y=sin2x+cos2x左移π/4个单位所得解析式?A.Y=cos2x+sin2xb:y=cos2x-sin2xc:y=sin2x-cos2xd:y=cosx-sinx

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y=sin2x+cos2x左移π/4个单位所得解析式?A.Y=cos2x+sin2xb:y=cos2x-sin2xc:y=sin2x-cos2xd:y=cosx-sinx

y=sin2x+cos2x左移π/4个单位所得解析式?A.Y=cos2x+sin2xb:y=cos2x-sin2xc:y=sin2x-cos2xd:y=cosx-sinx
y=sin2x+cos2x左移π/4个单位所得解析式?
A.Y=cos2x+sin2x
b:y=cos2x-sin2x
c:y=sin2x-cos2x
d:y=cosx-sinx

y=sin2x+cos2x左移π/4个单位所得解析式?A.Y=cos2x+sin2xb:y=cos2x-sin2xc:y=sin2x-cos2xd:y=cosx-sinx
y=sin2x+cos2x=根号下2sin(2x+π/4)
左移π/4个单位,所以y=根号下2sin[2(x+π/4)+π/4]=根号下2sin(2x+3π/4)
自己可以拆啊!
要自己算啊!
答案是D

y=sin2x+cos2x
=√2sin(2x+π/4)
=√2sin[2(x+π/8)]
左移π/4个单位
y'=√2sin[2(x+π/4+π/8)]=√2sin(2x+3π/4)