Answer
$sin(\frac{5\pi}{6})=\frac{1}{2}$
$cos(\frac{5\pi}{6})=-\frac{\sqrt 3}{2}$
$tan(\frac{5\pi}{6})=-\frac{\sqrt 3}{3}$
Work Step by Step
Let $θ'$ be the reference angle. We must have that $0\leqθ'\lt2\pi$
But, since $0\leqθ=\frac{5\pi}{6}\lt2\pi$, it is not necessary to find a reference angle.
$sin(\frac{5\pi}{6})=\frac{1}{2}$
$cos(\frac{5\pi}{6})=-\frac{\sqrt 3}{2}$
$tan(\frac{5\pi}{6})=-\frac{\sqrt 3}{3}$
