Question

Помогите пожалуйста (тригонометрия..)​

Answer 1

$\displaystyle\bf\\1)\\\\Sin(-45^\circ)+Cos(-45^\circ)+4Sin(-60^\circ)-3Cos (-30^\circ) =\\\\=-Sin45^\circ+Cos45^\circ-4Sin60^\circ-3Cos 30^\circ=\\\\=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2} -4\cdot\frac{\sqrt{3} }{2} -3\cdot\frac{\sqrt{3} }{2} =-2\sqrt{3} -1,5\sqrt{3} =-3,5\sqrt{3} \\\\\\2)\\\\5Sin(-30^\circ)+tg(-45^\circ)-3Ctg(-45^\circ)-5Cos (0^\circ) =\\\\=-5Sin30^\circ-tg45^\circ+3Ctg45^\circ-5Cos (0^\circ) =\\\\=-5\cdot\frac{1}{2} -1+3\cdot 1-5\cdot 1=-2,5-1+3-5=-5,5$

$\displaystyle\bf\\3)\\\\Sin\Big(-\frac{\pi }{2} \Big)-5Cos\Big(-\frac{\pi }{4} \Big)+5Sin\Big(-\frac{\pi }{4} \Big)-15Cos\frac{\pi }{2} =\\\\\\=-Sin\frac{\pi }{2} -5Cos\frac{\pi }{4} -5Sin\frac{\pi }{4} -15Cos\frac{\pi }{2} =\\\\\\=-1-5\cdot\frac{\sqrt{2} }{2} -5\cdot\frac{\sqrt{2} }{2} -15\cdot 0=-1-10\cdot\frac{\sqrt{2} }{2} =-1-5\sqrt{2} \\\\\\4)\\\\4tg\Big(-\frac{\pi }{4} \Big)\cdot Ctg\Big(-\frac{\pi }{4} \Big)+3Sin\Big(-\frac{\pi }{2} \Big)-5Cos\Big(-\frac{\pi }{6} \Big)=$

$\displaystyle\bf\\=-4tg\frac{\pi }{4} \cdot \Big(-Ctg\frac{\pi }{4} \Big)-3Sin\frac{\pi }{2} -5Cos\frac{\pi }{6} =-4\cdot 1\cdot (-1)-3\cdot 1 -5\cdot \frac{\sqrt{3} }{2}=\\\\\\=4-3-2,5\sqrt{3} =1-2,5\sqrt{3}$