sinx分之1减cosx等于什么
sinx = 2sin(x/2)cos(x/2) 1-cosx = 2[sin(x/2)]^2 两者除下来不就是 ctg(x/2)么绛旓細sinx = 2sin(x/2)cos(x/2) 1-cosx = 2[sin(x/2)]^2 涓よ呴櫎涓嬫潵涓嶅氨鏄 ctg(x/2)涔
绛旓細1.绛旀涓0.5 锛1-cosx锛(1+cosx)=sinx sinx/1-cosx=sinx(1+cosx)/sin*2=(1+cosx)/sinx=1/2 2.娌″杩囦斧~= =
绛旓細1-cosx绛変簬1/2(sinx/2)^2銆傚師寮=cos(2x)=(cosx)^2-(sinx)^2 =2(cosx)^2-1 =1-2(sinx)^2cos(2x)=1-2(sinx)^2 =1-2(sinx/2)^2 =1/2(sinx/2)^2 sin鍜宑os鐨勫叧绯绘槸锛歴in伪+cos伪=1锛泂inx=cos锛90-x锛夛紱tan伪=sin伪/cos伪锛泂in骞虫柟伪*cos骞虫柟伪=1銆俿in伪鏄寮︼紝cos...
绛旓細鎵浠sinx骞虫柟灏辩瓑浠蜂簬x^2 鑰1-cosx绛変环浜0.5x^2 浜庢槸寰楀埌 鍘熸瀬闄 =lim(x瓒嬩簬0) 0.5x^2 / x^2 =0.5 鏁呮瀬闄愬间负0.5
绛旓細鍋氫釜宸細=cosx/锛1-sinx锛-锛1+sinx锛/cosx 閫氬垎 涓や釜寮忓瓙宸负0鎵浠 cosx/1-sinx=1+sinx/cosx
绛旓細cosx锛sinx=鈭2/2cosx-鈭2/2sinx=鈭2(1/2cosx-1/2sinx)=鈭2(cos蟺/4cosx-sin蟺/4sinx)=鈭2cos(x+蟺/4)cosx涓巗inx鍙仛浣欏鸡鍑芥暟鍜屾寮﹀嚱鏁帮紝灞炰簬涓夎鍑芥暟锛屾槸鍩烘湰鍒濈瓑鍑芥暟涔嬩竴銆
绛旓細1/2[ln|(1+cosx)/(1-cosx)|] + C = 1/2[ln|2sin²(x/2)|] + C 鍥犱负 sin²(x/2) 鍙互杩涗竴姝ョ畝鍖栦负 1-cosx锛屾垜浠渶缁堝緱鍒扮Н鍒嗙殑缁撴灉锛1/2[ln|2(1-cosx)|] + C = ln|sqrt(2)sin(x/2)| + C 杩欏氨鏄眰瑙 鈭(sinx)/(1+cosx) dx 鐨勫畬鏁磋繃绋嬶紝閫氳繃浠ｆ崲鍜...
绛旓細x鈮0 f(x)=0 x=0 f(x)=(x-cosxsinx)/xsinx=(x-½sin2x)/xsinx lim(x鈫0)f(x)=lim(x鈫0)(1-cosx)/x x鈫0鏃 sinx锝瀤 =lim(x鈫0)sinx/1 娲涘繀杈炬硶鍒 =0 鍦▁=0澶勶紝宸﹀鏁=鍙冲鏁=鍑芥暟鍊硷紝鍑芥暟杩炵画銆
绛旓細cosx锛1-sinx鏄竴涓笁瑙掓柟绋 瑙ｆ硶濡備笅:渚涘弬鑰冿紝璇风瑧绾炽
绛旓細sinx-cosx=鏍瑰彿5/5 骞虫柟 sin^2x -2sinxcosx+cos^2x=1/5 1-2sinxcosx=1/5 2sinxcosx=1-1/5=4/5 sinxcosx=2/5
