Question

Помогите пожалуйста срочно

Answer 1

$\displaystyle\bf\\3)\\\\Cos\Big(\frac{\pi }{4} -\alpha \Big)-Cos\Big(\frac{\pi }{4} +\alpha \Big)=-2Sin\frac{\dfrac{\pi }{4} -\alpha +\dfrac{\pi }{4}+\alpha }{2} \cdot Sin\frac{\dfrac{\pi }{4} -\alpha -\dfrac{\pi }{4}-\alpha }{2}=\\\\\\=-2Sin\frac{\dfrac{\pi }{2} }{2} \cdot Sin\frac{-2\alpha }{2} =2Sin\frac{\pi }{4} \cdot Sin\alpha =2\cdot\frac{\sqrt{2} }{2} \cdot Sin\alpha =\sqrt{2} Sin\alpha \\\\\\4)$

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