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Question Number 157974 by yeti123 last updated on 30/Oct/21

$$\mathrm{which}\mathrm{one}\mathrm{do}\mathrm{you}\mathrm{prefer}?$$$$$$$${\sin }^{-1}\left(x\right)\mathrm{or}\arcsin \left(x\right)$$

Commented by puissant last updated on 30/Oct/21

$${sin}^{-1}\left(x\right)=arcsin\left(x\right)$$$$i{don}^{\prime }tunderstandyourproblem..$$

Commented by MJS_new last updated on 30/Oct/21

$$\mathrm{I}\mathrm{prefer}\arcsin \left(x\right)\mathrm{because}{\sin }^2\left(x\right)={\left(\sin x\right)}^2$$$$\mathrm{but}{\sin }^{-1}\left(x\right)\ne {\left(\sin \left(x\right)\right)}^{-1}\mathrm{which}\mathrm{is}\mathrm{confusing}.$$$$\mathrm{what}\mathrm{is}{\sin }^{-2}\left(x\right)?$$

Commented by tounghoungko last updated on 31/Oct/21

$$\sin {}^{-2}\left(x\right)={csc}^2\left(x\right)$$

Commented by MJS_new last updated on 31/Oct/21

$$u^{-2}=v^2\Rightarrow u^{-1}=\pm v$$$$$$$${\sin }^{-2}\left(x\right)={\csc }^2\left(x\right)$$$$\Rightarrow$$$${\sin }^{-1}\left(x\right)=\pm \csc \left(x\right)$$$$\mathrm{there}\mathrm{you}\mathrm{have}\mathrm{it}...$$

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