求化简!
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求化简!求化简! 求化简!把sin(x+π/3)展开然后呢f(x)=2cosxsin(x+π/3)-√3÷2=2cosx(sinxcosπ/3-cosxsinπ/3)-√3/2=sinxco
求化简!
求化简!
求化简!
把sin(x+π/3)展开然后呢f(x)=2cosxsin(x+π/3)-√3÷2
=2cosx(sinxcosπ/3-cosxsinπ/3)-√3/2
=sinxcosx-√3cosxcosx-√3/2
=1/2sin2x-1/2cos2x-1/2-√3/2
=√2/2sin(2x-π/4)-(1-√3)/2=√2/2sin(2x-π/4)-(1+√3)/2对不起...
全部展开
把sin(x+π/3)展开
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收起