sin^2x+2sinxcosx+3cos^x

题目中最后一项应有平方吧.
用降幂公式,就是余弦二倍角公式倒用.(sinx)^2=(1-cos2x)/2 .(cosx)^2=(1+cos2x)/2
还有正弦二倍角公式：sin2x=2sinxcosx
原式化简为：
sin^2x+2sinxcosx+3cos^2x=(1-cos2x)/2 +sin2x+3(1+cos2x)/2
=sin2x+cos2x+2
=√2[√2/2sin2x+√2/4cos2x]+2
=√2[cosπ/4sin2x+sinπ/4cos2x]+2
=√2sin(2x+π/4)+2

原式=sin^2x+cos^2x+2cos^2x+2sinxcosx
=1+2cos^2x+sin2x
=1+(2cos^2x-1)+sin2x+1
=cos2x+sin2x+2
=√2sin(2x+π/4)+2