4是不是4次方?
f(x)=(cos²x+sin²x)(cos²x-sin²x)-2sinxcosx
=1*cos2x-sin2x
=-(sin2x-cos2x)
=-√2(√2/2*sin2x-√2/2cos2x)
=-√2(sin2xcosπ/4-cos2xsinπ/4)
=-√2sin(2x-π/4)
所以T=2π/2=π
4是不是4次方?
f(x)=(cos²x+sin²x)(cos²x-sin²x)-2sinxcosx
=1*cos2x-sin2x
=-(sin2x-cos2x)
=-√2(√2/2*sin2x-√2/2cos2x)
=-√2(sin2xcosπ/4-cos2xsinπ/4)
=-√2sin(2x-π/4)
所以T=2π/2=π