已知f(x)= 根号2 * sin(2x + pai/4 ) + 根号2 ,求F(x)=f(x)-f(x - pai/4)的最大值、最小值

问题描述:

已知f(x)= 根号2 * sin(2x + pai/4 ) + 根号2 ,求F(x)=f(x)-f(x - pai/4)的最大值、最小值

f(x)=√2*sin(2x+π/4)+√2F(x)=f(x)-f(x-π/4)=√2*sin(2x+π/4)+√2-√2*sin(2(x-π/4)+π/4)-√2=√2*sin(2x+π/4)-√2*sin(2x-π/4)=√2*sin2x*√2/2+√2*cos2x*√2/2-√2*sin2x*√2/2+√2*cos2x*√2/2=sin2x+co...