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Last edited by hbghlyj 2021-2-23 12:15$b>a>c,\frac\pi2>d>c>0,a+b<π,\frac{\sin{a}}{\sin{b}}=\frac{\sin{c}}{\sin{d}},$求证$b-a>d-c$
证明:$\because2a<a+b<π,\therefore a<\frac\pi2$.$\because 0<c<a<\frac\pi2,\therefore\sin a>\sin c,\therefore \sin b>\sin d.\because 0<b<\pi,0<d<\frac\pi2,\therefore b>d.$
$\because b>a>c,b>d>c,\therefore |a-d|<b-c,\because |a-d|,b-c\in(0,\pi),\therefore \cos(a-d)>\cos(b-c)$.
$\because \sin{a}\sin{d}=\sin{b}\sin{c},\therefore \cos(a+d)-\cos(b+c)=\cos(a-d)-\cos(b-c)>0,\because a+d,b+c\in(0,\pi),\therefore a+d<b+c$. |
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