Question

ctg450°-sin 5π/4= 0-1 =
(решить)​

Answer 1

$Ctg450^{o}-Sin\frac{5\pi }{4}=Ctg(360^{o}+90^{o})-Sin(\pi+\frac{\pi }{4})=Ctg90^{o}+Sin\frac{\pi }{4}=0+\frac{\sqrt{2} }{2}=\frac{\sqrt{2} }{2}$