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Question Number 40466 by naziri2013@ last updated on 22/Jul/18

$$$$$$$$$$$$$$Q..{\cos }^{-1}(1-2x^2)={2\sin }^{-1}x,prove$$$$please$$$$$$$$$$

Commented by maxmathsup by imad last updated on 22/Jul/18

$$letprovethatarccos(1-2x^2)=2arcsinxlet$$$$f\left(x\right)=arccos(1-2x^2)-2arcsinx{D}_f?$$$$x\in D_f\iff -1⩽x⩽1and-1⩽1-2x^2⩽1but$$$$-1⩽1-2x^2⩽1\iff -2⩽-2x^2⩽0\iff 0⩽2x^2⩽2\iff 0⩽x^2⩽1\iff -1⩽x⩽1\Rightarrow$$$$D_f=[-1,1]wehave$$$$f^{{}^{\prime }}\left(x\right)=\frac{-(-4x)}{\sqrt{1-{(1-2x^2)}^2}}-\frac{2}{\sqrt{1-x^2}}=\frac{4x\sqrt{1-x^2}-2\sqrt{1-(4x^4-4x^2+1)}}{\sqrt{1-x^2}\sqrt{1-{(1-2x^2)}^2}}$$$$=\frac{4x\sqrt{1-x^2}-2\sqrt{4x^2-4x^4}}{\sqrt{1-x^2}\sqrt{4x^2-4x^4}}=\frac{4x\sqrt{1-x^2}-4\mid x\mid \sqrt{1-x^2}}{\sqrt{1-x^2}\sqrt{4x^2-4x^4}}$$$$if0<x<1wehave{f}^{{}^{\prime }}\left(x\right)=0\Rightarrow f\left(x\right)=c\mathrm{\forall }x\in [0,1]$$$$f\left(x\right)=f\left(0\right)=0$$$$ifx\in ]-1,0[f^{{}^{\prime }}\left(x\right)=\frac{8x\sqrt{1-x^2}}{-2x(1-x^2)}=\frac{-4}{\sqrt{1-x^2}}\ne 0\Rightarrow fisnotconstant$$$$finallywehavearccos(1-2x^2)=2arcsinxifx\in [0,1].$$

Commented by prakash jain last updated on 22/Jul/18

$$\mathrm{This}\mathrm{is}\mathrm{not}\mathrm{true}\mathrm{in}\mathrm{general}$$$$\mathrm{range}\mathrm{of}\cos {}^{-1}\mathrm{is}[0,\pi ]$$$$\mathrm{range}\mathrm{of}\sin {}^{-1}\mathrm{is}[-\frac{\pi }{2},\frac{\pi }{2}]$$$$\mathrm{put}x=-\frac{1}{2}$$$${2\sin }^{-1}x=-\frac{\pi }{3}$$$${\cos }^{-1}(1-2x^2)={\cos }^{-1}\frac{1}{2}=\frac{\pi }{3}$$

Commented by prakash jain last updated on 22/Jul/18

$$\mathrm{also}\mathrm{note}\mathrm{that}$$$${\sin }^{-1}\left(\sin x\right)\mathrm{does}\mathrm{not}\mathrm{always}\mathrm{equal}$$$$x\mathrm{due}\mathrm{to}\mathrm{restriction}\mathrm{on}\mathrm{range}\mathrm{of}$$$${\sin }^{-1}x.$$$$\sin \left({\sin }^{-1}x\right)\mathrm{however}\mathrm{always}\mathrm{equals}x.$$

Commented by prof Abdo imad last updated on 23/Jul/18

$$ihaveprovedthatequalityistruein[0,1]and$$$$x=-\frac{1}{2}\notin [0,1]!$$

Commented by prakash jain last updated on 23/Jul/18

$$\mathrm{M}y\mathrm{comment}\mathrm{was}\mathrm{for}\mathrm{the}\mathrm{question}.\mathrm{The}$$$$\mathrm{questioner}\mathrm{seems}\mathrm{to}\mathrm{be}\mathrm{a}\mathrm{student}\mathrm{so}$$$$\mathrm{wanted}\mathrm{to}\mathrm{highlight}\mathrm{common}$$$$\mathrm{mistakes}\mathrm{made}\mathrm{while}\mathrm{using}$$$$\mathrm{inverse}\mathrm{trignimetric}\mathrm{functions}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jul/18

$$x=sin\theta$$$${cos}^{-1}(1-2{sin}^2\theta )$$$${cos}^{-1}\left(cos2\theta \right)=2\theta$$$$$$$$2{sin}^{-1}\left(sin\theta \right)$$$$2\theta$$$$$$

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