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Mentallic

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## Main Question or Discussion Point

What is wrong with my proof?

Let [tex]\theta=cos^{-1}x-\frac{\pi}{2}[/tex]

Then [tex]cos\theta=cos\left(cos^{-1}x-\frac{\pi}{2}\right)[/tex]

[tex]RHS=xcos\frac{\pi}{2}-sin(cos^{-1}x)sin\frac{\pi}{2}[/tex]

[tex]=-\sqrt{1-x^2}[/tex]

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

[tex]\theta=\pi-cos^{-1}\sqrt{1-x^2}[/tex]

Hence [tex]cos^{-1}x-\frac{\pi}{2}=\pi-cos^{-1}\sqrt{1-x^2}[/tex]

So finally, [tex]cos^{-1}x+cos^{-1}\sqrt{1-x^2}=\frac{3\pi}{2}[/tex]

Except this is untrue for all values except [itex]x=-1[/itex]. I'm guessing I probably made a substitution which is valid for only certain values. Inverse trig seems to do that a lot to me

Let [tex]\theta=cos^{-1}x-\frac{\pi}{2}[/tex]

Then [tex]cos\theta=cos\left(cos^{-1}x-\frac{\pi}{2}\right)[/tex]

[tex]RHS=xcos\frac{\pi}{2}-sin(cos^{-1}x)sin\frac{\pi}{2}[/tex]

[tex]=-\sqrt{1-x^2}[/tex]

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

[tex]\theta=\pi-cos^{-1}\sqrt{1-x^2}[/tex]

Hence [tex]cos^{-1}x-\frac{\pi}{2}=\pi-cos^{-1}\sqrt{1-x^2}[/tex]

So finally, [tex]cos^{-1}x+cos^{-1}\sqrt{1-x^2}=\frac{3\pi}{2}[/tex]

Except this is untrue for all values except [itex]x=-1[/itex]. I'm guessing I probably made a substitution which is valid for only certain values. Inverse trig seems to do that a lot to me

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