prove that sin^(−1) (tanhx)=tan^(−1) (sinhx)


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Question Number 76657 by Tony Lin last updated on 29/Dec/19

$p r o v e t h a t {s i n}^{- 1} (t a n h x) = {t a n}^{- 1} (s i n h x)$
	Answered by john santu last updated on 29/Dec/19

	$(1) \tan h x = \sin y$ $\sin y = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} \to e^{x} - e^{- x} = (e^{x} + e^{- x}) \sin y$ $(2) \tan^{- 1} (\sin h x) = t$ $\sin h x = \tan t \to \frac{e^{x} - e^{- x}}{2} = \tan t$ $e^{x} - e^{- x} = 2 \tan t \to (e^{x} + e^{- x}) \sin y = 2 \tan t$ $\frac{e^{x} + e^{- x}}{2} \times \sin y = \tan t$ $\cosh x \times \tanh x = \sinh x$ $\tanh x = \tanh x .$
	Commented by mr W last updated on 29/Dec/19

	$i {c a n}^{'} t s e e y o u r p r o o f t h a t y = t .$
	Commented by john santu last updated on 29/Dec/19

	$o o y e s, s o r r y s i r, {s o m e t h i n g}^{'} s m i s s e d$
	Answered by mr W last updated on 29/Dec/19

	$l e t L H S = {s i n}^{- 1} (t a n h x) = u$ $\Rightarrow \tanh x = \sin u$ $\Rightarrow \frac{\sinh x}{\cosh x} = \sin u$ $\Rightarrow \frac{\sinh^{2} x}{\cosh^{2} x} = \sin^{2} u$ $\Rightarrow \frac{\sinh^{2} x}{1 + \sinh^{2} x} = \sin^{2} u$ $\Rightarrow \sinh^{2} x = \frac{\sin^{2} u}{1 - \sin^{2} u} = \frac{\sin^{2} u}{\cos^{2} u} = \tan^{2} u$ $\Rightarrow \sinh x = \tan u$ $\Rightarrow \tan^{- 1} (\sinh x) = u$ $\Rightarrow \tan^{- 1} (\sinh x) = {s i n}^{- 1} (t a n h x)$
	Commented by Tony Lin last updated on 29/Dec/19

	$t h a n k y o u b o t h$
	Answered by mind is power last updated on 29/Dec/19
	$Hello happy too back withe you Since th(x) is bijection ]−∞,+∞[−{0}→]−1,1[ ⇒∀u∈]−1,+1[,∃!x∈R^∗ ∣u=th(x) sin^− (th(x))=sin^− (u) tan^− (sh(x))=tan^− (sh(th^− (u))) th(u)=((sh(u))/(ch(u))) ch^2 −sh^2 =1⇒ sh^2 (u)=((th^2 (u))/(1−th^2 (u))) ⇒sh^2 (th^− (u))=(u^2 /(1−u^2 )) sh(th^− (u))=(u/(√(1−u^2 ))) tan^− ((u/(√(1−u^2 ))))=?sin^− (u),u∈[−1,1],t→sint bijection [−(π/2),(π/2)]→[−1,1] u=sin(α)⇒(u/(√(1−u^2 )))=tan(α) tan^− (tan(α))=α sin^− (sinα)=α,since α is unique since we have bijection ⇒injection ⇒tan^− ((u/(√(1−u^2 ))))=sin^− (u) end of proof$
	$Hello happy too back withe you$ $Since$ $th (x) is bijection] - \infty, + \infty [- {0} \to] - 1, 1 [$ $\Rightarrow \forall u \in] - 1, + 1 [, \exists! x \in R^{*} ∣ u = th (x)$ $\sin^{-} (th (x)) = \sin^{-} (u)$ $\tan^{-} (sh (x)) = \tan^{-} (sh ({th}^{-} (u)))$ $th (u) = \frac{sh (u)}{ch (u)}$ ${ch}^{2} - {sh}^{2} = 1 \Rightarrow$ ${sh}^{2} (u) = \frac{{th}^{2} (u)}{1 - {th}^{2} (u)}$ $\Rightarrow {sh}^{2} ({th}^{-} (u)) = \frac{u^{2}}{1 - u^{2}}$ $sh ({th}^{-} (u)) = \frac{u}{\sqrt{1 - u^{2}}}$ $\tan^{-} (\frac{u}{\sqrt{1 - u^{2}}}) = ? \sin^{-} (u), u \in [- 1, 1], t \to sint bijection [- \frac{π}{2}, \frac{π}{2}] \to [- 1, 1]$ $u = \sin (α) \Rightarrow \frac{u}{\sqrt{1 - u^{2}}} = \tan (α)$ $\tan^{-} (\tan (α)) = α$ $\sin^{-} (\sin α) = α, since α is unique since we have bijection \Rightarrow injection$ $\Rightarrow \tan^{-} (\frac{u}{\sqrt{1 - u^{2}}}) = \sin^{-} (u)$ $end of proof$
	Commented by Tony Lin last updated on 31/Dec/19

	$t h a n k s s i r$
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