1加sinx减cosx分之1加sinx加cosx等于2分之一,求tanx等于多少 为什么sinx加cosx除于cosx减sinx等于1加tan...
sinx\u52a0cosx=\u8d1f\u4e8c\u5206\u4e4b\u4e00\u6c42tanx\u5229\u7528\u4e8c\u8005\u5e73\u65b9\u548c\u7b49\u4e8e\u4e00\uff0c\u7b49\u51fa\u7ed3\u679c\uff08-1+\u221a7\uff09/4\u548c\uff08-1-\u221a7\uff09/4
\u5206\u5b50\u5206\u6bcd\u540c\u9664\u4ee5cosx
1 - cosx＝2sin²(x/2)，1＋cosx＝2cos²(x/2)，
sinx＝2sin(x/2)cos(x/2)，
左边＝[2cos²(x/2)＋2sin(x/2)cos(x/2)] / [2sin²(x/2)＋2sin(x/2)cos(x/2)]
＝cos(x/2) / sin(x/2)
＝1 / tan(x/2)＝1/2，
所以 tan(x/2)＝2，
因此 tanx＝2tan(x/2) / [1 - tan²(x/2)]
＝ - 4/3。
绛旓細鏍瑰彿(1 + sin(x)cos(x)) / (1 - sin(x)cos(x)) + 鏍瑰彿(1 + sin(x)cos(x)) / (1 - sin(x)cos(x))棣栧厛锛屾垜浠彲浠ユ敞鎰忓埌涓や釜鍒嗗紡鐨勫垎姣嶇浉鍚岋紝鍥犳鍙互灏嗗畠浠悎骞朵负涓涓垎鏁帮細(鏍瑰彿(1 + sin(x)cos(x)) + 鏍瑰彿(1 + sin(x)cos(x))) / (1 - sin(x)cos(x))鐒跺悗锛...
绛旓細鍦-蟺/2鍒跋/2涔嬮棿鏄崟鎸戦掑鐨勶紝鑰屼粠蟺/2鍒3蟺/2鏄崟璋冮掑噺鐨勩傚墠闈㈠姞涓涓锛屼篃鏄竴鏍峰晩锛屾墍浠=1+sinx鐨勫崟璋冮掑鍖洪棿涓恒-蟺/2+2K蟺锛屜/2+2K蟺銆戯紝鍗曡皟閫掑噺鍖洪棿銆愊/2+2K蟺锛3蟺/2+2K蟺銆戙傚啀鐪嬭繖涓cosx,瀹冧粠0鍒跋鏄崟璋冮掑噺鐨勶紝鑰屼粠蟺鍒颁簩蟺鏄崟璋冮掑鐨勩傚墠闈㈠姞涓礋鍙风殑璇濓紝...
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~= =
绛旓細鈭1锛忥箼1+sinx锕歞x=锛峵an锕45掳锕锛2锕+C锛孋涓轰换鎰忓疄鏁般傗埆1锛忥箼1+sinx锕歞x =鈭1锛忥箼1+cos(90掳锛峹锕歞x锕欎护90掳锛峹=t锛屽垯dx=锛峝t锕=锛嶁埆1锛忥箼1锛媍ost锕歞t =锛峵an锕檛锛2锕氾紜C =锛峵an锕欙箼90掳锛峹锕氾紡2锕=锛峵an锕45掳锕锛2锕+C锛孋涓轰换鎰忓疄鏁般
绛旓細1+tanx鍒嗕箣涓鐨勭Н鍒嗚绠楀涓嬶細鈭 1/(1 + tanx) dx = 鈭 1/(1 + sinx/cosx) dx = 鈭 cosx/(cosx + sinx) dx = (1/2)鈭 [(cosx + sinx) + (cosx - sinx)]/(cosx + sinx) dx = (1/2)鈭 dx + (1/2)鈭 (cosx - sinx)/(cosx + sinx) dx = (1/2)鈭 dx + (1...
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绛旓細鏄 x->0+ 鍚э紵x->鈭 鏃舵瀬闄愬簲璇ヤ笉瀛樺湪銆傚綋 x->0+ 鏃讹紝鏋侀檺涓 1 銆
绛旓細lim<x鈫0>(1-sinx)/(1+sinx)=(1-0)/(1+0)=1
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