Question

已知向量a=(1+sin2x,sinx-cosx),b=(1,sinx+cosx)

设函数f(x)=a·b
(1)若x∈[π/4，π/2]，求f(x)的最大值与最小值
(2)若f(θ)=8/5,求sin4θ的值

Answer 1

解：f(x)=向量a.向量b.
=(1+sin2x)*1+(sinx-cosx)*(sinx+cosx).
=1+sin2x-(cos^2x-sin^2x).
=1+sin2x-cos2x.
=1+√2sin(2x-π/4).
∴f(x)=√2sin(2x-π/4)+1.
(1)∵x∈[π/4,π/2],∴(2x-π/4)∈[π/4,3π/4]
令2x-π/4=π/2, 即x=3π/8时，sin(2x-π/4)=1, f(x)取得最大值，即f(x)max=1+√2.
当2x-π/4=π/4或2x-π/4=3π/4时，sin(2x-π/4)=√2/2,f(x)取得最小值,即f(x)min=2.
(2) 若f(θ)=8/5时，求sin4θ的值。
f(θ)=√2sin(2θ-π/4)+1=8/5.
√2sin(2θ-π/4)=3/5.
√2sin2θcos(π/4)-√2cos2θsin(π/4)=3/5.
sin2θ-cos2θ=3/5.
(sin2θ-cos2θ)^2=(3/5)^2.
sin^2(2θ)-2sin2θcos2θ+cos^2(2θ)=9/25.
1-sin4θ=9/25.
sin4θ=1-9/25.
∴sin4θ=16/25.