Find the values: $\sin{\left(2\arccos{\frac{2}{5}}\right)}=?$, $\arcsin{a}+\arccos{a}=?$, $\arctan{a}+\arctan{\frac{1}{a}}=?$

The following exercises involve the simplification of numerical expressions involving inverse trigonometric functions. We use the co-function relationships and reciprocal relationships among trigonometric functions.

1. $\sin{\left(2\arccos{\frac{2}{5}}\right)}=?$ ¹ $$\begin{align} &=2\sin{\left(\arccos{\frac{2}{5}}\right)}\cos{\left(\arccos{\frac{2}{5}}\right)}% =2\cdot\frac{2}{5}\sin{\left(\arccos{\frac{2}{5}}\right)}\\% &=\frac{4}{5}\sin{\left(\arcsin{\frac{\sqrt{5^2-2^2}}{5}}\right)}\\ &=\frac{4}{5}\cdot\frac{\sqrt{21}}{5}\\ &=\frac{4}{25}\sqrt{21}.□ \end{align}$$
2. $\arcsin{a}+\arccos{a}=?$ ¹

   We observe that because sine and cosine functions are co-functions of each other, for any $a$ in the domain of inverse cosine and inverse sine functions, $$y \equiv \sin(\theta) = \cos{\left(\frac{\pi}{2} - \theta\right)} \Rightarrow \arccos{y} = \frac{\pi}{2} -\arcsin{y}.$$

   Thus, the given equals $$\arcsin{a} + \left(\frac{\pi}{2}-\arcsin{a}\right) = \frac{\pi}{2}.□$$

3. $\arctan{a}+\arctan{\frac{1}{a}}=?$ ¹

   We observe that because tangent and cotangent functions are co-functions of each other, for any $a$ in the domain of inverse tangent sine functions, $$\DeclareMathOperator{\arccot}{arccot} y \equiv \tan(\theta) = \arccot{\left(\frac{\pi}{2} - \theta\right)} \Rightarrow \arccot{y} = \frac{\pi}{2} -\arctan{y}.$$

   Thus, the given equals $$\arctan{a} + \left(\frac{\pi}{2}-\arccot{\frac{1}{a}}\right) = \frac{\pi}{2} + \arctan{a} -\arccot{\frac{1}{a}}.$$ Because tangent and cotangent are reciprocals of each other, $$\arccot{\frac{1}{a}} = \arctan{a},$$ and so, $$\frac{\pi}{2} + \arctan{a} -\arccot{\frac{1}{a}} = \frac{\pi}{2} + \arctan{a} -\arctan{a} = \frac{\pi}{2}.□$$

References

¹ University Tutorial Problems.