Question

2cos(-45) + sin 405 - tg(-540) 2si(-π/6) + cos 13π/3 +ctg(-5π/2)

Answer 1

$2Cos(-45)+Sin405-tg(-540)=2Cos45 +Sin(360+45)+$$tg(180*3+0)= 2Cos45 +Sin45 +tg0=2* \frac{ \sqrt{2} }{2} + \frac{ \sqrt{2} }{2}+0= \sqrt{2} + \frac{ \sqrt{2} }{2}$$= \frac{3 \sqrt{2} }{2}$

$2Sin(- \frac{ \pi }{6})+Cos \frac{13 \pi }{3}+Ctg(- \frac{5 \pi }{2})=-2Sin \frac{ \pi }{6} +Cos(4 \pi + \frac{ \pi }{3})-Ctg(4 \pi +$$\frac{ \pi }{2})= - 2* \frac{1}{2} +Cos \frac{ \pi }{3}-Ctg \frac{ \pi }{2}=-1+ \frac{1}{2}-0=-0,5$