ответ:
tg(3π/8)·tg(π/8)=(sin(3π/8)·sin(π/8))/(cos(3π/8)·cos(π/8))=
=(1/2)(cos(2π/8)–cos(4π/8))/(1/2)(cos(4π/8)+cos(2π/8))
так как cos(4π/8)=cos(π/2)=0,
cos(2π/8)=cos(π/4), то
tg(3π/8)·tg(π/8)+1=
=cos(π/4)/cos(π/4)+1=2
о т в е т. 2
пошаговое объяснение:
tg(3π/8)·tg(π/8)=(sin(3π/8)·sin(π/8))/(cos(3π/8)·cos(π/8))=
=(1/2)(cos(2π/8)–cos(4π/8))/(1/2)(cos(4π/8)+cos(2π/8))
так как cos(4π/8)=cos(π/2)=0,
cos(2π/8)=cos(π/4), то
tg(3π/8)·tg(π/8)+1=
=cos(π/4)/cos(π/4)+1=2
о т в е т. 2
