cos(a+π/3)=1/3,a∈(0,π/2),则sina=___
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cos(a+π/3)=1/3,a∈(0,π/2),则sina=___
∵a∈(0,π/2),∴a+π/3∈(π/3,π),∴sin(a+π/3)>0,又cos(a+π/3)=1/3,
∴sin(a+π/3)=√{1-[cos(a+π/3)]^2}=√[1-(1/3)^2]=2√2/3.
∴sina
=sin[(a+π/3)-π/3]
=sin(a+π/3)cos(π/3)-cos(a+π/3)sin(π/3)
=(2√2/3)×(1/2)-(1/3)×(√3/2)
=√2/3-√3/6.
∴sin(a+π/3)=√{1-[cos(a+π/3)]^2}=√[1-(1/3)^2]=2√2/3.
∴sina
=sin[(a+π/3)-π/3]
=sin(a+π/3)cos(π/3)-cos(a+π/3)sin(π/3)
=(2√2/3)×(1/2)-(1/3)×(√3/2)
=√2/3-√3/6.
cos(a+π/3)=1/3,a∈(0,π/2),则sina=___
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