha } =-\frac{tg\alpha }{\frac{1}{tg\alpha } } =-tg^{2} \alpha \\\\3)\\\\\frac{Cos\Big(\dfrac{3\pi }{2} +\alpha \Big)- tg(-\alpha )+ Ctg\Big(\dfrac{3\pi }{2}+\alpha \Big) }{Ctg\Big(\dfrac{\pi }{2} +\alpha \Big)+Cos(-\alpha )+Sin\Big(\dfrac{3\pi }{2}-\alpha \Big) } =\\\\\\=\frac{Sin\alpha +tg\alpha -tg\alpha }{-tg\alpha +Cos\alpha -Cos\alpha } =-\frac{Sin\alpha }{tg\alpha } =-\frac{Sin\alpha }{\frac{Sin\alpha }{Cos\alpha } }=$

$\displaystyle\bf\\=-\frac{Sin\alpha \cdot Cos\alpha }{Sin\alpha } =-Cos\alpha \\\\\\4)\\\\\frac{Sin\Big(\dfrac{3\pi }{2}-\alpha \Big) \cdot tg\Big(\dfrac{\pi }{2} +\alpha \Big)}{tg(360^\circ-\alpha )\cdot Sin(-360^\circ-\alpha )} =\frac{-Cos\alpha \cdot(-Ctg\alpha )}{-tg\alpha \cdot Sin(360^\circ+\alpha )} =$

$\displaystyle\bf\\=-\frac{Cos\alpha \cdot Ctg\alpha }{tg\alpha \cdot Sin\alpha } =\frac{Cos\alpha \cdot\dfrac{Cos\alpha }{Sin\alpha } }{\dfrac{Sin\alpha }{Cos\alpha } \cdot Sin\alpha } =-\frac{Cos^{2}\alpha \cdot Cos\alpha }{Sin^{2}\alpha \cdot Sin\alpha } =\\\\\\=-\frac{Cos^{3} \alpha }{Sin^{3} \alpha } =-Ctg^{3} \alpha$