已知函数f(x)=2sin(x+π/4)cos(x-π/4)
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已知函数f(x)=2sin(x+π/4)cos(x-π/4)
(1)求f(π/8) (2)若f(x)=1/2,在[0,π]上有2个根,X1、X2 且X1<X2求f(X2-X1)的值
(1)求f(π/8) (2)若f(x)=1/2,在[0,π]上有2个根,X1、X2 且X1<X2求f(X2-X1)的值
f(x)=2sin(x+π/4)cos(x-π/4)
=2sin(x+π/4)cos(π/4-x)
=2sin(x+π/4)cos[(π/2)-(x+π/4)]
=2sin²(x+π/4)
=1-cos(2x+π/2)
=sin(2x)+1
(1)f(π/8)=sin(π/4)+1=(√2/2)+1;
(2)令f(x)=1/2,
则sin(2x)+1=1/2,
sin2x= -1/2,
又x∈[0,π],2x∈[0,2π],
∴2x=7π/6,或11π/6,
由题意,x1=7π/6,x2=11π/6,
∴x2-x1=2π/3,
f(x2-x1)=f(11π/6)=sin(11π/3)+1= sin(5π/3)+1=(-√3/2)+1.
=2sin(x+π/4)cos(π/4-x)
=2sin(x+π/4)cos[(π/2)-(x+π/4)]
=2sin²(x+π/4)
=1-cos(2x+π/2)
=sin(2x)+1
(1)f(π/8)=sin(π/4)+1=(√2/2)+1;
(2)令f(x)=1/2,
则sin(2x)+1=1/2,
sin2x= -1/2,
又x∈[0,π],2x∈[0,2π],
∴2x=7π/6,或11π/6,
由题意,x1=7π/6,x2=11π/6,
∴x2-x1=2π/3,
f(x2-x1)=f(11π/6)=sin(11π/3)+1= sin(5π/3)+1=(-√3/2)+1.
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