作业帮已知向量a=(cos3x/2,sin3x/2）,b=（cosx/2,—sinx/2）,且x∈[0,π/2],f（x）=a·b-2λ│a+b│求：（1）a*b及│a+b│(2)若f（x）的最小值是-3/2,求实数λ的值
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![已知向量a=(cos3x/2,sin3x/2）,b=（cosx/2,—sinx/2）,且x∈[0,π/2],f（x）=a·b-2λ│a+b│求：（1）a*b及│a+b│(2)若f（x）的最小值是-3/2,求实数λ的值](/uploads/image/z/351254-38-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%28cos3x%2F2%2Csin3x%2F2%EF%BC%89%2Cb%3D%EF%BC%88cosx%2F2%2C%E2%80%94sinx%2F2%EF%BC%89%2C%E4%B8%94x%E2%88%88%5B0%2C%CF%80%2F2%5D%2Cf%EF%BC%88x%EF%BC%89%3Da%C2%B7b-2%CE%BB%E2%94%82a%2Bb%E2%94%82%E6%B1%82%EF%BC%9A%EF%BC%881%EF%BC%89a%2Ab%E5%8F%8A%E2%94%82a%2Bb%E2%94%82%282%29%E8%8B%A5f%EF%BC%88x%EF%BC%89%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E6%98%AF-3%2F2%2C%E6%B1%82%E5%AE%9E%E6%95%B0%CE%BB%E7%9A%84%E5%80%BC)
已知向量a=(cos3x/2,sin3x/2）,b=（cosx/2,—sinx/2）,且x∈[0,π/2],f（x）=a·b-2λ│a+b│求：（1）a*b及│a+b│(2)若f（x）的最小值是-3/2,求实数λ的值
已知向量a=(cos3x/2,sin3x/2）,b=（cosx/2,—sinx/2）,且x∈[0,π/2],f（x）=a·b-2λ│a+b│
求：
（1）a*b及│a+b│
(2)若f（x）的最小值是-3/2,求实数λ的值
已知向量a=(cos3x/2,sin3x/2）,b=（cosx/2,—sinx/2）,且x∈[0,π/2],f（x）=a·b-2λ│a+b│求：（1）a*b及│a+b│(2)若f（x）的最小值是-3/2,求实数λ的值
(1)因为向量a=(cos3x/2,sin3x/2）,b=（cosx/2,—sinx/2）,所以：
|a|＝|b|＝1
且a*b=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)
=cos(3x/2 +x/2)
=cos2x
则|a+b|²=|a|²+2a*b+|b|²
=2+2cos2x
=2(1+cos2x)
=4cos²x
因为x∈[0,π/2],所以：
|a+b|=2cosx
(2)由（1）可得：
f（x）=a·b-2λ│a+b│
＝cos2x－2λ*2cosx
=2cos²x-4λcosx-1
=2(cosx-λ)²-2λ²-1
因为x∈[0,π/2],所以cosx∈[0,1]
若λ
ab=cos(3x/2)*cos(x/2)-sin(3x/2)*sin(x/2)
=cos[(3x+x)/2]
=cos(2x).
a+b=(cos(3x/2)+cos(x/2),sin(3x/2)-sin(x/2)),
|a+b|=√[(cos(3x/2)+cos(x/2))^2+(sin(3x/2)-sin(x/2))^2]
=√[2(1+co...
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ab=cos(3x/2)*cos(x/2)-sin(3x/2)*sin(x/2)
=cos[(3x+x)/2]
=cos(2x).
a+b=(cos(3x/2)+cos(x/2),sin(3x/2)-sin(x/2)),
|a+b|=√[(cos(3x/2)+cos(x/2))^2+(sin(3x/2)-sin(x/2))^2]
=√[2(1+cos2x)]
=2*|cosx|,
因为,x∈[-π/3,π/4]。则有,cosx>0,
即,
|a+b|=2*|cosx|=2cosx.
2.若f（x）=a*b-|a+b|。则有,
f(x)=cos2x-2cosx,
=2cos^2x-1-2cosx
=2(cosx-1/2)^2-3/2.
而,x∈[-π/3,π/4]。则有,
1)当X=0时,cos0=1,则f(x)=2(1-1/2)^2-3/2=-1.
2)当X=π/4时,cosπ/4=√2/2,则f(x)=2*(√2/2-1/2)^2-3/2=-√2.
则,f(x)最大值=-1,f(x)最小值=-√2.
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