Question Number 146465 by mathdanisur last updated on 13/Jul/21
$${arccos}\left({cos}\:\mathrm{9}\right)\:=\:? \\ $$
Answered by gsk2684 last updated on 13/Jul/21
$$\mathrm{5}\frac{\Pi}{\mathrm{2}}<\mathrm{9}<\:\mathrm{3}\Pi \\ $$$$\left(\mathrm{5}\frac{\Pi}{\mathrm{2}}−\mathrm{2}\Pi\right)<\left(\mathrm{9}−\mathrm{2}\Pi\right)<\left(\mathrm{3}\Pi−\mathrm{2}\Pi\right) \\ $$$$\frac{\Pi}{\mathrm{2}}<\left(\mathrm{9}−\mathrm{2}\Pi\right)<\Pi \\ $$$${then} \\ $$$$\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{9}\right) \\ $$$$=\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{cos}\:\left(\mathrm{9}−\mathrm{2}\Pi\right)\right) \\ $$$$=\mathrm{9}−\mathrm{2}\Pi \\ $$
Commented by mathdanisur last updated on 13/Jul/21
$${thank}\:{you}\:{Ser} \\ $$
Commented by mathmax by abdo last updated on 13/Jul/21
$$\mathrm{but}\:\mathrm{9}\:\mathrm{can}\:\mathrm{be}\:\mathrm{9}^{\mathrm{0}} …! \\ $$
Commented by mathdanisur last updated on 13/Jul/21
$${how}\:{Ser} \\ $$
Commented by gsk2684 last updated on 13/Jul/21
$${then}\: \\ $$$${using}\:{the}\:{formula}\: \\ $$$$\mathrm{cos}^{−\mathrm{1}} \mathrm{cos}\:\theta=\theta\:\:{if}\:\:\mathrm{0}^{\mathrm{0}} \leqslant\theta\leqslant\mathrm{180}^{\mathrm{0}} \\ $$$$\therefore\mathrm{cos}^{−\mathrm{1}} \mathrm{cos}\:\mathrm{9}^{\mathrm{0}} =\mathrm{9}^{\mathrm{0}} \\ $$