已知cos(2A-π╱3)+2sin(A-π╱4)sin(A+π╱4)=1╱2.求cos(2π╱3-2a)
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已知cos(2A-π╱3)+2sin(A-π╱4)sin(A+π╱4)=1╱2.求cos(2π╱3-2a)
原式===> cos(2a-π/3)-[cos2a-cos(π/4)]=1/2
===> cos2acos(π/3)+sin2asin(π/3)-coa2a+√2/2=1/2
===> (1/2)cos2a+(√3/2)sin2a-cos2a=(1-√2)/2
===> (√3/2)sin2a-(1/2)cos2a=(1-√2)/2
稍等一下. 再答: ===> -cos(2a+π/3)=(1-√2)/2
===> cos[π-(2a+π/3)]=(1-√2)/2
===> cos(2π/3-2a)=(1-√2)/2
===> cos2acos(π/3)+sin2asin(π/3)-coa2a+√2/2=1/2
===> (1/2)cos2a+(√3/2)sin2a-cos2a=(1-√2)/2
===> (√3/2)sin2a-(1/2)cos2a=(1-√2)/2
稍等一下. 再答: ===> -cos(2a+π/3)=(1-√2)/2
===> cos[π-(2a+π/3)]=(1-√2)/2
===> cos(2π/3-2a)=(1-√2)/2
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