急 、COSπ/9*COS2π/9*COS4π/9=?同样,
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急 、COSπ/9*COS2π/9*COS4π/9=?同样,
![急 、COSπ/9*COS2π/9*COS4π/9=?同样,](/uploads/image/z/19685987-35-7.jpg?t=%E6%80%A5+%E3%80%81COS%CF%80%2F9%2ACOS2%CF%80%2F9%2ACOS4%CF%80%2F9%3D%3F%E5%90%8C%E6%A0%B7%2C)
π/9=20°2π/9=40°4π/9=80°
cos20cos40cos80
=sin20cos20cos40cos80/sin20
=1/2*sin40cos40cos80/sin20
=1/4*sin80cos80/sin20
=1/8*sin160/sin20
=1/8*sin20/sin20
=1/8
cos20cos40cos80
=sin20cos20cos40cos80/sin20
=1/2*sin40cos40cos80/sin20
=1/4*sin80cos80/sin20
=1/8*sin160/sin20
=1/8*sin20/sin20
=1/8
已知cos2θ=3/9,求sin4次方θ+cos4次方θ的值
[sin4α/(1+cos4α)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]=?
高中三角函数化简求值化简:√2-sin^2+cos4 (根号下) cos(π/9)*cos(2π/9)*cos(3π/9
高一数学必修4 化简COSπ/7COS2π/7COS4π/7,化简,是不是可以看成COSa/2COSaCOS2a来做,请
已知sin4次方-cos4次方=5/9,求cos4
急.设f(θ)=[2cos2θ+sin2(2π-θ)+sin(π/2+θ)-3]/[2+2cos2(π+θ)+cos(-
已知sin⁴θ-cos⁴θ=9分之5,求cos4θ的值
已知sin^4θ-cos^4θ=5/9,求cos4θ的值.
COSπ/5乘以COS2π/5
求COSπ/5*COS2π/5
已知α∈(π/2,π),且sinα=3/5,则sin^2α/2+(sin4αcos2α)/(1+cos4α)的值为多
cos π/15 * cos 2π/15 * cos4π/15 * cos8π/15