sin（π/4-3x）=sin[π/2-(π/4+3x)]
=cos(π/4+3x)
sin（π/4-3x）cos(π/3-3x)- sin（π/4+3x）sin(π/3-3x)
=cos(π/4+3x)cos(π/3-3x)- sin（π/4+3x）sin(π/3-3x)
=cos(π/4+3x+π/3-3x)=cos(π/4+π/3)
=√2/2*1/2-√2/2*√3/2
=(√2-√6)/4
sin（π/4-3x）cos(π/3-3x)- sin（π/4+3x）sin(π/3-3x)=
sin（π/4-3x）cos(π/3-3x)- sin（π/4+3x）sin(π/3-3x)=
数学人气：309 ℃时间：2019-12-19 10:11:12
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