【答案】
分析:先根據題意設△ABC的角B,C的對邊分別為b,c,進而利用三角形面積公式表示出三角形面積,進而根據
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求得bccosA=3,進而利用同角三角函數(shù)的基本關系,利用平方關系聯(lián)立方程求得sinA和cosA,進而利用cosB的值和同角三角基本函數(shù)的關系式,求sinB,最后根據兩角和公式求得cos(A+B),利用三角形內角和可知,cosC=cos(π-A-B),利用誘導公式整理求得答案.
解答:解:由題意,設△ABC的角B,C的對邊分別為b,c,則S=
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bcsinA=
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>0
∴A∈(0,
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),cosA=3sinA.
又sin
2A+cos
2A=1,
∴sinA=
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,cosA=
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由題意cosB=
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,則sinB=
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=
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∴cos(A+B)=cosAcosB+sinAsinB=
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∴cosC=cos(π-A-B)=-cos(A+B)=-
點評:本題主要考查了三角形中的幾何計算,同角三角函數(shù)的基本關系和兩角和的化簡求職.考查了學生對基礎知識的正握和基本運算能力的考查.