三角函数证明题？要过程
\u4e09\u89d2\u51fd\u6570\u8bc1\u660e\u9898\uff1f\u8981\u8fc7\u7a0b\uff01\u8bc1\u660e\u89c1\u56fe\uff1a
=1-sin²θ+sin²θsin²φ+cos²φsin²θ
=1+sin²θsin²φ-sin²θ(1-cos²φ)
=1+sin²θsin²φ-sin²θsin²φ
=1
绛旓細鎴戜滑鍙互鍒╃敤涓夎鍑芥暟鐨勫嶈鍏紡鍜屽崐瑙掑叕寮忔潵璇佹槑杩欎釜绛夊紡銆傞鍏堬紝鎴戜滑鏈変互涓嬩笁瑙掑嚱鏁板叕寮忥細tan(2胃) = 2 tan(胃)/(1 - tan²(胃)) 锛堝嶈鍏紡锛塼an(胃/2) = sin(胃) / (1 + cos(胃)) 锛堝崐瑙掑叕寮忥級灏單稿垎鍒瓑浜10掳鍜20掳浠ｅ叆鍗婅鍏紡锛屽緱鍒帮細tan 5掳 = sin 10掳 / (...
绛旓細=4sin2acos2acos4a/(cos5acos3a)(cos2acos4a)=4sin2a/(cos5acos3a)鍙宠竟=4(tan5a-tan3a)=4锛坰in5acos3a-cos5asin3a锛/(cos5acos3a)=4sin(5a-3a)/(cos5acos3a)=4sin2a/(cos5acos3a)=宸﹁竟 鈭达紙tan5a锛媡an3a锛/锛坈os2acos4a锛夛紳4锛坱an5a锛峵an3a锛夈傝璇佹槑涓夎鍑芥暟鐨勬亽绛夊紡鏃讹紝涔...
绛旓細=-tan(B+C)*tanB*tanC =tanA*tanB*tanC 蹇呰鏉′欢锛歵anA+tanB+tanC=tanA*tanB*tanC tanA+tanB+tanC=tanB+tanC-(tanB+tanC)/(1-tanB*tanC)tanA+tanB+tanC=tanB+tanC-tan(B+C)tanA=-tan(B+C)A+B+C=n蟺 鍗冲緱A+B+C=n蟺锛坣鈭圸锛夌殑鍏呰鏉′欢鏄痶anA+tanB+tanC=tanA*tanB*tanC ...
绛旓細17锛1锛璇佹槑锛氱敱浣欏鸡瀹氱悊鍙緱锛2ccosA-2acosC=2c[b^+c^2-a^2)/2bc]-2a[a^2+b^2-c^2)/2ab]=(b^2+c^2-a^2)/b-(a^2+b^2-c^2)/b =(2c^2-2a^2)/b,鍥犱负 2c^2-2a^2=b^2,鎵浠 2ccosA-2acosC=b^2/b =b銆傦紙2锛夎В锛氬洜涓 tanA=1/3锛屾墍浠 sinA/co...
绛旓細=1/2*cos2b 鍙坈os2a=1/2*cos2b 鍗虫湁:宸﹁竟=鍙宠竟 [1-(tana/2)^2]/[1+(tana/2)^2]=[1-(tanb)^2]/2[1+(tanb)^2]涓囪兘鍏紡璇佹槑:cosa=[(cosa/2)^2-(sina/2)^2]/[(cosa/2)^2+(sina/2)^2] ,(鍒嗗瓙,鍒嗘瘝鍚岄櫎(cosa/2)^2寰 =[1-(tana/2)^2]/[1+(tana/2)^2]...
绛旓細璇佹槑锛氾紙1锛変綔鈻矨BC鐨勯珮BD 鍦ㄢ柍ABD涓紝鍥犱负sinA=BD/AB 鎵浠D=ABsinA=csinA 鎵浠モ柍ABC闈㈢Н=1/2AC.BD=1/2.b.csinA=1/2bcsinA 鍚岀悊鍙璇佹槑寰楋細涓夎褰㈤潰绉疉BC=1/2bcsinA=1/2acsinB=1/2absinA 锛2锛夋牴鎹紙1锛夛細鐢1/2bcsinA=1/2acsinB鍙樺舰鍙緱锛歛/sinA=b/sinB 鍚岀悊鍙緱锛歜/sinB=...
绛旓細璇佹槑娆茶瘉tanxsinx/(tanx-sinx)=(tanx+sinx)/tanxsinx 鍗抽渶璇乼an^2xsin^2x=(tanx-sinx)(tanx+sinx)鍗抽渶璇乼an^2xsin^2x=tan^2x-sin^2x 鍗抽渶璇乼an^2x-tan^2xsin^2x=sin^2x 鍗抽渶璇乼an^2(1-sin^2x)=sin^2x 鍗抽渶璇乼an^2xcos^2x=sin^2x 鍗抽渶璇乻in^2x/cos^2脳cos^2x=sin^2x 鍗抽渶...
绛旓細杩欑鏂规硶鏈绠鍗曪紝姣旇緝瀹规槗鐞嗚В锛屾垜鐚滀綘鎯抽棶鐨勬槸濡備綍鎶婄瓑寮忓乏渚х殑寮忓瓙鍖栫畝鎴愬彸闈㈢殑寮忓瓙锛岃繖涓氨闇瑕佺粡楠屽拰鎶宸э紝浠ユ垜涓鐨勭粡楠岋紝鏄渶瑕佸仛涓瀹氶噺鐨棰樼洰骞舵荤粨涓浜涜寰嬬殑锛屾瘮濡傝鎶1鎹㈡垚cos骞虫柟+sin骞虫柟锛屾垨鑰呭垎瀛愬垎姣嶅悓涔樻煇涓紡瀛愮瓑锛岃繖涓杩囩▼姣旇緝闅撅紝甯屾湜浣犺兘澶熺悊瑙ｃ傝繖閲屾垜鍙槸鎻愪緵鑷繁鐨勫缓璁紝骞堕潪鏈浣...
绛旓細璇佹槑锛歴in(2伪+尾)/sin伪-2cos(伪+尾)= [sinacos(伪+尾)+cosasin(伪+尾)]/sina-2cos(伪+尾)= (cosa/sina)(sinacosb+cosasinb)-cosacosb+sinasinb= cosacosb+(cosa^2/sina)sinb-cosacosb+sinasinb= (1-sina^2)sinb/sina+sinasinb= sinb/sina-sina/sinb+sinasinb= sinb/sina 鍗宠瘉...
绛旓細璁锯柍ABC涓洪攼瑙掆柍锛璇佹槑鈭2<sinA+sinB+sinC鈮(3/2)鈭3 璇佹槑锛氱敱浜巗inA+sinB+sinC+sin60⁰=2sin[(A+B)/2]cos[(A-B)/2]+2sin[(C+60⁰)/2]cos[(C-60⁰)/2]鈭0<cos[(A-B)/2]鈮1锛屻愬綋A=B鏃禼os[(A-B)/2]=cos0=1锛涘綋A鈮燘鏃禼os[(A-B)/2]>0銆0...
