Sum of inverse functions

Let $\begin{cases} \alpha = \arcsin (e) & \implies \sin \alpha = e \\ \beta = \arccos (e) & \implies \cos \beta = e \end{cases}$

$\begin{aligned} \implies \sin \alpha & = \cos \beta \\ \sin \alpha & = \sin \left(\frac \pi 2 - \beta \right) \\ \implies \alpha & = \frac \pi 2 - \beta \\ \alpha + \beta & = \frac \pi 2 \\ \implies \arcsin e + \arccos e & = \frac \pi 2 \approx \boxed{1.571} \end{aligned}$

As it turns out, whether $e$ is in the domain of either of those functions is of no importance to solving the problem.

Note that if we define $\sin{\phi} = a$ as $\dfrac{e^{i\phi}-e^{-i\phi}}{2i} \implies 2ai = e^{i\phi}-e^{-i\phi}$ (this is valid).

Multiplying by $e^{i\phi}$ gives:

$e^{2i\phi}-2aie^{i\phi}-1 = 0$ . Hence, $e^{i\phi} = \dfrac{2ai + \sqrt{-4a^2+4}}{2}$ .

So $e^{i\phi} = ai + \sqrt{1-a^2}$ .

$\therefore \phi = -i\ln{(ai + \sqrt{1-a^2})}$ .

But since $\phi = \arcsin{a}$ , we see that $\arcsin{a} = -i\ln{(ai + \sqrt{1-a^2})}$ .

For $\cos{\phi} = a$ , we see that $\phi = -i\ln{(a + \sqrt{a^2-1})}$ so $\arccos{a} = -i\ln{(a + \sqrt{a^2-1})}$ if we repeat the steps taken with

$\sin{\phi}$ for $\cos{\phi} = \dfrac{e^{i\phi}+e^{-i\phi}}{2}$ .

Thus, $\arcsin{a} + \arccos{a} = -i\ln{(ai + \sqrt{1-a^2})} - i\ln{(a + \sqrt{a^2-1})}$

If we let $v = a + \sqrt{a^2-1}$ we see that $\arcsin{a} + \arccos{a} = -i\ln{(iv)} - i\ln{(v)} = -i\ln{\left( \dfrac{iv}{v} \right)} = -i\ln{i}$ .

Since $\ln{i} = i\dfrac{\pi}{2}$ in the principal branch and $i^2 = -1$ , we see that $\arcsin{a} + \arccos{a} = \dfrac{\pi}{2} \approx \boxed{1.571}$ (to three decimal places).

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Note that this result gives no regards to the type or size of number that $'a'$ is. As such, $a$ may be any complex number and it will always satisfy the equation $\arcsin{a} + \arccos{a} = \dfrac{\pi}{2}$ for the principal branch.