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arccos-cos-9-




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

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