Inverse Trigonometric Functions practice questions — A-Level Maths

For any $x\in [-1,1]$, $\arcsin x+\arccos x={\displaystyle \frac{\pi }{2}}$ (a standard identity, since $\arccos x={\displaystyle \frac{\pi }{2}}-\arcsin x$).

${\sin }^2\theta =1-{\displaystyle \frac{4}{9}}={\displaystyle \frac{5}{9}}$, so $\sin \theta ={\displaystyle \frac{\sqrt{5}}{3}}$. Hence $\tan \theta ={\displaystyle \frac{\sin \theta }{\cos \theta }}={\displaystyle \frac{\sqrt{5}\mathrm{/}3}{2\mathrm{/}3}}={\displaystyle \frac{\sqrt{5}}{2}}$.

${\cos }^2\theta =1-{\sin }^2\theta =1-{\displaystyle \frac{9}{25}}={\displaystyle \frac{16}{25}}$, and $\cos \theta \ge 0$, so $\cos \theta ={\displaystyle \frac{4}{5}}$.

${\displaystyle \frac{2\pi }{3}}\in [0,\pi ]$, so $\arccos \text{\hspace{0.17em}⁣}(\cos {\displaystyle \frac{2\pi }{3}})={\displaystyle \frac{2\pi }{3}}$.

$\tan {\displaystyle \frac{\pi }{3}}=\sqrt{3}$, and ${\displaystyle \frac{\pi }{3}}\in \text{\hspace{0.17em}⁣}(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}})$, so $\arctan \sqrt{3}={\displaystyle \frac{\pi }{3}}$.

$\sin {\displaystyle \frac{7\pi }{6}}=-{\displaystyle \frac{1}{2}}$. Then $\arcsin \text{\hspace{0.17em}⁣}(-{\displaystyle \frac{1}{2}})=-{\displaystyle \frac{\pi }{6}}$ (in the principal range).

$\cos {\displaystyle \frac{\pi }{2}}=0$, so $\arccos 0={\displaystyle \frac{\pi }{2}}$.

$\tan {\displaystyle \frac{\pi }{4}}=1$, so $\arctan 1={\displaystyle \frac{\pi }{4}}$.

$\sin {\displaystyle \frac{\pi }{6}}={\displaystyle \frac{1}{2}}$, and ${\displaystyle \frac{\pi }{6}}$ is in the principal range, so $\arcsin {\displaystyle \frac{1}{2}}={\displaystyle \frac{\pi }{6}}$.

We need $\theta \in [0,\pi ]$ with $\cos \theta =\frac{1}{2}$.

From the unit circle, $\cos \text{\hspace{0.17em}⁣}\left(\frac{\pi }{3}\right)=\frac{1}{2}$.

$\frac{\pi }{3}$ lies in $[0,\pi ]$, so $\arccos \text{\hspace{0.17em}⁣}\left(\frac{1}{2}\right)={\displaystyle \frac{\pi }{3}}$.