Question

помогите пожалуйста решить 1 вариант

Answer 1

α - угол третьей четверти поэтому Sinα < 0

$Sinalpha=- sqrt{1-Cos^{2}alpha}=-sqrt{1-(-0,6)^{2}}=-sqrt{1-0,36}=-sqrt{0,64}=-0,8\tg(frac{pi }{2}+alpha)=-Ctgalpha=-frac{Cosalpha }{Sinalpha }=-frac{-0,6}{-0,8}=-0,75\2)frac{2Sin^{2}alpha*Ctgalpha}{Cos^{2}alpha-Sin^{2}alpha}=frac{2Sin^{2}alpha*frac{Cosalpha }{Sinalpha }}{Cos2alpha} =frac{2Sinalpha Cosalpha}{Cos2alpha }=frac{Sin2alpha }{Cos2alpha}=tg2alpha\\tg2alpha=tg2alpha \frac{2Sin^{2} alpha }{tg2alpha*tgalpha}=Cos^{2}alpha-Sin^{2}alpha$

$frac{2Sin^{2}alpha}{tg2alpha*frac{Sinalpha}{Cosalpha }}=Cos2alpha\frac{2Sinalpha Cosalpha}{tg2alpha }=Cos2alpha\frac{Sin2alpha }{frac{Sin2alpha }{Cos2alpha} }=Cos2alpha \Cos2alpha=Cos2alpha$

$3)Sin240^{0} =Sin(180^{0}+60^{0} )=-Sin60^{0}=-frac{sqrt{3}}{2} \Cosfrac{3pi }{4}=Cos(pi-frac{pi }{4})=-Cosfrac{pi }{4}=-frac{sqrt{2}}{2}\ Ctg(-frac{pi }{6})=-Ctgfrac{pi }{6} =-sqrt{3}$