Respuesta :

LammettHash
First take [tex]y=\sin^{-1}x[/tex] and [tex]z=\cos^{-1}x[/tex]. Then

[tex]\cos(2\sin^{-1}x+\cos^{-1}x)=\cos(2y+z)=\cos2y\cos z-\sin2y\sin z[/tex]
[tex]=(\cos^2y-\sin^2y)\cos z-2\sin y\cos y\sin z[/tex]

Now

[tex]\sin y=\sin(\sin^{-1}x)=x[/tex]
[tex]\cos y=\cos(\sin^{-1}x)=\sqrt{1-x^2}[/tex]

both provided that [tex]-\dfrac\pi2\le y\le\dfrac\pi2[/tex], and

[tex]\sin z=\sin(\cos^{-1}x)=\sqrt{1-x^2}[/tex]
[tex]\cos z=\cos(\cos^{-1}x)=x[/tex]

both provided that [tex]0\le z\le\pi[/tex].

Then

[tex]\cos(2\sin^{-1}x+\cos^{-1}x)=(1-2x^2)x-2x(1-x^2)[/tex]
[tex]\implies -x=\cos\pi\implies -x=-1\implies x=1[/tex]

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