Question Number 157974 by yeti123 last updated on 30/Oct/21
$$\mathrm{which}\:\mathrm{one}\:\mathrm{do}\:\mathrm{you}\:\mathrm{prefer}? \\ $$$$ \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left({x}\right)\:\:\:\:\:\:\:\mathrm{or}\:\:\:\:\:\:\:\mathrm{arcsin}\left({x}\right) \\ $$
Commented by puissant last updated on 30/Oct/21
$$\:{sin}^{−\mathrm{1}} \left({x}\right)\:=\:{arcsin}\left({x}\right)\: \\ $$$${i}\:{don}'{t}\:{understand}\:{your}\:{problem}.. \\ $$
Commented by MJS_new last updated on 30/Oct/21
$$\mathrm{I}\:\mathrm{prefer}\:\mathrm{arcsin}\:\left({x}\right)\:\mathrm{because}\:\mathrm{sin}^{\mathrm{2}} \:\left({x}\right)\:=\left(\mathrm{sin}\:{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{but}\:\mathrm{sin}^{−\mathrm{1}} \:\left({x}\right)\:\neq\:\left(\mathrm{sin}\:\left({x}\right)\right)^{−\mathrm{1}} \:\mathrm{which}\:\mathrm{is}\:\mathrm{confusing}. \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{sin}^{−\mathrm{2}} \:\left({x}\right)\:? \\ $$
Commented by tounghoungko last updated on 31/Oct/21
$$\mathrm{sin}\:^{−\mathrm{2}} \left({x}\right)=\:{csc}^{\mathrm{2}} \left({x}\right) \\ $$
Commented by MJS_new last updated on 31/Oct/21
$${u}^{−\mathrm{2}} ={v}^{\mathrm{2}} \:\Rightarrow\:{u}^{−\mathrm{1}} =\pm{v} \\ $$$$ \\ $$$$\mathrm{sin}^{−\mathrm{2}} \:\left({x}\right)\:=\mathrm{csc}^{\mathrm{2}} \:\left({x}\right) \\ $$$$\Rightarrow \\ $$$$\mathrm{sin}^{−\mathrm{1}} \:\left({x}\right)\:=\pm\mathrm{csc}\:\left({x}\right) \\ $$$$\mathrm{there}\:\mathrm{you}\:\mathrm{have}\:\mathrm{it}... \\ $$