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设函数f(x)=sinxsin(π/2+x)+cos²x,在△ABC中,角A B C的对边分别为abc
(1)求f(x)的最大值
(2)若f(A)=1,A/B=7π/12,b=根号6,求A和a
人气:329 ℃ 时间:2020-06-06 04:40:52
解答
(1)
f(x)=sinxsin(π/2+x)+cos²x
=sinxcosx+cos²x
=1/2sin2x +(cos2x+1)/2
=1/2(sin2x+cos2x)+1/2
=√2/2(√2/2sin2x+√2/2cos2x)+1/2
=√2/2sin(2x+π/4)+1/2
当sin(2x+π/4)=1时得最大值√2/2+1/2
(2)
f(A)=√2/2sin(2A+π/4)+1/2=1
√2/2sin(2A+π/4)=1/2
sin(2A+π/4)=√2/2
∵B=7π/12
∴0
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