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)
(1) f(x)=ab=2√2cosx+sinxcosx+2√2sinx-sinxcosx
=2√2(sinx+cosx) = 2√2 × √2/2sin(x+π/4) = 2sin(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】.
(2)若ab=1,则2sin(x+π/4)=1,sin(x+π/4)=1/2.
因为x属于【0,π】,(x+π/4)属于【π/4,5π/4】,所以 x+π/4=5π/6,x=7π/12.
所以x+7π/12=7π/6,cos(x+7π/12)= cos7π/6 = -√3/2.