（1）展开f(x)=sin x*cos(7π/4)+cos x*sin(7π/4)+cos x*cos(7π/4)-sin x*sin(7π/4) =√2 *(sin x-cos x) =√2 *(sin x+sin (x+π/2)) 和差化积 =2√2 * sin(x+π/4) *cos(-π/4) =2sin(x+π/4) 最小正周期2π，最小值 -2 （2）cos(β-α)=cosβcosα+sinβsinα=4/5 cos(β+α)=cosβcosα-sinβsinα=-4/5 则cosβcosα=0 sinβsinα=4/5 又0<α>

对4x/(1+)求导得4(-x^2+1)/[(x^2+1)^2]=-4(x^2-1)/[(x^2-1)^2+4(x^2-1)+4]

x^2-1=0时，即x=1或x=-1时，原式=0;

(x^2-1)不等于0时，分子分母同时除以(x^2-1)，可得

原式=-4/[(x^2-1)+4/(x^2-1)+4]，又由于（x^2-1)大于等于-1，根据重要不等式，（x^2-1)+4/(x^2-1)大于等于4且在x^2-1=2是渠道最小值4，此时分母最小，原式=-1/2，将（x^2-1)=-1代入原式，得原式=4，故斜率范围为[-1/2,4]。

解：y=1+sin(2x)+2cos^2(x) =1+sin(2x)+1+cos(2x) =2+sin(2x)+cos(2x) =2+√2sin(2x+π/4) 所以周期为π (2) -π/2+2kπ<=2x+π/4<=π/2+2kπ -3π/4+2kπ<=2x<=π/4+2kπ -3π/8+kπ<=x<=π/8+kπ (3)y=2+√2sin(2x+π/4) =2+√2sin(2(x+π/8)) 所以是先左移π/8,然后上升2得到

答案为一