填空题：在区间[0,π/2]上,sinx及x+cosx满足柯西微分中值定理的中的ξ=?求大神指导.

设函数f(x)=sinx,F(x)=x+cosx,
∵f(x),F(x)在区间[0,π/2]是连续的,且在(0,π/2)均是可导,根据柯西中值定理,
[f(π/2)-f(0)]/[F(π/2)-F(0)]=f'(ξ)/F'(ξ),
f(π/2)=sin(π/2)=1,
f(0)=sin0=0,
F(π/2)=π/2+cosπ/2=π/2.
F(0)=0+cos0=1,
f'x)=cosx,
F'(x)=1-sinx,
(1-0)/(π/2-1)=cosξ/(1-sinξ),
2/(π-2)=cosξ/(1-sinξ),
∵(1-sinξ)(1+sinξ)=(cosξ)^2,
∴cosξ/(1-sinξ)=(1+sinξ)/cosξ,
设2/(π-2)=k,
cosξ/(1-sinξ)=k=(1+sinξ)/cosξ,
(1+sinξ)^2/(cosξ)^2=k^2,
1+2sinξ+(sinξ)^2=k^2[1-(sinξ)^2]
(sinξ)^2(1+k^2)+2sinξ+1-k^2=0,
sinξ=(-1±k^2)/(1+k^2),
因负根不在区间内,所以舍去,
∴sinξ=(-1+k^2)/(1+k^2),
将k=2/(π-2)代入,
sinξ=(-π^2+4π)/(π^2-4π+8),
∴ξ=arcsin[(-π^2+4π)/(π^2-4π+8)]≈0.5095.