The Unit Circle practice questions — A-Level Maths

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

$\tan \text{\hspace{0.17em}⁣}(\pi +{\displaystyle \frac{\pi }{6}})=\tan {\displaystyle \frac{\pi }{6}}={\displaystyle \frac{1}{\sqrt{3}}}$.

${\displaystyle \frac{7\pi }{4}}\in \text{\hspace{0.17em}⁣}({\displaystyle \frac{3\pi }{2}},2\pi )$, which is the fourth quadrant.

$\cos \text{\hspace{0.17em}⁣}(-{\displaystyle \frac{\pi }{3}})=\cos {\displaystyle \frac{\pi }{3}}={\displaystyle \frac{1}{2}}$.

$\sin \text{\hspace{0.17em}⁣}(2\pi -{\displaystyle \frac{\pi }{6}})=-\sin {\displaystyle \frac{\pi }{6}}=-{\displaystyle \frac{1}{2}}$.

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

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

${\displaystyle \frac{\pi }{3}}={60}^{\circ }$, and $\cos {60}^{\circ }={\displaystyle \frac{1}{2}}$.

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

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