=2sin^B+cos(90°-A/2-2B)=(2sin^B-1)+1+cos(60°- 2B)---------[这步不太明白]
减1,加1,这没问题吧,即：
(2sin^B-1)+1
cos(90°-A/2-2B)=cos(90°-60°/2-2B)
=cos(90°-30°-2B)=cos(60°-2B)
即：=(2sin^B-1)+1+cos(60°-2B)
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=1+[(-1/2)·cos2B + sin120°·sin2B]----------------[这步不太明白]
上一步：1-cos2B +[(1/2)·cos2B + sin60°·sin2B]
=1+(-cos2B +(1/2)·cos2B)+sin60°·sin2B
=1+(-1/2)cos2B+sin60°sin2B
注意：sin60°=sin120°,cos120°=-1/2
即：=1+cos120°cos2B+sin120°sin2B