Why does simplifying $\arcsin(x) -\arcsin(y) = \frac{\pi}{2}$ to $y=-\cos(\arcsin(x))$ change the graph?

Question

I simplified $$\arcsin(x) -\arcsin(y) = \frac{\pi}{2}$$ as follows: $$\arcsin(y) = \arcsin(x) -\frac{\pi}{2} \tag{1}$$ Taking $\sin$ on both sides: $$\sin(\arcsin(y)) =-\sin\left(\frac{\pi}{2}-\arcsin(x)\right)$$ $$y=-\cos(\arcsin(x)) \tag{2}$$

The graph of $(1)$ (as shown in this Desmos graph) is the portion of the unit circle in the fourth quadrant. The graph of $(2)$ (Desmos) is a semicircle in the lower quadrants.

Since two graphs are different, both functions should be different. But I didn't alter the original function, I just simplified it. There must be some step (during simplification) where I changed the function unknowingly. But I'm not able to find the error I made. Please tell where I changed the function and how ?

Answer 1

Usually, arcsin has values in $[-\pi / 2, + \pi /2]$, which implies that your first equation is defined only for $0\leq x \leq 1$. If it is not clear, just remember that when taking sin on both sides , the equality obtained is not equivalent to the initial one, since $x \mapsto \sin (x)$ is only injective on some restriction of its domain.

Answer 2

As for real $z,-1\le z\le1,$ using the principal value of $$-\dfrac\pi2\le\arcsin z\le\dfrac\pi2$$

We need $$-\dfrac\pi2\le\arcsin x-\dfrac\pi2\le\dfrac\pi2\iff0\le\arcsin x\le\pi\implies x\ge0$$

Similarly, $$-\dfrac\pi2\le\arcsin y+\dfrac\pi2\le\dfrac\pi2\iff-\pi\le\arcsin y\le0\implies y\le0$$

But as $-\dfrac\pi2\le\arcsin x\le-\dfrac\pi2$

$$y=-\cos(\arcsin x)=-\sqrt{1-x^2}$$

So, $1-x^2\ge0\iff-1\le x\le1$

but as $\sqrt{1-x^2}\ge0, y\le0$