如何用向量的方法证明正弦和余弦定理? 如何用向量的方法证明正弦和余弦定理?
\u7528\u5411\u91cf\u7684\u65b9\u6cd5\u8bc1\u660e\u6b63\u5f26\u5b9a\u7406\u6b65\u9aa41
\u8bb0\u5411\u91cfi \uff0c\u4f7fi\u5782\u76f4\u4e8eAC\u4e8eC\uff0c\u25b3ABC\u4e09\u8fb9AB,BC,CA\u4e3a\u5411\u91cfa,b,c
\u2234a+b+c=0
\u5219i\uff08a+b+c\uff09
=i\u00b7a+i\u00b7b+i\u00b7c
=a\u00b7cos(180-(C-90))+b\u00b70+c\u00b7cos(90-A)
=-asinC+csinA=0
\u63a5\u7740\u5f97\u5230\u6b63\u5f26\u5b9a\u7406
\u5176\u4ed6
\u6b65\u9aa42.
\u5728\u9510\u89d2\u25b3ABC\u4e2d\uff0c\u8bbeBC=a,AC=b,AB=c\u3002\u4f5cCH\u22a5AB\u5782\u8db3\u4e3a\u70b9H
CH=a\u00b7sinB
CH=b\u00b7sinA
\u2234a\u00b7sinB=b\u00b7sinA
\u5f97\u5230a/sinA=b/sinB
\u540c\u7406\uff0c\u5728\u25b3ABC\u4e2d\uff0c
b/sinB=c/sinC
\u6b65\u9aa43.
\u8bc1\u660ea/sinA=b/sinB=c/sinC=2R\uff1a
\u4efb\u610f\u4e09\u89d2\u5f62ABC,\u4f5cABC\u7684\u5916\u63a5\u5706O.
\u4f5c\u76f4\u5f84BD\u4ea4\u2299O\u4e8eD. \u8fde\u63a5DA.
\u56e0\u4e3a\u76f4\u5f84\u6240\u5bf9\u7684\u5706\u5468\u89d2\u662f\u76f4\u89d2,\u6240\u4ee5\u2220DAB=90\u5ea6
\u56e0\u4e3a\u540c\u5f27\u6240\u5bf9\u7684\u5706\u5468\u89d2\u76f8\u7b49,\u6240\u4ee5\u2220D\u7b49\u4e8e\u2220C.
\u6240\u4ee5c/sinC\uff1dc/sinD=BD=2R
\u7c7b\u4f3c\u53ef\u8bc1\u5176\u4f59\u4e24\u4e2a\u7b49\u5f0f\u3002
\u25b3ABC\u4e3a\u9510\u89d2\u4e09\u89d2\u5f62\uff0c\u8fc7\u70b9A\u4f5c\u5355\u4f4d\u5411\u91cfj\u5782\u76f4\u4e8e\u5411\u91cfAC\uff0c\u5219j\u4e0e\u5411\u91cfAB\u7684\u5939\u89d2\u4e3a90\u00b0-A\uff0cj\u4e0e\u5411\u91cfCB\u7684\u5939\u89d2\u4e3a90\u00b0-C
\u2235AC+CB=AB
\u5728\u5411\u91cf\u7b49\u5f0f\u4e24\u8fb9\u540c\u4e58\u5411\u91cfj\uff0c\u5f97\uff1a
j\u00b7\uff08AC+CB\uff09=j\u00b7AB
\u2234\u2502j\u2502\u2502AC\u2502cos90\u00b0+\u2502j\u2502\u2502CB\u2502cos(90\u00b0-C)
=\u2502j\u2502\u2502AB\u2502cos(90\u00b0-A)
\u2234asinC=csinA \uff08AB\u7684\u6a21=c\uff0ccos\uff0890º-C\uff09=sinC\uff09\uff08CB\u7684\u6a21=a\uff0ccos\uff0890º-A\uff09=sinA
\u2234a/sinA=c/sinC
\u540c\u7406\uff0c\u8fc7\u70b9C\u4f5c\u4e0e\u5411\u91cfCB\u5782\u76f4\u7684\u5355\u4f4d\u5411\u91cfj\uff0c\u53ef\u5f97
c/sinC=b/sinB
\u6269\u5c55\u8d44\u6599\u5229\u7528\u4e09\u89d2\u5f62\u7684\u9762\u79ef\u8bc1\u660e\u4f59\u5f26\u5b9a\u7406
\u5df2\u77e5\u25b3ABC\uff0c\u8bbeBC=a\uff0cCA=b\uff0cAB=c\uff0c\u4f5cAD\u22a5BC\uff0c\u5782\u8db3\u4e3aD.
\u5219Rt\u25b3ADB\u4e2d\uff0csinB=AD/AB
\u2234AD=AB\u00b7sinB=csinB
\u2234S\u25b3ABC=1/2a\u00b7AD=1/2acsinB
\u540c\u7406\u53ef\u8bc1S\u25b3ABC=1/2absinC=1/2bcsinA
\u5373S\u25b3ABC=1/2acsinB=1/2absinC=1/2bcsinA
\u2234absinC=bcsinA=acsinB
\u5728\u7b49\u5f0f\u4e24\u7aef\u540c\u65f6\u9664\u4ee5ABC\uff0c\u53ef\u5f97sinC/c=sinB/b=sinA/a
△ABC为锐角三角形,过点A作单位向量j垂直于向量AC,则j与向量AB的夹角为90°-A,j与向量CB的夹角为90°-C
∵AC+CB=AB
在向量等式两边同乘向量j,得:
j·(AC+CB)=j·AB
∴│j││AC│cos90°+│j││CB│cos(90°-C)
=│j││AB│cos(90°-A)
∴asinC=csinA (AB的模=c,cos(90º-C)=sinC)(CB的模=a,cos(90º-A)=sinA
∴a/sinA=c/sinC
同理,过点C作与向量CB垂直的单位向量j,可得
c/sinC=b/sinB
扩展资料
利用三角形的面积证明余弦定理
已知△ABC,设BC=a,CA=b,AB=c,作AD⊥BC,垂足为D.
则Rt△ADB中,sinB=AD/AB
∴AD=AB·sinB=csinB
∴S△ABC=1/2a·AD=1/2acsinB
同理可证S△ABC=1/2absinC=1/2bcsinA
即S△ABC=1/2acsinB=1/2absinC=1/2bcsinA
∴absinC=bcsinA=acsinB
在等式两端同时除以ABC,可得sinC/c=sinB/b=sinA/a
绛旓細涓嬮潰a銆乥銆乧閮借〃绀哄悜閲,|a|銆亅b|銆亅c|琛ㄧず鍚戦噺鐨妯 鍥犱负a=b-c 鎵浠^2锛(b-c)^2 = b^2 +c^2 -2*bc 鎵浠a|^2锛潀b|^2 + |c|^2 -2*|b|*|c|*cosa 鍏跺畠浠ユ绫绘帹銆傚鍥1锛屸柍ABC涓洪攼瑙掍笁瑙掑舰锛岃繃鐐笰浣滃崟浣嶅悜閲廽鍨傜洿浜庡悜閲廇C锛屽垯j涓庡悜閲廇B鐨勫す瑙掍负90掳锛岮锛宩涓庡悜閲廋B...
绛旓細姝ラ1 璁鍚戦噺i 锛屼娇i鍨傜洿浜嶢C浜嶤锛屸柍ABC涓夎竟AB,BC,CA涓哄悜閲廰,b,c 鈭碼+b+c=0 鍒檌锛坅+b+c锛 =i路a+i路b+i路c =a路cos(180-(C-90))+b路0+c路cos(90-A) =-asinC+csinA=0 鎺ョ潃寰楀埌姝e鸡瀹氱悊 鍏朵粬 姝ラ2. 鍦ㄩ攼瑙掆柍ABC涓紝璁綛C=a,AC=b,AB=c銆
绛旓細cos(伪) = (b^2 + c^2 - a^2) / (2bc)涓嬮潰鎴戜滑鍙互浣跨敤鍚戦噺鐨鍚戦噺绉潵璇佹槑浣欏鸡瀹氱悊銆傛垜浠彲浠ュ皢涓夎褰㈢殑涓変釜杈瑰悜閲忚〃绀轰负鍚戦噺OA=a, OB=b鍜孫C=c锛屽叾涓璒涓轰换鎰忕偣銆傜幇鍦紝鎴戜滑鍙互浣跨敤鍚戦噺鐨勫悜閲忕Н瀹氫箟鏉ヨ绠楀す瑙払AC鐨姝e鸡鍊硷紝濡備笅鎵绀猴細|a 脳 b| = absin(伪)鍏朵腑锛宎 脳 b琛ㄧず鍚戦噺a...
绛旓細鍦ㄤ笁瑙掑舰ABC骞抽潰涓婂仛涓鍗曚綅鍚戦噺i,i鈯C,鍥犱负 BA+AC+CB=0鎭掓垚绔嬶紝涓よ竟涔樹互i寰 i*BA+i*AC=0鈶 鏍规嵁鍚戦噺鍐呯Н瀹氫箟锛宨*BA=c*cos(i,AB)=c*sinB,鍚岀悊 i*AC=bcos(i,AC)=b(-sinC)=-bsinC浠e叆鈶犲緱 csinB-bsinC=0 鎵浠/sinB=c/sinC 绫讳技鍦帮紝鍋氬彟澶栦袱杈圭殑鍗曚綅鍨傜洿鍚戦噺鍙璇a/sinA=b/...
绛旓細骞抽潰鍚戦噺鐨鏁伴噺绉湪姝d綑寮瀹氱悊鐨勬帹瀵煎拰璇佹槑杩囩▼涓彂鎸ヤ簡鍏抽敭浣滅敤銆傛浣欏鸡瀹氱悊鍦ㄥ畾鐞嗙殑鍙戠幇銆佸簲鐢ㄤ腑娌℃湁娑夊強骞抽潰鍚戦噺鐨勫奖瀛愩姝e鸡瀹氱悊 姒傝堪 a/sinA=b/sinB=c/sinC=2R 姝e鸡瀹氱悊(Sine theorem) 锛1锛夊凡鐭ヤ笁瑙掑舰鐨勪袱瑙掍笌涓杈癸紝瑙d笁瑙掑舰 锛2锛夊凡鐭ヤ笁瑙掑舰鐨勪袱杈瑰拰鍏朵腑涓杈规墍瀵圭殑瑙掞紝瑙d笁瑙掑舰 锛3锛杩愮敤...
绛旓細鍦ㄤ笁瑙掑舰ABC骞抽潰涓婂仛涓鍗曚綅鍚戦噺i,i鈯C,鍥犱负 BA+AC+CB=0鎭掓垚绔,涓よ竟涔樹互i寰 i*BA+i*AC=0鈶 鏍规嵁鍚戦噺鍐呯Н瀹氫箟,i*BA=c*cos(i,AB)=c*sinB,鍚岀悊 i*AC=bcos(i,AC)=b(-sinC)=-bsinC浠e叆鈶犲緱 csinB-bsinC=0 鎵浠/sinB=c/si...
绛旓細姝ラ1 璁鍚戦噺i 锛屼娇i鍨傜洿浜嶢C浜嶤锛屸柍ABC涓夎竟AB,BC,CA涓哄悜閲廰,b,c 鈭碼+b+c=0 鍒檌锛坅+b+c锛=i路a+i路b+i路c =a路cos(180-(C-90))+b路0+c路cos(90-A)=-asinC+csinA=0 鎺ョ潃寰楀埌姝e鸡瀹氱悊 鍏朵粬 姝ラ2.鍦ㄩ攼瑙掆柍ABC涓紝璁綛C=a,AC=b,AB=c銆備綔CH鈯B鍨傝冻涓虹偣H CH=a路...
绛旓細鎵璋撶殑娉曞悜閲忓嵆涓哄瀭鐩翠簬骞抽潰鐨勪竴涓悜閲忋傦紙鍗浠浠绘剰骞抽潰鍐呴兘瀛樺湪鏃犳暟鏉℃硶鍚戦噺銆傦級娉曞悜閲忎笌鍏堕暱搴︽棤鍏充絾鍏舵ā涓嶈兘涓0銆1銆佹枩绾夸笌骞抽潰鎵鎴愮殑瑙掞紝鍙敤鏂滅嚎鎵鍦ㄥ悜閲忎笌骞抽潰鐨勬硶鍚戦噺鐨澶硅鐨浣欏鸡鐨勭粷瀵瑰煎嵆涓虹洿绾夸笌骞抽潰鎵鎴愯鐨姝e鸡鍊笺2銆佷簩闈㈣姹傝В鍑轰袱涓钩闈㈢殑娉曞悜閲忓垯涓ゆ硶鍚戦噺鐨勫す瑙掍笌浜岄潰瑙掔殑骞抽潰瑙...
绛旓細姝ラ1 璁鍚戦噺i 锛屼娇i鍨傜洿浜嶢C浜嶤锛屸柍ABC涓夎竟AB,BC,CA涓哄悜閲廰,b,c 鈭碼+b+c=0 鍒檌锛坅+b+c锛=i路a+i路b+i路c =a路cos(180-(C-90))+b路0+c路cos(90-A)=-asinC+csinA=0 鎺ョ潃寰楀埌姝e鸡瀹氱悊 鍏朵粬 姝ラ2.鍦ㄩ攼瑙掆柍ABC涓紝璁綛C=a,AC=b,AB=c銆備綔CH鈯B鍨傝冻涓虹偣H CH=a路...
绛旓細鎴戜滑鐭ラ亾涓夎褰㈠唴瑙掑拰涓180搴︺傚亣璁句笁瑙掑舰ABC涓紝鈭燙涓洪挐瑙掓垨鐩磋銆傚熀浜庢锛屽彲浠ラ氳繃涓夎褰㈢殑瑙掑害涓庤竟闀跨殑鍏崇郴鏉ユ帹瀵浣欏鸡瀹氱悊銆傚埄鐢ㄤ笁瑙掑嚱鏁扮殑瀹氫箟鍜屾ц川锛屽姝e鸡銆佷綑寮︿笌杈归暱涔嬮棿鐨勫叧绯伙紝閫氳繃鏁板鎺ㄥ锛屽彲浠ュ緱鍒颁綑寮﹀畾鐞嗙殑琛ㄨ揪寮忋傝繖绉璇佹槑鏂规硶涓昏渚濊禆浜庝笁瑙掑嚱鏁扮殑鍩烘湰鎬ц川銆2. 鍒╃敤鍚戦噺璇佹槑鐨勬柟娉锛氬悜閲忎綔涓...