已知函数f (x )=sin x cos x +cos²x -1/2(1)若x是锐角三角形ABC 的内角,求f (x )的值域我是文科
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已知函数f (x )=sin x cos x +cos²x -1/2(1)若x是锐角三角形ABC 的内角,求f (x )的值域我是文科
已知函数f (x )=sin x cos x +cos²x -1/2
(1)若x是锐角三角形ABC 的内角,求f (x )的值域我是文科
已知函数f (x )=sin x cos x +cos²x -1/2(1)若x是锐角三角形ABC 的内角,求f (x )的值域我是文科
由f(x)=sinxcosx+cos²x-1/2
=(1/2)sin2x+(cos2x+1)/2-1/2
=(1/2)sin2x+(1/2)cos2x
=√2/2[(√2/2)sin2x+(√2/2)cos2x]
=√2/2(sinπ/4sin2x+cosπ/4cos2x0
=√2/2cos(2x-π/4)
①当2x-π/4=0时:
x=π/8,有最大值f(x)=√2/2.
②由x<π/2,∴2x<π,2x-π/4<3π/4
∴f(x)<(√/2)×(-√2/2)=-1/2
∴值域:1/2<y≤√2/2.
f (x )=sin x cos x +cos²x -1/2
=1/2sin2x+1/2+1/2cos2x-1/2
=1/2sin2x+1/2cos2x
=根号2/2sin(2x+pi/4)
0
由sin(2x+pi/4)得值域:(-1/2,根号2/2]0
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f (x )=sin x cos x +cos²x -1/2
=1/2sin2x+1/2+1/2cos2x-1/2
=1/2sin2x+1/2cos2x
=根号2/2sin(2x+pi/4)
0
由sin(2x+pi/4)得值域:(-1/2,根号2/2]
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