HSF.TF.A.2 math practice. Learn by doing.

On the unit circle, $\sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$.

On the unit circle, $\tan\left(\dfrac{\pi}{3}\right) = \dfrac{\sin\left(\dfrac{\pi}{3}\right)}{\cos\left(\dfrac{\pi}{3}\right)} = \dfrac{\sqrt{3}/2}{1/2} = \sqrt{3}$.

The reference angle for $\dfrac{5\pi}{6}$ is $\dfrac{\pi}{6}$. Cosine is negative in Quadrant II, so $\cos\left(\dfrac{5\pi}{6}\right) = -\cos\left(\dfrac{\pi}{6}\right) = -\dfrac{\sqrt{3}}{2}$.

The reference angle for $\dfrac{3\pi}{4}$ is $\dfrac{\pi}{4}$. Tangent is negative in Quadrant II, so $\tan\left(\dfrac{3\pi}{4}\right) = -\tan\left(\dfrac{\pi}{4}\right) = -1$.

The angle $\dfrac{5\pi}{6}$ is in Quadrant II with reference angle $\dfrac{\pi}{6}$. Sine is positive there, so $\sin\left(\dfrac{5\pi}{6}\right) = \sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$.

On the unit circle, $\cos\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}$.

At $\dfrac{3\pi}{2}$ the terminal side points to $(0, -1)$ on the unit circle, so $\sin\left(\dfrac{3\pi}{2}\right) = -1$.

The reference angle for $\dfrac{7\pi}{4}$ is $\dfrac{\pi}{4}$. Tangent is negative in Quadrant IV, so $\tan\left(\dfrac{7\pi}{4}\right) = -\tan\left(\dfrac{\pi}{4}\right) = -1$.