(1-cosx)/sinx 还能再怎么化简 sinx/1－cosx应该怎么继续分解？
sinx/(1-cosx)\u600e\u4e48\u5316\u7b80
\u7b54\uff1a\u4e8c\u500d\u89d2\u516c\u5f0f\uff1asin2a=2sinacosa\uff0ccos2a=2cos²a-1=1-2sin²asinx/(1-cosx)=2sin(x/2)cos(x/2)/[2sin²(x/2)]=cot(x/2)
sinx/1\uff0dcosx
\uff1dsinx\uff081\uff0bcosx\uff09/\uff081\u4e00cosx\uff09\uff081\uff0bcosx\uff09
\uff1dsinx\uff081\uff0bcosx\uff09/\uff081\u4e00cos²x\uff09
\uff1dsinx\uff081\uff0bcosx\uff09/sin²x
\uff1d\uff081\uff0bcosx\uff09/sinx
1-（1-2sinx的平方））/2sinxcosx=2sinx的平方/2sinxcosx=sinx/cosx=tanx
注：我这里的X是你原题上X的一半，也就是半角
(1-cosx)/sinx
=sinx(1-cosx)/sin²x
=sinx(1-cosx)/(1-cos²x)
=sinx(1-cosx)/(1+cosx)(1-cosx)
=sinx/(1+cosx)
=2sin(x/2)cos(x/2)/2cos²(x/2)
=sin(x/2)/cos(x/2)
=tan(x/2)
(1-cosx)/sinx =[1-(1-2·cosx/2^2)]/(2·sinx/2·cosx/2)=2·cosx/2^2/(2·sinx/2·cosx/2)=cosx/2/sinx/2=ctgx/2
tanx/2
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绛旓細澶归煎噯鍒欑殑璇,浣犺鍖栫畝涓嬪晩 1 - cosX = 2sin(X/2)^2 寰堟槑鏄2 sin(X/2)^2 > 0 鑰屼笖,鍦 [0,蟺/2] 涓,cos鏄皬浜1澶т簬0鐨,鏁, sin(X/2)^2 < sin(X/2)鎵浠 2sin(X/2)^2 < 2 sin(X/2)鍗 0 < 1 - cosX < 2 sin(X/2)2sin(X/2)瓒嬩簬0,鎵浠1 - cosX 瓒嬩簬0.....
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