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
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![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](/uploads/image/z/2476656-0-6.jpg?t=%E9%94%90%E8%A7%92%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E4%B8%89%E4%B8%AA%E5%86%85%E8%A7%92A%2CB%2CC%2C%E4%B8%A4%E5%90%91%E9%87%8FP%3D%282-2sinA%2CcosA%2BsinA%29+Q%3D%EF%BC%88sinA-cosA%2C1%2BsinA%EF%BC%89%2CP%E4%B8%8EQ%E6%98%AF%E5%85%B1%E7%BA%BF%E5%90%91%E9%87%8F%E6%B1%82%E6%95%B0y%3D2sin%5EB%2Bcos%E3%80%90%EF%BC%88C-3B%29%2F2%E3%80%91%E5%8F%96%E6%9C%80%E5%A4%A7%E5%80%BC%E6%97%B6%2C%E8%A7%92B%E7%9A%84%E5%A4%A7%E5%B0%8F%2C%E5%B7%B2%E7%AE%97%E5%87%BA%E8%A7%92A%3D60%C2%B0y%3D2sin%5EB%2Bcos%28C-3B%29%2F2%3D2sin%5EB%2Bcos%28180%C2%B0-A-4B%29%2F2%3D2sin%5EB)
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
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,
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
=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)
-------------------------------------------------
=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