Jeżeli cosβ=- 1/3 i β∈(π, 3/2π ), to wartość wyrażenia sin(β- 1/3π ) jest równa

$\displaystyle \frac{-2\sqrt{2}+\sqrt{3}}{6}$\displaystyle \frac{-2\sqrt{2}+\sqrt{3}}{6}

$\displaystyle \frac{2\sqrt{6}+1}{6}$\displaystyle \frac{2\sqrt{6}+1}{6}

$\displaystyle \frac{2\sqrt{2}+\sqrt{3}}{6}$\displaystyle \frac{2\sqrt{2}+\sqrt{3}}{6}

$\displaystyle \frac{1-2\sqrt{6}}{6}$\displaystyle \frac{1-2\sqrt{6}}{6}