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Prove that in triangle ABC, 2cos A cosB ...

Prove that in `triangle ABC, 2cos A cosB cos Cle(1)/(8)`.

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`2 cos A cos B cos C`
`[cos(A+B)+cos(A-B)]cos C`
`=[-cosC+cos(A-B)]cosC`
`=cos(A-B)cosC-cos^(2)C`
`le cos C-cos^(2)C` [As `cos (A-B)le1`]
`=1//4-(cosC-1//2)^(2)le1//4`
So, `cos A cos B cos C le1//8`
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