设函数f(x)=sin2x+cos2x+1
设△ABC的内角A、B、C的对应边分别为a、b、c,若a=1,b=2根号2,f(C)=2,求边长c及sinA的值
人气:239 ℃ 时间:2019-12-11 22:40:27
解答
f(x)=sin2x+cos2x+1=√2sin(2x+π/4)+1f(C)=√2sin(2C+π/4)+1=2∴sin(2C+π/4)=√2/20<C<π,∴π/4<2C+π/4<2π+π/4∴2C+π/4=3π/4 ∴C=π/4由余弦定理知cosC=(a²+b²-c²)/2ab=√2/2∴(9-c...
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