Question

设f(x)={sinπx(x<0),f(x-1)+1(x≥0〕g(x)=｛cosπx(x<1/2),

设f(x)={sinπx(x<0),f(x-1)+1(x≥0〕g(x)=｛cosπx(x<1/2),g(x-1)-1(x≥1/2｝求g(1/4)+f(1/3)+g(5/6)+f(3/4)的值

Answer 1

解g(1/4)=cosπ(1/4)=-√2/2
g(5/6)=g(5/6-1)=g(-1/6)=cos(-1/6)π=-√3/2
f(1/3)=f(1/3-1)+1
=f(-2/3)+1
=sin(-2/3)π+1
=-sin2π/3+1
=-√3/2+1
f(3/4)=f(3/4-1)+1
=f(-1/4)+1
=sin(-1/4)π+1
=-sinπ/4+1
=-√2/2+1
故原式
=-√2/2-√3/2+1-√3/2-√2/2+1
=-√2-√2+2