已知函数FX=(2COS^2X-1)SIN2X+1/2COS4X,若a=(π/2,π),且F(a/4)=根号2/4,求COSa
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![已知函数FX=(2COS^2X-1)SIN2X+1/2COS4X,若a=(π/2,π),且F(a/4)=根号2/4,求COSa](/uploads/image/z/3859977-57-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0FX%3D%282COS%5E2X-1%29SIN2X%2B1%2F2COS4X%2C%E8%8B%A5a%3D%28%CF%80%2F2%2C%CF%80%EF%BC%89%2C%E4%B8%94F%EF%BC%88a%2F4%29%3D%E6%A0%B9%E5%8F%B72%2F4%2C%E6%B1%82COSa)
已知函数FX=(2COS^2X-1)SIN2X+1/2COS4X,若a=(π/2,π),且F(a/4)=根号2/4,求COSa
已知函数FX=(2COS^2X-1)SIN2X+1/2COS4X,若a=(π/2,π),且F(a/4)=根号2/4,求COSa
已知函数FX=(2COS^2X-1)SIN2X+1/2COS4X,若a=(π/2,π),且F(a/4)=根号2/4,求COSa
已知函数FX=(2COS^2X-1)SIN2X+1/2COS4X,若a=(π/2,π),且F(a/4)=根号2/4,求COSa
F(x)=(2COS^2X-1)SIN2X+1/2COS4X=COS2XSIN2X+1/2COS4X
=1/2SIN4X+1/2COS4X=√2/2sin(4x+π/4)
F(a/4)=√2/2sin(a+π/4)=√2/4==>sin(a+π/4)=1/2==>a+π/4=5π/6==>a=7π/12
所以cosa=cos(π/2+π/12)=-sin(π/12)=-(√6-√2)/4
解
f(x)=(2cos²-1)sin2x+1/2cos4x
=cos2xsin2x+1/2cos4x
=1/2sin4x+1/2cos4x
=√2/2sin(4x+π/4)
∴
T=2π/4=π/2为其最小周期
最大值为:√2/2
f(a)=√2/2sin(4a+π/4)=√2/2
∴4a+π/4=π/2+2kπ<...
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解
f(x)=(2cos²-1)sin2x+1/2cos4x
=cos2xsin2x+1/2cos4x
=1/2sin4x+1/2cos4x
=√2/2sin(4x+π/4)
∴
T=2π/4=π/2为其最小周期
最大值为:√2/2
f(a)=√2/2sin(4a+π/4)=√2/2
∴4a+π/4=π/2+2kπ
∴a=π/16+kπ/2
∵a∈(π/2,π)
∴k=1时,a=9π/16∈(π/2,π)
∴a=9π/16
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