scalpelpp
幼苗
共回答了18个问题采纳率:88.9% 举报
√3sin2x+cos2x
=2[√3/2sin2x+1/2cos2x]
=2[cosπ/6 sin2x+sinπ/6 cos2x]
=2sin(2x+π/6)
f(x)=sinxcosx-(1/2)cos2x
=1/2(sin2x-cos2x)
=√2/2 [√2/2sin2x-√2/2cos2x]
=√2/2 [cosπ/4 sin2x-sinπ/4 cos2x]
=√2/2sin(2x-π/4)
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1年前
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