证明:因为θ是锐角,即0<θ<π/2所以0<cosθ<1<π/2,0<sinθ<1<π/2,先证sin(cosθ)<cosθsin(cosθ)<cosθ<=>sin(cosθ)<sin(π/2-θ)<=>cosθ<π/2-θ<=>sin(π/2-θ)<π/2-θ设t=π/2-θ,0&l 展开
证明:因为θ是锐角,即0<θ<π/2所以0<cosθ<1<π/2,0<sinθ<1<π/2,先证sin(cosθ)<cosθsin(cosθ)<cosθ<=>sin(cosθ)<sin(π/2-θ)<=>cosθ<π/2-θ<=>sin(π/2-θ)<π/2-θ设t=π/2-θ,0<t<π/2,f(t)=sint-tf'(t)=cost-1<0,所以f(t)=sint-t是单调减的,在t=0处取得最大值f(0)=sin0-0=0,所以当0<t<π/2时,f(t)<0,即sint-t<0,sint<t即sin(π/2-θ)<π/2-θ成立.所以sin(cosθ)<cosθ再证cosθ<cos(sinθ)因为cosθ在(0,π/2)是单调减的所以cosθ<cos(sinθ)<=>θ>sinθ由前面证的结论可知θ>sinθ恒成立所以cosθ<cos(sinθ)所以sin(cosθ)<cosθ<cos(sinθ) 收起
