在锐角三角形ABC中，c=√2bsinC 求sinC+cosA的取值范围

c=√2bsinC
c=2RsinC→R=√2b/2
b=2RsinB→b=√2bsinB→sinB=√2/2→∠B=45°→∠A+∠C=135°

∴sinC+cosA=sin(A+B)+cosA=sin(A+45°)+cosA=sin(A+45°)+sin(90°-A)
=2sin(135°/2)cos[(2A-45°)/2]
=2sin67.5°cos(A-22.5°)

∵∠A<90°[∠C<90°]
∴cos(A-22.5°)>cos(90°-22.5°)=cos67.5°
又∵cos(A-22.5°)≤1 [∠A=22.5°是取等号]

∴cos67.5°<cos(A-22.5°)≤1
∴sin135°<2sin67.5°cos(A-22.5°)≤2sin67.5°
√2/2 <2sin67.5°cos(A-22.5°)≤√(2+√2)

亦即：√2/2 <sinC+cosA≤√(2+√2)

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c=√2bsinC
c/sinC=b/sinB
sinB=√2/2
B=π/4
∴sinC+cosA=sin(A+B)+cosA=sin(A+45°)+cosA=sin(A+45°)+sin(90°-A)
=2sin(135°/2)cos[(2A-45°)/2]
=2sin67.5°cos(A-22.5°)

∵∠A<90°[∠C<90°]
∴cos(A-22.5°)>cos(90°-22.5°)=cos67.5°
又∵cos(A-22.5°)≤1 [∠A=22.5°是取等号]

∴cos67.5°<cos(A-22.5°)≤1
∴sin135°<2sin67.5°cos(A-22.5°)≤2sin67.5°

亦即：sin135°<sinC+cosA≤2sin67.5°