MHB Question related to inverse sine functions

[gsn57iaf]

Please guide why answers are different in following
two cases and which one is correct?
Case 1. sin^-1 ( – 1/2 ) – sin^-1 (– 1) = 7π/6 – 3π/2 = – π/3
Case 2. sin^-1 ( – 1/2 ) – sin^-1 (– 1)
= – sin^-1 ( 1/2 ) + sin^-1 (1)

= – π/6 + π/2 = π/3

 

[Prove It]

Please guide why answers are different in following
two cases and which one is correct?
Case 1. sin^-1 ( – 1/2 ) – sin^-1 (– 1) = 7π/6 – 3π/2 = – π/3
Case 2. sin^-1 ( – 1/2 ) – sin^-1 (– 1)
= – sin^-1 ( 1/2 ) + sin^-1 (1)

= – π/6 + π/2 = π/3

In order to have an inverse function, a function needs to be one-to-one on its domain. As $\displaystyle \begin{align*} y = \sin{(x)} \end{align*}$ is not, its domain is restricted. By convention, the domain chosen is $\displaystyle \begin{align*} -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \end{align*}$. That means that the inverse sine function is defined to give an output restricted to $\displaystyle \begin{align*} -\frac{\pi}{2} \leq \arcsin{(x)} \leq \frac{\pi}{2} \end{align*}$. So with this knowledge, which one do you think gives the correct answer?

 

[gsn57iaf]

In order to have an inverse function, a function needs to be one-to-one on its domain. As $\displaystyle \begin{align*} y = \sin{(x)} \end{align*}$ is not, its domain is restricted. By convention, the domain chosen is $\displaystyle \begin{align*} -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \end{align*}$. That means that the inverse sine function is defined to give an output restricted to $\displaystyle \begin{align*} -\frac{\pi}{2} \leq \arcsin{(x)} \leq \frac{\pi}{2} \end{align*}$. So with this knowledge, which one do you think gives the correct answer?

Thanks sir. The key $\displaystyle \begin{align*} -\frac{\pi}{2} \leq \arcsin{(x)} \leq \frac{\pi}{2} \end{align*}$ I ignored in search of answer.