how to find $ x^3+y^3$ if $(\sin^{-1}x)^2+(\cos^{-1}y)^2 = \frac{\pi^2}{2}$ I have to find  $ x^3+y^3$ if $$(\sin^{-1}x)^2+(\cos^{-1}x)^2 = \frac{\pi^2}{2}$$
My book has arbitrarily declared that $x=y$, I have an idea, but I have no way to check if it's conceptually right as we just assumed $x=y$ even when we did this in class a few months ago.
I'd appreciate any help
Idea:- let $x= \sin(\theta)$
so
$$\theta^2 +y^2= \frac{\pi^2}{2}$$
which leaves us with two variables, beyond which, I'm stuck
what do I do from here

You're idia would only work if it's sin(x) and cos(x)
but there are arcsin(x) and arccos(x)...
Trigonometric Way
We can use the trigonometric identity that relates arcsin and arccos:
$$ \arccos(x) = \frac{\pi}{2} - \arcsin(x) $$
Now we can say:
$$
\begin{align*}
\sin^{-1}(x)^{2} + \cos^{-1}(x)^{2} &= \frac{\pi^{2}}{2}\\
\arcsin(x)^{2} + \arccos(x)^{2} &= \frac{\pi^{2}}{2}\\
\arcsin(x)^{2} + (\frac{\pi}{2} - \arcsin(x))^{2} &= \frac{\pi^{2}}{2}\\
\arcsin(x)^{2} + (\frac{\pi}{2})^{2} - 2 \cdot \frac{\pi}{2} \cdot \arcsin(x) + \arcsin(x)^{2} &= \frac{\pi^{2}}{2}\\
\arcsin(x)^{2} + (\frac{\pi}{2} - \arcsin(x))^{2} &= \frac{\pi^{2}}{2}\\
2 \cdot \arcsin(x)^{2} + \frac{\pi^{2}}{4} - \pi \cdot \arcsin(x) &= \frac{\pi^{2}}{2} \quad\mid\quad y := \arcsin(x)\\
2 \cdot y^{2} + \frac{\pi^{2}}{4} - \pi \cdot y &= \frac{\pi^{2}}{2} \quad\mid\quad -(\frac{\pi^{2}}{2})\\
2 \cdot y^{2} - \pi \cdot y + \frac{\pi^{2}}{4} -\frac{\pi^{2}}{2} &= 0 \quad\mid\quad \div 2\\
y^{2} - \frac{\pi}{2} \cdot y + \frac{\pi^{2}}{8} -\frac{\pi^{2}}{4} &= 0 \quad\mid\quad \text{pq-Formula}\\
p = -\frac{\pi}{2} ~&\text{ and }~ q = \frac{\pi^{2}}{8} -\frac{\pi^{2}}{4}\\
y_{1, ~2} &= -\frac{p}{2} \pm \sqrt{(\frac{p}{2})^{2} - q}\\
y_{1, ~2} &= -\frac{-\frac{\pi}{2}} \pm \sqrt{(\frac{-\frac{\pi}{2}}{2})^{2} - (\frac{\pi^{2}}{8} -\frac{\pi^{2}}{4})}\\
y_{1, ~2} &= \frac{\pi}{4} \pm \sqrt{(\frac{\pi}{4})^{2} - \frac{\pi^{2}}{8} + \frac{\pi^{2}}{4}} \quad\mid\quad y := \arcsin(x)\\
\arcsin(x) &= \frac{\frac{\pi}{2}}{2} \pm \sqrt{(\frac{\pi}{4})^{2} - \frac{\pi^{2}}{8} + \frac{\pi^{2}}{4}} \quad\mid\quad \sin()\\
x &= \sin(\frac{\frac{\pi}{2}}{2} \pm \sqrt{(\frac{\pi}{4})^{2} - \frac{\pi^{2}}{8} + \frac{\pi^{2}}{4}})\\
x &= \sin(\frac{\pi}{4} \pm \sqrt{(\frac{\pi}{4})^{2} + \frac{-\pi^{2} + 2 \cdot \pi^{2}}{8}})\\
x &= \sin(\frac{\pi}{4} \pm \sqrt{(\frac{\pi}{4})^{2} + \frac{\pi^{2}}{8}})\\
x &= \sin(\frac{\pi}{4} \pm \sqrt{\frac{\pi^{2}}{16} + \frac{\pi^{2}}{8}})\\
x &= \sin(\frac{\pi}{4} \pm \sqrt{\frac{\pi^{2} + 2 \cdot \pi^{2}}{16}})\\
x &= \sin(\frac{\pi}{4} \pm \sqrt{\frac{3 \cdot \pi^{2}}{16}})\\
&\Rightarrow x^{3} = \sin(\frac{\pi}{4} \pm \sqrt{\frac{3 \cdot \pi^{2}}{16}})^{3}
\end{align*}
$$
Complex Way
I would try to solve the problem using the complex logarithm forms of the arc functions:
$$
\begin{align*}
\sin^{-1}(x) = \arcsin(x) &= -\mathrm{i} \cdot \ln(x \cdot \mathrm{i} + \sqrt{1 - x^{2}})\\
\cos^{-1}(x) = \arccos(x) &= -\mathrm{i} \cdot \ln(x + \sqrt{1 - x^{2}} \cdot \mathrm{i})\\
\end{align*}
$$
Now we can say:
$$
\begin{align*}
\sin^{-1}(x)^{2} + \cos^{-1}(x)^{2} &= \frac{\pi^{2}}{2}\\
(-\mathrm{i} \cdot \ln(x \cdot \mathrm{i} + \sqrt{1 - x^{2}}))^{2} + (-\mathrm{i} \cdot \ln(x + \sqrt{1 - x^{2}} \cdot \mathrm{i}))^{2} &= \frac{\pi^{2}}{2}\\
\ln(x \cdot \mathrm{i} + \sqrt{1 - x^{2}})^{2} + \ln(x + \sqrt{1 - x^{2}} \cdot \mathrm{i})^{2} &= \frac{\pi^{2}}{2}\\
...
\end{align*}
$$
But that won't work that fast...
