在平面直角坐标系xoy中,已知A(6/5,0),P(cosα,sinα),其中0＜a＜π/2

1.证：向量PA=(6/5-cosa,0-sina)=(6/5-cosa,-sina).
向量PO=(0-cosa,0-sina)=(-cosa,-sina).
向量PA.向量PO=(6/5)(-cosa+(-cosa)(-cosa)+(-sina)(-sina).
=(6/5)(-cosa)+cos^2a+sin^2a.
=(6/5)(-5/6)+1.【cosa=5/6】
=-1+1.
=0.
∴向量PA⊥向量PO.
2.|PA|^2=(6/5-cosa)^2+(-sina)^2.
=(36/25-(12/5)cosa+cos^2a+sin^2a.
=(36/25)-(12/5)cosa+1.
|PO|^2=(-cosa)^2+(-sina)^2.
=cos^2a+sin^2.
=1.
题设|PA|=|PO|,----> |PA|^2=|PO|^2.
即,（36/25)-(12/5)cosa+1=1.
cosa=(36/25)*(5/12).
∴cosa=3/5.
sina=√(1-cos^2a)=4/5.
sin(2a+π/4)=sin2acos(π/4)+cos2asin(π/4).
=2sinacosa*(√2/2)+cos2a*(√2/2).【cos2a=2cos^2a-1=2*(3/5)^2-1=-7/25】
=(√2/2)[2*(4/5(*(3/5)-7/25].
=(√2/2)*(14/25).
=7√2/25.----即为所求.