顶丞琳
幼苗
共回答了17个问题采纳率:94.1% 举报
(1)
先化简f(x),得
f(x)=a*(1+cos2x)+b*(1/2)*sin2x
=a+a*cos2x+(b/2)*sin2x
∵f(0)=2
∴a+a=2,得a=1
∵f(π/3)=1/2+√3/2
∴a-(a/2)+b*√3/4=1/2+b*√3/4
=1/2+√3/2
∴b=2
∴f(x)
=1+cos2x+sin2x
=1+√2*sin(2x+π/4)
从而f(x)=sin2x+cos2x+1=√2sin(2x+π/4)+1≥1-√2
所以f(x)最小值为1-√2 同理最大值1+√2
(2)由f(α)=f(β)得sin(2α+π/4)=sin(2β+π/4)
∵α-β≠kπ,(k∈Z)
∴2α+π/4=(2k+1)π-(2β+π/4)
即α+β=kπ+π/4
∴tan(α+β)=1
1年前
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