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  • 2019河南模考题

    $$cos left( A-B ight) =frac {2sin Asin B} {sin C}
    Rightarrow sin C cdot cos left( A-B ight) =cos left( A-B ight) -cos left( A+B ight) ,$$

    $$cos left( A+B ight) =left( 1-sin C ight) cdot cosleft( A-B ight) Rightarrow -cos C=left( 1-sin C ight) cdot cos left( A-B ight) $$

    $$Rightarrow sin ^{2}frac {C} {2}-cos ^{2}frac {C} {2} =left( sin ^{2}frac {C} {2}-2cdot sin frac {C} {2}cos frac {C} {2}+cos ^{2}frac {C} {2} ight) cdot cos left( A-B ight) $$
    $$Rightarrow sin ^{2}frac {C} {2}-cos ^{2}frac {C} {2} =left( sin frac {C} {2}-cos frac {C} {2} ight) ^{2}cdot cos left( A-B ight) $$

    $frac { sin ^{2}frac {C} {2}-cos ^{2}frac {C} {2} } {left( sin frac {C} {2}-cos frac {C} {2} ight) ^{2}}=cos left( A-B ight) Rightarrow frac {sin frac {C} {2}+cosfrac {C} {2}} {sin frac {C} {2}-cos frac {C} {2}}=cosleft( A-B ight) $

    $frac {sin frac {C} {2}+cos frac {C} {2}} {left| sin frac {C} {2}-cos frac {C} {2} ight| }>1$

    $C_{ riangle ABC}=a+b+sqrt {a^{2}+b^{2}}$

    $left( a^{2}+b^{2} ight) left( 4^{2}+3^{2} ight) geq left( 4a+3b ight) ^{2}$

    $a+b+sqrt {a^{2}+b^{2}}geq a+b+frac {4a+3b} {5}-frac {9} {5}a+frac {8} {5}b$

    $left( frac {8} {5}a+frac {9} {5}b ight) cdot 1-left( frac {9} {5}a+frac {8} {5}b ight) cdot left( frac {2} {a}+frac {1} {b} ight) =frac {2b} {5}+frac {1b} {5}cdot frac {b} {a}+frac {9} {5}cdot frac {b} {a}geq frac {2b} {5}+2cdot sqrt {frac {16} {5} imes frac {9} {5} imes frac {b} {a} imes frac {a} {b}}=frac {26} {5}+frac {24} {5}=10$

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  • 原文地址:https://www.cnblogs.com/wangshixi12/p/12459880.html
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