若乔1030
幼苗
共回答了17个问题采纳率:100% 举报
是求sinA+cosA吧?
f(x)=cos²(x+π/12)+1/2sin2x
=1/2[1+cos(2x+π/6)+sin2x]
=1/2(1+cosπ/6cos2x-sinπ/6sin2x+sin2x)
=1/2(1+√3/2cos2x+1/2sin2x)
=1/2[1+cos(2x-π/6)]
=1/2cos(2x-π/6)+1/2
故T=2π/2=π
f(A)+f(A-π/3)
=1/2cos(2A-π/6)+1/2cos(2A-2π/3-π/6)+1
=1/2cos(2A-π/6)-1/2cos(2A+π/6)+1
=sinπ/6sin2A+1
=1/2sin2A+1
=37/25
故sin2A=24/25
故(sinA+cosA)²=1+sin2A=49/25
因sin2A=2sinAcosA=24/25>0,
A为△ABC内角
故sinA>0
故cosA>0
故sinA+cosA>0
故sinA+cosA=7/5
1年前
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