Question

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)

求高手解答！！本道题的答案为5！！本人求步骤！详细清楚！！
答案为5哦~~~题目如图所示！！

Answer 1

直接代公式呀
g(1/4)中，x=1/4,因此g(1/4)=cosπ/4=√2/2
f(1/3)中，x=1/3,因此f(1/3)=f(x-1)+1=f(-2/3)+1=sin(-2π/3)+1=1-sin(2π/3)=1-√3/2
g(5/6)中，x=5.6>1/2,因此g(5/6)=g(x-1)-1=g(-1/6)-1=cos(-π/6)-1=√3/2-1
所以
g(1/4)+f(1/3)+g(5/6)＝√2/2+1-√3/2+√3/2-1=√2/2

Answer 2

过程如下
g(1/4)=cosπ/4=√2/2 f(1/3)=f(1/3-1)+1=f(-2/3)+1 =sin(-2π/3)+1=√3/2 +1 g(5/6)=+√3/2 +1
f(1/4)=f(1/4-10+1=f(-3/4)+1=sin(-3π/4)+1=-√2/2+1 把这些式子加起来就可以了 答案是5