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Question Number 110287 by mathdave last updated on 28/Aug/20

Commented by bemath last updated on 28/Aug/20

$${\sin }^{-1}\left(5\right)=\frac{\pi }{2}-{\cos }^{-1}\left(5\right)$$$$\sin \left({\sin }^{-1}\left(5\right)\right)=\sin (\frac{\pi }{2}-{\cos }^{-1}\left(5\right))$$$$\Rightarrow 5=\cos \left({\cos }^{-1}\left(5\right)\right)$$$$\Rightarrow 5=5\left(true\right)$$

Commented by udaythool last updated on 28/Aug/20

$$\mathrm{its}\mathrm{straight}\mathrm{forward}...$$$$\mathrm{assume}\mathrm{lhs}\mathrm{as}\theta \mathrm{then}\mathrm{take}\sin \mathrm{of}$$$$\mathrm{both}\mathrm{sides},\mathrm{use}\mathrm{trigonometric}$$$$\mathrm{relation};\sin (A+B)\mathrm{and}\sin$$$$\left({\cos }^{-1}x\right)=\sqrt{1-x^2}\mathrm{we}\mathrm{get}$$$$25+\sqrt{-24}\sqrt{-24}=1=\sin \theta \Rightarrow \theta =\frac{\pi }{2}$$

Answered by $@y@m last updated on 28/Aug/20

$$Let{\sin }^{-1}5=\theta$$$$\sin \theta =5$$$$\Rightarrow \cos (\frac{\pi }{2}-\theta )=5$$$$\Rightarrow {\cos }^{-1}5=\frac{\pi }{2}-\theta$$$$\Rightarrow \theta +{\cos }^{-1}5=\frac{\pi }{2}$$$$\Rightarrow {\sin }^{-1}5+{\cos }^{-1}5=\frac{\pi }{2}$$$$$$

Commented by $@y@m last updated on 28/Aug/20

$$Itisliterallycorrectbuttheoretically$$$$wrongas\sin \theta \ne 5forany\theta$$

Answered by Her_Majesty last updated on 28/Aug/20

$${sin}^{-1}z+{cos}^{-1}z=\pi /2\mathrm{\forall }z\in \mathbb{C}$$

Commented by $@y@m last updated on 28/Aug/20

Commented by Her_Majesty last updated on 28/Aug/20

$$domainsareonlyrelevantwithin\mathbb{R}$$$$rememberthedomainof\sqrt{x}is{\mathbb{R}}^{+}butthe$$$$nextstepistodefinei:=\sqrt{-1}andnowthe$$$$domainis\mathbb{R}(x>0\Rightarrow \sqrt{-x}=i\sqrt{x})$$$$similarfor{sin}^{-1},{cos}^{-1}...$$

Commented by $@y@m last updated on 29/Aug/20

$$Canyougivemeanexample(i.e.$$$$valueof\theta )forwhich{\sin }^{-1}5=\theta ?$$

Commented by Her_Majesty last updated on 29/Aug/20

$$x=siny\iff y={sin}^{-1}x$$$$siny=x=\frac{e^{iy}-e^{-iy}}{2i}$$$$2ix=e^{iy}-e^{-iy}$$$$t=e^{iy}\iff y=-ilnt$$$$2ix=t+1/t$$$$t^2-2ixt-1=0$$$$t=ix\pm \sqrt{{\left(ix\right)}^2+1}=ix\pm i\sqrt{x^2-1}=i(x\pm \sqrt{x^2-1})$$$$y=-ilnt=-iln\left(i(x\pm \sqrt{x^2-1})\right)=$$$$=-i(lni+ln(x\pm \sqrt{x^2-1}))=$$$$=\frac{\pi }{2}-iln(x\pm \sqrt{x^2-1})$$$$\Rightarrow$$$${sin}^{-1}x=\frac{\pi }{2}-iln(x\pm \sqrt{x^2-1})$$

Commented by Her_Majesty last updated on 29/Aug/20

$$similarweget$$$${cos}^{-1}x=-iln(x\pm \sqrt{x^2-1})$$

Answered by 1549442205PVT last updated on 29/Aug/20

$$\mathrm{Put}\phi ={\sin }^{-1}5,{\cos }^{-1}5=\theta \mathrm{we}\mathrm{have}$$$$\sin \phi =5,\cos \theta =5\Rightarrow \sin \phi =\cos \theta$$$$\iff \sin \phi =\sin (\frac{\pi }{2}-\theta )\Rightarrow \phi =\frac{\pi }{2}-\theta +2\mathrm{k}\pi$$$$\iff \phi +\theta =\frac{\pi }{2}+2\mathrm{k}\pi$$$$\mathrm{But}\mathrm{bythe}\mathrm{definition}\mathrm{of}\mathrm{the}\mathrm{function}$$$${\sin }^{-1}\left(\mathrm{x}\right),{\cos }^{-1}\left(\mathrm{x}\right)\mathrm{we}\mathrm{have}\frac{-\pi }{2}⩽\phi ⩽\frac{\pi }{2},$$$$\mathrm{and}0⩽\theta ⩽\pi \mathrm{Hence}\frac{-\pi }{2}⩽\phi +\theta ⩽\frac{3\pi }{2},\mathrm{so}$$$$-\frac{\pi }{2}⩽\phi +\theta =\frac{\pi }{2}+2\mathrm{k}\pi ⩽\frac{3\pi }{2}\Rightarrow \mathrm{k}=0$$$$\Rightarrow \mathit{\phi }+\mathit{\theta }=\frac{\mathit{\pi }}{2}\Rightarrow {\mathbf{sin}}^{-1}5+{\mathbf{cos}}^{-1}5=\frac{\mathit{\pi }}{2}(\mathbf{Q}.\mathbf{E}.\mathbf{D})$$

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