Question

sin^2(-П3)+cos^2(-n6)
б) 4sin n6· sin^n4 · tg^2 n3

Answer 1

$sin^2(- pi /3)+cos^2(- pi /6)=(-sin pi /3)^2+cos^2 pi /6=\=( sqrt{3}/2)^2+ ( sqrt{3}/2)^2= 2*3/4=3/2=1,5\\4sin( pi /6)*sin^2( pi /4)*tg^2( pi /3)=4*1/2*( sqrt{2}/2)^2*( sqrt{3})^2=\=2*2/4*3=3$