tan(α-β）等于多少？公式，注意是减！！！ 如何由tan(α±β)公式推导到tanα±tanβ公式？？高...
tan(\u03b1-\u03b2)\u600e\u4e48\u7b97\u7684 \u8fd8\u6709tan(\u03b1+\u03b2)tan(\u03b1-\u03b2)=(tan\u03b1-tan\u03b2)/(1+tan\u03b1tan\u03b2)\u3002
tan(\u03b1+\u03b2)=(tan\u03b1+tan\u03b2)/(1-tan\u03b1tan\u03b2)\u3002
\u5206\u6790\u8fc7\u7a0b\u5982\u4e0b\uff1a
tan(\u03b1-\u03b2)\u53ef\u4ee5\u770b\u6210\uff1asin(\u03b1-\u03b2)/cos(\u03b1-\u03b2)
=(sin\u03b1cos\u03b2-cos\u03b1sin\u03b2)/(cos\u03b1cos\u03b2+sin\u03b1sin\u03b2)
\u4e0a\u4e0b\u9664\u4ee5cos\u03b1cos\u03b2
=(tan\u03b1-tan\u03b2)/(1+tan\u03b1tan\u03b2)
tan(\u03b1+\u03b2)\u540c\u7406
=(sin\u03b1cos\u03b2+cos\u03b1sin\u03b2)/(cos\u03b1cos\u03b2-sin\u03b1sin\u03b2)
=(tan\u03b1+tan\u03b2)/(1-tan\u03b1tan\u03b2)
\u6269\u5c55\u8d44\u6599\uff1a
\u79ef\u7684\u5173\u7cfb\uff1a
\uff081\uff09sin\u03b1 = tan\u03b1 \u00d7 cos\u03b1\uff08\u5373sin\u03b1 / cos\u03b1 = tan\u03b1\uff09\u3002
\uff082\uff09cos\u03b1 = cot\u03b1 \u00d7 sin\u03b1 \uff08\u5373cos\u03b1 / sin\u03b1 = cot\u03b1\uff09\u3002
\uff083\uff09tan\u03b1 = sin\u03b1 \u00d7 sec\u03b1 (\u5373 tan\u03b1 / sin\u03b1 = sec\u03b1)\u3002
\u5012\u6570\u5173\u7cfb\uff1a
\uff081\uff09tan\u03b1 \u00d7 cot\u03b1 = 1\u3002
\uff082\uff09sin\u03b1 \u00d7 csc\u03b1 = 1\u3002
\uff083\uff09cos\u03b1 \u00d7 sec\u03b1 = 1\u3002
\u5546\u7684\u5173\u7cfb\uff1a
\uff081\uff09sin\u03b1 / cos\u03b1 = tan\u03b1 = sec\u03b1 / csc\u03b1\u3002
\u548c\u89d2\u516c\u5f0f\uff1a
\uff081\uff09sin ( \u03b1 \u00b1 \u03b2 ) = sin\u03b1 \u00b7 cos\u03b2 \u00b1 cos\u03b1 \u00b7 sin\u03b2\u3002
\uff082\uff09sin ( \u03b1 + \u03b2 + \u03b3 ) = sin\u03b1 \u00b7 cos\u03b2 \u00b7 cos\u03b3 + cos\u03b1 \u00b7 sin\u03b2 \u00b7 cos\u03b3 + cos\u03b1 \u00b7 cos\u03b2 \u00b7 sin\u03b3 - sin\u03b1 \u00b7 sin\u03b2 \u00b7 sin\u03b3\u3002
\uff083\uff09cos ( \u03b1 \u00b1 \u03b2 ) = cos\u03b1 cos\u03b2 ∓ sin\u03b2 sin\u03b1\u3002
\u53c2\u8003\u8d44\u6599\uff1a\u767e\u5ea6\u767e\u79d1-\u6b63\u5f26
sin(\u03b1+\u03b2)=sin\u03b1cos\u03b2+cos\u03b1sin\u03b2
cos(\u03b1+\u03b2)=cos\u03b1cos\u03b2-sin\u03b1sin\u03b2
\u4e24\u5f0f\u76f8\u9664
tan(\u03b1+\u03b2)=sin(\u03b1+\u03b2)/cos(\u03b1+\u03b2)=(sin\u03b1cos\u03b2+cos\u03b1sin\u03b2)/(cos\u03b1cos\u03b2-sin\u03b1sin\u03b2)
\u5206\u5b50\u5206\u6bcd\u540c\u9664\u4ee5cos\u03b1cos\u03b2
=(tan\u03b1+tan\u03b2)/(1-tan\u03b1\u00b7tan\u03b2)
将tan(a-b)看成tan (a+(-b))就行了，这样比较容易记
绛旓細tan(伪-尾)=(tan伪-tan尾)/(1+tan伪tan尾锛夊綋tan(伪-尾)=0鏃讹紝伪-尾锛0锛屛憋紳尾锛屽洜姝an伪-tan尾涓嶄竴瀹绛変簬0锛屽洜涓1+tan伪tan尾鍦ㄥ垎姣嶄笂锛屽洜姝(tan伪-tan尾)/(1+tan伪tan尾锛変笉涓瀹氫负0锛庡綋tan伪-tan尾锛0鏃讹紝鍚岀悊鍙帹鍑簍an(伪-尾)=0 鎵浠ワ紝蹇呰涓嶅厖鍒 ...
绛旓細姝ｅ垏绛変簬瀵硅竟姣旈偦杈, 路涓夎鍑芥暟鎭掔瓑鍙樺舰鍏紡 路涓よ鍜屼笌宸殑涓夎鍑芥暟: cos(伪+尾)=cos伪路cos尾-sin伪路sin尾 cos(伪-尾)=cos伪路cos尾+sin伪路sin尾 sin(伪卤尾)=sin伪路cos尾卤cos伪路sin尾 tan(伪+尾)=(tan伪+tan尾)/(1-tan伪路tan尾) tan(伪-尾)=(tan伪-tan尾)/(1+tan伪路tan尾) ...
绛旓細浣犲ソ锛佷笁瑙掑嚱鏁鍏紡鎬荤粨濡備笅锛氫袱瑙掑拰鍏紡 sin(A+B)= sinAcosB+cosAsinB sin(A-B)= sinAcosB-cosAsinB cos(A+B)= cosAcosB-sinAsinB cos(A-B)= cosAcosB+sinAsinB tan(伪+尾)=(tan伪+tan尾)/(1-tan伪路tan尾)tan(伪-尾)=(tan伪-tan尾)/(1+tan伪路tan尾)鍗婅鍏紡 tan(A/2)=(1...
绛旓細姝ｅ垏绛変簬瀵硅竟姣旈偦杈, 路涓夎鍑芥暟鎭掔瓑鍙樺舰鍏紡 路涓よ鍜屼笌宸殑涓夎鍑芥暟: cos(伪+尾)=cos伪路cos尾-sin伪路sin尾 cos(伪-尾)=cos伪路cos尾+sin伪路sin尾 sin(伪卤尾)=sin伪路cos尾卤cos伪路sin尾 tan(伪+尾)=(tan伪+tan尾)/(1-tan伪路tan尾) tan(伪-尾)=(tan伪-tan尾)/(1+tan伪路tan尾) ...
绛旓細tan锛埼锛嵨憋級=(tan尾-tan伪)/(1+tan尾tan伪)tan(-伪锛尾)=tan(尾-伪)=(tan尾-tan伪)/(1+tan尾tan伪)
绛旓細璁ㄨ:1銆 伪銆佄插強 伪尾涓轰换鎰忓疄鏁般2銆佸綋伪鎴栁=0鏃讹紝鍒櫸蔽=1锛tan(伪尾)=1 3銆佸綋伪=90掳鏃讹紝尾=1掳鎴栁=1掳鏃讹紝尾=9o掳:鍒櫸蔽=90掳锛tan伪尾=鈭炪傚彲浠ユ帹骞垮埌浠绘剰瀹炴暟2n蟺+90掳銆4銆佸悓鐞單=180掳 尾=1掳鎴栁=1 尾=180掳鏃讹紝鍒櫸蔽=180掳锛宼an伪尾=0銆備篃鍙帹骞垮埌...
绛旓細1銆乼an锛埼憋紜尾锛夛紳锛tan伪锛媡an尾锛/锛1锛峵an伪路tan尾锛夈2銆tan锛埼憋紞尾锛锛濓紙tan伪鈥攖an尾锛/锛1+tan伪路tan尾锛夈傚湪Rt鈻矨BC锛堢洿瑙掍笁瑙掑舰锛変腑锛屸垹C=90掳锛孉B鏄垹C鐨勫杈筩锛孊C鏄垹A鐨勫杈筧锛孉C鏄垹B鐨勫杈筨锛屾鍒囧嚱鏁板氨鏄痶anB=b/a锛屽嵆tanB=AC/BC銆傛鍒囧嚱鏁板浘鍍忕殑鎬ц川锛氬畾涔夊煙锛...
绛旓細tan锛埼卤尾锛夌瓑浜锛坱an锛坅lpha锛+tan锛坆eta锛夛級/锛坱an锛坅lpha锛*tan锛坆eta锛+1锛 鎴栬咃紙tan锛坅lpha锛-tan锛坆eta锛夛級/锛-tan锛坅lpha锛*tan锛坆eta锛+1锛夈傛鍒囧嚱鏁版槸鐩磋涓夎褰腑涓涓鐨勫杈逛笌閭昏竟鐨勬瘮鍊笺倀an锛埼甭蔽诧級鐨鍏紡鏄痶an锛堬紙伪卤尾锛夛級=锛坱an锛埼憋級卤tan锛埼诧級锛/锛1卤tan锛...
绛旓細姝ｅ垏鍑芥暟鐨勫姞鍑忔硶鍏紡鍙互鐢ㄦ潵璁＄畻涓や釜姝ｅ垏鍑芥暟鐨勫拰鎴栧樊銆傚叿浣撶殑鍔犲噺娉曞叕寮忓涓嬶細1. 姝ｅ垏鍑芥暟鐨勫姞娉曞叕寮忥細tan(x + y) = (tan(x) + tan(y)) / (1 - tan(x) * tan(y))2. 姝ｅ垏鍑芥暟鐨勫噺娉曞叕寮忥細tan(x - y) = (tan(x) - tan(y)) / (1 + tan(x) * tan(y))鍏朵腑锛寈鍜寉涓...
绛旓細sin3伪=4sin伪路sin(蟺/3+伪)sin(蟺/3-伪) cos3伪=4cos伪路cos(蟺/3+伪)cos(蟺/3-伪) tan3a = tan a 路 tan(蟺/3+a)路 tan(蟺/3-a) 涓夊嶈鍏紡鎺ㄥ sin(3a) =sin(a+2a) =sin2acosa+cos2asina =2sina(1-sin²a)+(1-2sin²a)sina =3sina-4sin^3a cos3a =cos(2a+a)...
