灵鹊做题网已知△ABC的三内角A、B、C同时满足：①2cos²A/2-3cosA=0,②sin²B/2+sin
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已知△ABC的三内角A、B、C同时满足：①2cos²A/2-3cosA=0,②sin²B/2+sin²C/2＝1/2
试判断△ABC的形状.
试判断△ABC的形状.
2cos²A/2 -1 = cosA
1即
cosA +1 -3cosA =0
cosA = 1/2
A = 60°
2即
-2sin²B/2 -2sin²C/2 = -2 *1/2 = -1
1-2sin²B/2 +1-2sin²C/2 = -1 +2
cosB + cosC = 1
2cos(B/2+C/2)cos(B/2-C/2) = 1
2cos((180°-A)/2)cos(B/2-C/2) = 1
2sinA/2 cos(B/2-C/2)=1
cos(B/2-C/2)=1
B/2-C/2 =0
B=C
B=C = (180°-A)/2 = 60°
△ABC是等边三角形
1即
cosA +1 -3cosA =0
cosA = 1/2
A = 60°
2即
-2sin²B/2 -2sin²C/2 = -2 *1/2 = -1
1-2sin²B/2 +1-2sin²C/2 = -1 +2
cosB + cosC = 1
2cos(B/2+C/2)cos(B/2-C/2) = 1
2cos((180°-A)/2)cos(B/2-C/2) = 1
2sinA/2 cos(B/2-C/2)=1
cos(B/2-C/2)=1
B/2-C/2 =0
B=C
B=C = (180°-A)/2 = 60°
△ABC是等边三角形
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