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When deducing the formula for $\sin^{-1}x+\sin^{-1}y$, we eventually obtain an expression $$\sin(A+B)=x\sqrt{1-y^2}+y\sqrt{1-x^2}=k$$ where $A=\sin^{-1}x$ and $B=\sin^{-1}y$, and then deduce $A+B=\sin^{-1}k$.

But for inverse to exist, $A$ and $B$ both must be within the range $[-\frac{\pi}{2},\frac{\pi}{2}]$, but here $A+B$ is in the range $[-\pi,\pi]$. So how can we take $\sin$ to the other side and write $A+B=\sin^{-1}k$ whence $A+B$ should have been in the range $[-\frac{\pi}{2},\frac{\pi}{2}]$?

Blue
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a_i_r
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1 Answers1

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Trig function of inverse trig functions and combinations will be automatically taken care as you are dealing with identities for arguments b with inverse circular angle function arguments $ (-\infty,\infty) $ including co-terminal angles.

It is natural to be concerned about the unbounded domain limits beyond $ (\pm \pi/2).$

The co-terminal angles are readily absorbed.

$$\sin(\sin^{-1}x+\sin^{-1}y)==x\sqrt{1-y^2}+y\sqrt{1-x^2} $$ $$\sin^{-1}x+\sin^{-1}==\sin ^{-1}( x\sqrt{1-y^2}+y\sqrt{1-x^2}) $$

are true for all real $ |x|<1|, \;|y|<1 $.

The following 3d plot in WA link neatly fits inside a bounded cube of side 2 units. The entire cube can be shifted up or down by an arbitrary amount.

ArcSin Range Shifted by 1000 $\pi$ angle units

Inverse Trig graph

Narasimham
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  • Yes you are right that $\sin(\sin^{-1}x+\sin^{-1}y)=x\sqrt{1-y^2}+y\sqrt{1-x^2}$ but now how can you take $\sin^{-1}$ on both sides?Since $\sin^{-1}x+\sin^{-1}$ are in the range $[-\pi,\pi]$ and not $[-\frac{\pi}{2},\frac{\pi}{2}]$. – a_i_r Jul 23 '21 at 07:19
  • Don't follow what the problem is. We can add or remove a thousand semi revolutions in range for inverse functions. – Narasimham Jul 23 '21 at 12:29