<s>[INST] 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
 [/INST] 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...
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