(1)根据三角函数的定义,且α+[π/4]为钝角,得
sin(α+[π/4])=t=[4/5],cos(α+[π/4])=-[3/5],
即
2
2(sinα+cosα)=[4/5],
2
2(cosα-sinα)=-[3/5],
cosα=
2
10,sinα=
7
2
10;
(2)∵f(x)=cos([πx/2]+α),
∴f(1)=cos([π/2]+α)=-sinα=-
7
2
10,
f(2)
(1)根据三角函数的定义,且α+[π/4]为钝角,得
sin(α+[π/4])=t=[4/5],cos(α+[π/4])=-[3/5],
即
2
2(sinα+cosα)=[4/5],
2
2(cosα-sinα)=-[3/5],
cosα=
2
10,sinα=
7
2
10;
(2)∵f(x)=cos([πx/2]+α),
∴f(1)=cos([π/2]+α)=-sinα=-
7
2
10,
f(2)