已知函数f(x)=cos²x+cos²(x+α)+cos²(x+β)
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已知函数f(x)=cos²x+cos²(x+α)+cos²(x+β)
其中α,β为常数,且满足0
其中α,β为常数,且满足0
要使f(x)是与x无关的定值则要求
cos^2(x+a)+cos^2 (x+b) =sin^2 x +k (k为常数)
cos^2 x cos^2 a +sin^2 x sin^2 a +2cosxsinx cosasina
+cos^2 x cos^2 b +sin^2 x sin^2 b+2cosx sinx cosbsinb =sin^2 x +k
cos^2 x(cos^2 a+cos^2 b)+sin^2 x(sin^2 a+sin^2 b)
+2cosxsinx (cosasina +cosbsinb) =sin^2 x+k
cosasina+cosbsinb=0 ...(1)
cos^2a +cos^2b -sin^2 x (cos^2a+cos^2 b) +sin^2 x(sin^2 a+sin^2 b)=sin^2 x+k
sin^2 x (sin^2 a+sin^2 b -cos^2 a-cos^2 b) +cos^2 a+cos^2 b =sin^2 x+k
则要sin^2 a+sin^2 b-cos^2 a-cos^2 b =1 ...(2)
-cos2a -cos2b =1 ...(3)
sin2a +sin2b=0 ...(4)
由(3)(4)得
sin^2 2a =sin^2 2b 1-cos^2 2a =1-cos^2 2b cos^2 2a=cos^2 2b
cos2a =-1-cos2b
cos^2 2a =1+2cos2b +cos^2 2b
1+2cos2b =0
cos2b =-1/2 ...(5) cos2a=-1/2 ...(6)
b=60 sin2b =sin120=sin60=根号3/2
sin2a =-根号3/2 a=150
因此只要a=150 b=60 就可以符合条件.
cos^2(x+a)+cos^2 (x+b) =sin^2 x +k (k为常数)
cos^2 x cos^2 a +sin^2 x sin^2 a +2cosxsinx cosasina
+cos^2 x cos^2 b +sin^2 x sin^2 b+2cosx sinx cosbsinb =sin^2 x +k
cos^2 x(cos^2 a+cos^2 b)+sin^2 x(sin^2 a+sin^2 b)
+2cosxsinx (cosasina +cosbsinb) =sin^2 x+k
cosasina+cosbsinb=0 ...(1)
cos^2a +cos^2b -sin^2 x (cos^2a+cos^2 b) +sin^2 x(sin^2 a+sin^2 b)=sin^2 x+k
sin^2 x (sin^2 a+sin^2 b -cos^2 a-cos^2 b) +cos^2 a+cos^2 b =sin^2 x+k
则要sin^2 a+sin^2 b-cos^2 a-cos^2 b =1 ...(2)
-cos2a -cos2b =1 ...(3)
sin2a +sin2b=0 ...(4)
由(3)(4)得
sin^2 2a =sin^2 2b 1-cos^2 2a =1-cos^2 2b cos^2 2a=cos^2 2b
cos2a =-1-cos2b
cos^2 2a =1+2cos2b +cos^2 2b
1+2cos2b =0
cos2b =-1/2 ...(5) cos2a=-1/2 ...(6)
b=60 sin2b =sin120=sin60=根号3/2
sin2a =-根号3/2 a=150
因此只要a=150 b=60 就可以符合条件.
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