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已知三角形ABC,角A,B,C所对的边分别为a,b,c且满足a(3a-2b)=3(c+b)(c-b),求sin2C+cos(2A+2B)的值
人气:424 ℃ 时间:2020-05-20 05:57:31
解答
a(3a-2b)=3(c+b)(c-b),
变形得(a²+b²-c²)/2ab=1/3
就是cosC=1/3
则sinC=2√2/3
sin2C+cos(2A+2B)
=sin2C+cos2C
=2sinCcosC+cos²C-sin²C
=(4√2-7)/9
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