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What is arccos(cos7)=?
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are we using the definition of \(\arccos x\) whose image is \([0,\pi)\)?
yes
in that case recognize \(2\pi\lt7\lt3\pi\) hence \(\arccos(\cos7)=\arccos(\cos(7-2\pi))=7-2\pi\)
3pi-7 too?
i guess 3pi will change the angle?
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@oldrin.bataku ?
@DLS \(\cos(3\pi-7)=\cos(2\pi+\pi-7)=\cos(\pi-7)=\cos(7-\pi)\ne\cos(7)\)
okay thanks!
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