已知函数f(x)=cos(π/6-2x)+cos(2x+π/6)+sin(2x+π/3)-sin(π/3-2x).
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已知函数f(x)=cos(π/6-2x)+cos(2x+π/6)+sin(2x+π/3)-sin(π/3-2x).
⑴:求函数f(x)的最小正周期.
⑵:在锐角△ABC中,角ABC所对的边分别为abc,且f(A)=1,a=1,试求△ABC的面积S的最大值
⑴:求函数f(x)的最小正周期.
⑵:在锐角△ABC中,角ABC所对的边分别为abc,且f(A)=1,a=1,试求△ABC的面积S的最大值
f(x)
=cos(π/6-2x)+cos(2x+π/6)+sin(2x+π/3)-sin(π/3-2x)
=sin(π/2-(π/6-2x))+cos(2x+π/6)+sin(2x+π/3)-cos(π/2-(π/3-2x))
=sin(2x+π/3)+cos(2x+π/6)+sin(2x+π/3)-cos(2x+π/6)
=2sin(2x+π/3)
最小正周期是2π/2=π
(2)
f(A)=1
2sin(2A+π/3)=1
2A+π/3=5π/6或π/6(舍去,∵A>0)
∴A=π/4
余弦定理
cosA=(b^2+c^2-a^2)/(2bc)
∴√2bc=b^2+c^2-1
1+√2bc=b^2+c^2
∵b^2+c^2>=2bc
∴1+√2bc>=2bc
bc
=cos(π/6-2x)+cos(2x+π/6)+sin(2x+π/3)-sin(π/3-2x)
=sin(π/2-(π/6-2x))+cos(2x+π/6)+sin(2x+π/3)-cos(π/2-(π/3-2x))
=sin(2x+π/3)+cos(2x+π/6)+sin(2x+π/3)-cos(2x+π/6)
=2sin(2x+π/3)
最小正周期是2π/2=π
(2)
f(A)=1
2sin(2A+π/3)=1
2A+π/3=5π/6或π/6(舍去,∵A>0)
∴A=π/4
余弦定理
cosA=(b^2+c^2-a^2)/(2bc)
∴√2bc=b^2+c^2-1
1+√2bc=b^2+c^2
∵b^2+c^2>=2bc
∴1+√2bc>=2bc
bc
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