已知两个向量a=(cosx,sinx),b=(2根号2+sinx,2根号2-cosx),f(x)=ab,x属于【0,π】1)求f(x)的值域(2)若ab=1,求cos(x+7π/12)

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已知两个向量a=(cosx,sinx),b=(2根号2+sinx,2根号2-cosx),f(x)=ab,x属于【0,π】1)求f(x)的值域(2)若ab=1,求cos(x+7π/12)

已知两个向量a=(cosx,sinx),b=(2根号2+sinx,2根号2-cosx),f(x)=ab,x属于【0,π】1)求f(x)的值域(2)若ab=1,求cos(x+7π/12)
已知两个向量a=(cosx,sinx),b=(2根号2+sinx,2根号2-cosx),f(x)=ab,x属于【0,π】
1)求f(x)的值域
(2)若ab=1,求cos(x+7π/12)

已知两个向量a=(cosx,sinx),b=(2根号2+sinx,2根号2-cosx),f(x)=ab,x属于【0,π】1)求f(x)的值域(2)若ab=1,求cos(x+7π/12)
(1) f(x)=ab=2√2cosx+sinxcosx+2√2sinx-sinxcosx
=2√2(sinx+cosx) = 2√2 × √2sin(x+π/4) = 4sin(x+π/4)
x属于【0,π】,(x+π/4)属于【π/4,5π/4】,sin(x+π/4)属于【-√2/2,1】.
所以 f(x)的值域为【-2√2,4】.
(2)若ab=1,则4sin(x+π/4)=1,sin(x+π/4)=1/4.
由cos(x+π/4)=±√(1-1/16)=-√15/4(余弦在第2,3象限取负),
∴cos(x+7π/12)
=cos(x+π/4+π/3)
=cos(x+π/4)cosπ/3-sin(x+π/4)sinπ/3
=(-√15/4)×(1/2)-(1/4)×(√3/2)
=-(√15+√3)/8.