∵y=2sin(x+
π
4 )cos(x-
π
4 )
=2sin(x-
π
4 +
π
2 )cos(x-
π
4 )
=2cos(x-
π
4 )cos(x-
π
4 )
=cos[2(x-
π
4 )]+1
=cos(2x-
π
2 )+1
=sin2x+1,
若y=2sin(x+
π
4 )cos(x-
π
4 )=
1
2 ,
∴2x=2kπ+
3π
2 ±
π
3 (k∈N),即x=kπ+
3π
4 ±
π
6 (k∈N),
则|P 2P 6|=2π.
故选B
∵y=2sin(x+
π
4 )cos(x-
π
4 )
=2sin(x-
π
4 +
π
2 )cos(x-
π
4 )
=2cos(x-
π
4 )cos(x-
π
4 )
=cos[2(x-
π
4 )]+1
=cos(2x-
π
2 )+1
=sin2x+1,
若y=2sin(x+
π
4 )cos(x-
π
4 )=
1
2 ,
∴2x=2kπ+
3π
2 ±
π
3 (k∈N),即x=kπ+
3π
4 ±
π
6 (k∈N),
则|P 2P 6|=2π.
故选B