f（x）=（sinx＋cosx）²＋2cos²x化简
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f(x)=(sinx+cosx)²+2cos²x
=sin²x+cos²x+2sinxcosx+1+cos2x
=1+sin2x+1+cos2x
=sin2x+cos2x+2
=√2[(√2/2)sin2x+(√2/2)cos2x]+2
=√2(sinxcosπ/4 +cosxsinπ/4)+2
=√2sin(x+ π/4)+2
=sin²x+cos²x+2sinxcosx+1+cos2x
=1+sin2x+1+cos2x
=sin2x+cos2x+2
=√2[(√2/2)sin2x+(√2/2)cos2x]+2
=√2(sinxcosπ/4 +cosxsinπ/4)+2
=√2sin(x+ π/4)+2
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引用xuzhouliuying的回答：
f(x)=(sinx+cosx)²+2cos²x
=sin²x+cos²x+2sinxcosx+1+cos2x
=1+sin2x+1+cos2x
=sin2x+cos2x+2
=√2[(√2/2)sin2x+(√2/2)cos2x]+2
=√2(sinxcosπ/4 +cosxsinπ/4)+2
=√2sin(x+ π/4)+2
f(x)=(sinx+cosx)²+2cos²x
=sin²x+cos²x+2sinxcosx+1+cos2x
=1+sin2x+1+cos2x
=sin2x+cos2x+2
=√2[(√2/2)sin2x+(√2/2)cos2x]+2
=√2(sinxcosπ/4 +cosxsinπ/4)+2
=√2sin(x+ π/4)+2
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fx=√2sin(2x+∏/4)+2
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