[ Inverse trigonometric functions ]

Calculate:

a) $\arccos \left[\sin \left(-\frac{\pi}{7}\right)\right]$

b) $\arcsin \left(\cos \frac{33 \pi}{5}\right)$.

## Answer

a) $\frac{9 \pi}{14} ;$ b) - $\frac{\pi}{10}$.

$$
\operatorname{arctan} x+\operatorname{arccot} x=\frac{\pi}{2} ; \quad \arcsin x+\arccos x=\frac{\pi}{2}
$$

#
Based on the definition we have:

$$
-\frac{\pi}{2} \leqslant \operatorname{arctg} x \leqslant \frac{\pi}{2}, \quad 0 \leqslant \operatorname{arcctg} x \leqslant \pi
$$

From this,

$$
-\frac{\pi}{2} \leqslant \operatorname{arctg} x+\operatorname{arcctg} x \leqslant \frac{3 \pi}{2}
$$

It remains to verify the equality

$$
\sin (\operatorname{arctg} x+\operatorname{arcctg} x)=0
$$

To prove the second equality, it is sufficient to note that

$$
-\frac{\pi}{2} \leqslant \arcsin x+\arccos x \leqslant \frac{3 \pi}{2}
$$

and find

$$
\sin (\arcsin x+\arccos x)
$$

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