Question

Дано: cosa= -0,6, п/2<а<п; sinB= -0,6, 3п/2<В<2п. Найти: sin(a-B)

Answer 1

$cosalpha=-0,6,;;;fracpi2 textless alpha textless pi\sinbeta=-0,6,;;;frac{3pi}2 textless beta textless 2pi\sin(alpha-beta)=sinalphacosbeta-sinbetacosalpha\sinalpha=sqrt{1-cos^2alpha}=sqrt{1-0,36}=sqrt{0,64}=pm0,8\fracpi2 textless alpha textless piRightarrowsinalpha textgreater 0,;sinalpha=0,8\cosbeta=sqrt{1-sin^2beta}=sqrt{1-0,36}=sqrt{0,64}=pm0,8\frac{3pi}2 textless beta textless 2piRightarrowcosbeta textgreater 0,;cosbeta=0,8\sin(alpha-beta)=0,8cdot0,8-(-0,6)cdot(-0,6)=0,64-0,36=0,28$