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Question Number 40107 by maxmathsup by imad last updated on 15/Jul/18
![prove the relations 1) ∀t ∈]0,1] arctan(((√(1−t^2 ))/t))=arccost 2) ∀ t∈[−1,1] 2 arccos(√((1+t)/2)) =arccost](Q40107.png)
$${prove}\:{the}\:{relations} \\ $$$$\left.\mathrm{1}\left.\right)\left.\:\forall{t}\:\in\right]\mathrm{0},\mathrm{1}\right]\:\:\:{arctan}\left(\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{{t}}\right)={arccost} \\ $$$$\left.\mathrm{2}\right)\:\forall\:{t}\in\left[−\mathrm{1},\mathrm{1}\right]\:\:\:\:\mathrm{2}\:{arccos}\sqrt{\frac{\mathrm{1}+{t}}{\mathrm{2}}}\:={arccost} \\ $$
Answered by math khazana by abdo last updated on 26/Jul/18
![1) let t =cosθ with θ ∈[0,(π/2)[we have arccos(t)=θ and arctan{((√(1−t^2 ))/t)} =arctan{ ((√(1−cos^2 θ))/(cosθ))} =arctan(((sinθ)/(cosθ)))=arctan(tanθ) =θ ⇒ arctan(((√(1−t^2 ))/t))=arccost 2) let t=cosθ with θ ∈[0,π] we have arccoss(t)=θ and 2arccos(√((1+t)/2))= 2 arcos(√((1+cost)/2)) =2 arccos(cos((θ/2))=2.(θ/2) =θ ⇒ 2 arccos(√((1+t)/2))=arccost ∀ t ∈[−1,1] .](Q40679.png)
$$\left.\mathrm{1}\right)\:{let}\:{t}\:={cos}\theta\:\:{with}\:\theta\:\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\left[{we}\:{have}\:\right.\right. \\ $$$${arccos}\left({t}\right)=\theta\:\:{and}\: \\ $$$${arctan}\left\{\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{{t}}\right\}\:={arctan}\left\{\:\frac{\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} \theta}}{{cos}\theta}\right\} \\ $$$$={arctan}\left(\frac{{sin}\theta}{{cos}\theta}\right)={arctan}\left({tan}\theta\right)\:=\theta\:\Rightarrow \\ $$$${arctan}\left(\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{{t}}\right)={arccost}\: \\ $$$$\left.\mathrm{2}\right)\:{let}\:{t}={cos}\theta\:\:{with}\:\theta\:\in\left[\mathrm{0},\pi\right]\:\:{we}\:{have} \\ $$$${arccoss}\left({t}\right)=\theta\:{and} \\ $$$$\mathrm{2}{arccos}\sqrt{\frac{\mathrm{1}+{t}}{\mathrm{2}}}=\:\mathrm{2}\:{arcos}\sqrt{\frac{\mathrm{1}+{cost}}{\mathrm{2}}} \\ $$$$=\mathrm{2}\:{arccos}\left({cos}\left(\frac{\theta}{\mathrm{2}}\right)=\mathrm{2}.\frac{\theta}{\mathrm{2}}\:=\theta\:\Rightarrow\right. \\ $$$$\mathrm{2}\:{arccos}\sqrt{\frac{\mathrm{1}+{t}}{\mathrm{2}}}={arccost}\:\:\forall\:{t}\:\in\left[−\mathrm{1},\mathrm{1}\right]\:. \\ $$$$ \\ $$