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

Решите задачу:

$\cos\alpha=-0,6,\;\;\;\frac\pi2\ \textless \ \alpha\ \textless \ \pi\\sin\beta=-0,6,\;\;\;\frac{3\pi}2\ \textless \ \beta\ \textless \ 2\pi\\\sin(\alpha-\beta)=\sin\alpha\cos\beta-\sin\beta\cos\alpha\\\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-0,36}=\sqrt{0,64}=\pm0,8\\\frac\pi2\ \textless \ \alpha\ \textless \ \pi\Rightarrow\sin\alpha\ \textgreater \ 0,\;\sin\alpha=0,8\\\cos\beta=\sqrt{1-\sin^2\beta}=\sqrt{1-0,36}=\sqrt{0,64}=\pm0,8\\\frac{3\pi}2\ \textless \ \beta\ \textless \ 2\pi\Rightarrow\cos\beta\ \textgreater \ 0,\;\cos\beta=0,8\\\sin(\alpha-\beta)=0,8\cdot0,8-(-0,6)\cdot(-0,6)=0,64-0,36=0,28$