锐角三角形ABC中,三个内角A,B,C,两向量P=(2-2sinA,cosA+sinA) Q=（sinA-cosA,1+

锐角三角形ABC中,三个内角A,B,C,两向量P=(2-2sinA,cosA+sinA) Q=（sinA-cosA,1+sinA）,P与Q是共线向量求
数y=2sin^B+cos【（C-3B)/2】取最大值时,角B的大小,已算出角A=60°
y=2sin^B+cos(C-3B)/2=2sin^B+cos(180°-A-4B)/2
=2sin^B+cos(90°-A/2-2B)=(2sin^B-1)+1+cos(60°- 2B)[[[[[[[[[[[这步不太明白]]]]]]]]]]]]]]]
=1-cos2B +(cos60°·cos2B + sin60°·sin2B)
=1-cos2B +[(1/2)·cos2B + sin60°·sin2B]
=1+[(-1/2)·cos2B + sin120°·sin2B][[[[[[这步不太明白]]]]]]]]]]]]]]]
=1+[cos120°·cos2B + sin120°·sin2B]
=1+cos(120°-2B)
由于B是锐角120°-2B∈(-60°,120°)
-1/2＜cos(120°-2B)≤1;
y=2sin^B+cos(C-3B)/2
=1+cos(120°-2B)的最大值是2.
两个地方不太明白,希望能告知原因,