搜狗做题网f(x)=[4(cosx)^4-2cos2x-1]/[tan(pai/4+x)sin^2(pai/4-x)] 求f(-1
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f(x)=[4(cosx)^4-2cos2x-1]/[tan(pai/4+x)sin^2(pai/4-x)] 求f(-17pai/12)=
![f(x)=[4(cosx)^4-2cos2x-1]/[tan(pai/4+x)sin^2(pai/4-x)] 求f(-1](/uploads/image/z/4711617-9-7.jpg?t=f%28x%29%3D%5B4%28cosx%29%5E4-2cos2x-1%5D%2F%5Btan%28pai%2F4%2Bx%29sin%5E2%28pai%2F4-x%29%5D+%E6%B1%82f%28-1)
分子=4(cosx)^4-2(2cos²x-1)-1
=4(cosx)^4-4cos²x+1
=(2cos²x-1)²
=cos²2x
sin²(π/4-x)
=cos²[π/2-(π/4-x)]
=cos²(π/4+x)
所以分母=[sin(π/4+x)/cos(π/4+x)]*cos²(π/4+x)]
=sin(π/4+x)cos(π/4+x)
=1/2*sin[2(π/4+x)]
=1/2*sin(π/2+2x)
=1/2*cos2x
所以f(x)=2cos²2x/cos2x=2cos2x
所以f(-17π/12)
=2cos(-17π/6)
=2cos(7π/6)
=2cos(π+π/6)
=-2cos(π/6)
=-√3
=4(cosx)^4-4cos²x+1
=(2cos²x-1)²
=cos²2x
sin²(π/4-x)
=cos²[π/2-(π/4-x)]
=cos²(π/4+x)
所以分母=[sin(π/4+x)/cos(π/4+x)]*cos²(π/4+x)]
=sin(π/4+x)cos(π/4+x)
=1/2*sin[2(π/4+x)]
=1/2*sin(π/2+2x)
=1/2*cos2x
所以f(x)=2cos²2x/cos2x=2cos2x
所以f(-17π/12)
=2cos(-17π/6)
=2cos(7π/6)
=2cos(π+π/6)
=-2cos(π/6)
=-√3
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