Question

已知向量a=(sinx,sinx+cosx)b=(2cosx,cosx-sinx),设f(x)=a*b

（1）当x∈[0，π/2]时，求函数f(x)的值域；
（2）[-π/8，0]，f(θ)=2/5，求sin(2θ+3π/4)的值。
要详细过程，谢谢

Answer 1

解：
f(x)=a*b =2sinxcosx+(sinx+cosx)(cosx-sinx)=sin2x+cos2x=√2sin(2x+π/4)
1）当x∈[0，π/2]时 2x+π/4 ∈[π/4,π/4 +π] 当2x+π/4=π/2 f(x)取到最大值
当2x+π/4=-π/2时 f(x)取到最小值
所以f(x)的值域为 [-√2,√2]
2) θ ∈[-π/8，0] 则 2θ+3π/4∈[π/2，3π/4]
sin(2θ+3π/4)=sin(2θ+π/4+π/2)=cos(2θ+π/4)
又f(θ)=2/5=√2sin(2θ+π/4)
则 sin(2θ+π/4)=√2/5
所以cos(2θ+π/4) =√23/5=sin(2θ+3π/4)
望你采纳 ~~

Answer 2

f(x)=sian2x+cos2x=√(2)sin(
x[0，π/2].2x+π/4∈[π/4,5π/4],sin(2θ+3π/4)∈[--(√(2)/2),1],f(x)∈[-1,√(2)]