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)
答案:2 悬赏:10 手机版
解决时间 2021-01-25 10:22
- 提问者网友：姑娘长的好罪过
- 2021-01-24 12:14
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
最佳答案
- 五星知识达人网友：廢物販賣機
- 2019-06-01 01:12
g(1/4)=cos（π/4）
f(1/3)=f(-2/3)+1=sin(-2π/3)+1=1-sin(2π/3)
g(5/6)=g(-1/6)-1=cos(-π/6)-1=cos(π/6)-1
f(3/4)=f(-1/4)+1=sin(-π/4)+1
所以g(1/4)+f(1/3)+g(5/6)+f(3/4)=cos（π/4）+1-sin(2π/3)+cos(π/6)-1+sin(-π/4)+1
=1
f(1/3)=f(-2/3)+1=sin(-2π/3)+1=1-sin(2π/3)
g(5/6)=g(-1/6)-1=cos(-π/6)-1=cos(π/6)-1
f(3/4)=f(-1/4)+1=sin(-π/4)+1
所以g(1/4)+f(1/3)+g(5/6)+f(3/4)=cos（π/4）+1-sin(2π/3)+cos(π/6)-1+sin(-π/4)+1
=1
全部回答
- 1楼网友：怀裏藏嬌
- 2019-12-16 08:16
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
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