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In a plane triangle find the maximum value of \( \cos A \cdot \cos B \cdot \cos C \). |
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Answer» \(\because\) A.M. \(\geq\) G.M. \(\therefore\) \(\frac{cos A+cos B+cos C}3\geq(cos A. cos B. cos C)^{1/3}\) But we know that Cos A + Cos B + Cos C \(\leq\) 3/2 \(\therefore\) (cos A. Cos B. Cos C)1/3 \(\leq\) \(\frac{3/2}3=\frac12\) \(\therefore\) Cos A. Cos B. Cos C \(\leq\) \((\frac1{2})^3\) ⇒ Cos A. Cos B. Cos C \(\leq\) \(\frac18\) Hence, maximum value of Cos A. Cos B. Cos C is 1/8 |
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