The unit circle and circular functions practice questions — IB Maths AA

$\sin\!\left(2\pi - \dfrac{\pi}{6}\right) = -\sin\dfrac{\pi}{6} = -\dfrac{1}{2}$.

$\dfrac{7\pi}{4} \in \!\left(\dfrac{3\pi}{2}, 2\pi\right)$, which is the fourth quadrant.

$\sin x = -\dfrac{1}{2}$ at $x = \dfrac{7\pi}{6}$ (Q3) and $x = \dfrac{11\pi}{6}$ (Q4). Two solutions on $[0, 2\pi)$.

$\tan\!\left(\pi + \dfrac{\pi}{6}\right) = \tan\dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}}$.

$\cos\!\left(-\dfrac{\pi}{3}\right) = \cos\dfrac{\pi}{3} = \dfrac{1}{2}$.

$\dfrac{\pi}{6} = 30^{\circ}$, and $\sin 30^{\circ} = \dfrac{1}{2}$.

A standard unit-circle value: $\sin\dfrac{\pi}{6} = \dfrac{1}{2}$.

In the second quadrant ($\pi/2 < x < \pi$), $\sin x > 0$.

$\cos\dfrac{\pi}{4} = \sin\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}$.

$\tan\dfrac{\pi}{3} = \dfrac{\sin(\pi/3)}{\cos(\pi/3)} = \dfrac{\sqrt{3}/2}{1/2} = \sqrt{3}$.