已知向量a=(cosA，sinA)，向量b=(√3，-1)，则|2a-b|的最大值、最小值分别是？答案详解

a = (cosa,sina)，b = (√3, -1)
那么2a-b = (2cosa-√3, 2sina+1)
|2a-b| = √ ((2cosa-√3)^2 + (2sina+1)^2)
= √ (4cos^2 - 4√3cos + 3 + 4sin^2 + 4sin + 1)
= √(4sina - 4√3cosa + 8)
= 2√2(1/2 sina - √3/2 cosa + 1)
= 2√2(sin(a-π/6) +1)
因为sin(a-π/6)取值范围为[ -1, 1]
所以sin(a-π/6) + 1 取值范围为[0, 2]
所以最大值为2√2*2 = 4
最小值= 2√2*0 = 0
请采纳。

a = (cosa,sina)，b = (√3, -1)
那么2a-b = (2cosa-√3, 2sina+1)
|2a-b| = √ ((2cosa-√3)^2 + (2sina+1)^2)
= √ (4cos^2 - 4√3cos + 3 + 4sin^2 + 4sin + 1)
= √(4sina - 4√3cosa + 8)
= 2√2(1/2 sina - √3/2 cosa + 1)
= 2√2(sin(a-π/6) +1)
因为sin(a-π/6)取值范围为[ -1, 1]
所以sin(a-π/6) + 1 取值范围为[0, 2]
所以最大值为2√2*2 = 4
最小值= 2√2*0 = 0