Question

数学函数问题！！

函数若图 ，求g(1/4)+f(2/3)+g(5/6)+f(1/4)

Answer 1

g(1/4)=cos(π/4)=(根号2)/2
g(5/6)=g(5/6-1)+3/2=g(-1/6)+3/2=cos(-π/6)+3/2=（根号3)/2 +3/2
f(2/3)=f(2/3-1)+1/2=f(-1/3)+1/2=sin(-π/3)+1/2=-（根号3)/2 +1/2
f(1/4)=f(1/4-1)+1/2=f(-3/4)+1/2=sin(-3π/4) +1/2=-（根号2)/2 +1/2
那么g(1/4)+f(2/3)+g(5/6)+f(1/4)
=1/2+1/2+3/2=5/2
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Answer 2

g(1/4)+f(2/3)+g(5/6)+f(1/4)

2/3>0
所以f(2/3)
=f(2/3-1)+1/2
=f(-1/3)+1/2
=sin(-π/3）+1/2
=-√3/2+1/2
=(1-√3)/2

同理
f(1/4)
=f(1/4-1)+1/2
=sin(-3π/4)+1/2
=(1-√2)/2

1/4<1/2
g(1/4)=cosπ/4=√2/2

5/6>1/2
g(5/6)
=g(5/6-1)+3/2
=g(-1/6)+3/2
=cos(-π/6)+3/2
=(√3+3)/2

原式=(√2+1-√3+√3+3+1-√2)/2=5/2

Answer 3

g(1/4)=cosπ/4=√2/2
f(2/3)=f(-1/3)+1/2=sin(-π/3)+1/2=(1-√3)/2
g(5/6)=g(-1/6)+3/2=cos(-π/6)+3/2=(3+√3)/2
f(1/4)=f(-3/4)+1/2=sin(-3π/4)+1/2=(1-√2)/2
g(1/4)+f(2/3)+g(5/6)+f(1/4)=5/2