我在职高，请问一题数学题 关于职高的数学题
\u4e00\u5230\u56db\u9898\u600e\u4e48\u505a\uff0c\u662f\u804c\u9ad8\u91cc\u7684\u6570\u5b66\u9898sin? \u03b8+cos?\u03b8=(sin 2 \u03b8+cos 2\u03b8) 2-2sin 2\u03b8cos 2\u03b8=1-2sin 2\u03b8cos 2\u03b8=5/9 2sin 2\u03b8cos 2\u03b8=4/9 (sin \u03b8cos\u03b8) 2=2/9 \u56e0\u4e3a\u03b8\u4e3a\u7b2c\u4e09\u8c61\u9650\u89d2,\u6240\u4ee5sin \u03b8cos\u03b8\uff1d\u221a 2/3
\u89e3\uff1aa²+b²+1/2-a-b=\uff08a-1/2)²+\uff08b-1/2)² \u56e0\u4e3aa\uff0cb\u2208R\uff0c\uff08a-1/2)²\u22650\uff0c\uff08b-1/2)²\u22650\uff0c
\u6240\u4ee5a²+b²+1/2-a-b\u22650 \u5373a²+b²+1/2\u2265a+b
过切点,且垂直于直线3x+4y-2=0的直线为y+1=(4/3)(x-2),即4x-3y-11=0
圆心(a,b)在此直线上,且到点P的距离d=r,即：
①4a-3b-11=0,②|3a-4b-2|/√(3²+4²)=r
将y=0代入圆方程,得：(x-a)²+b²=r²,x1=a+√(r²-b²),x2=a-√(r²-b²)
圆截x轴正半轴所得弦长8=|x1-x2|=2√(r²-b²),即r²-b²=16 ③
①②③联立解得：r=5,a=5,b=3
所以圆方程为(x-5)²+(y-3)²=25
自己看吧，下班了@@@
绛旓細鍦嗘埅x杞存鍗婅酱鎵寰楀鸡闀8=|x1-x2|=2鈭(r²-b²),鍗硆²-b²=16 鈶 鈶犫憽鈶㈣仈绔嬭В寰楋細r=5,a=5,b=3 鎵浠ュ渾鏂圭▼涓(x-5)²+(y-3)²=25 鑷繁鐪嬪惂锛屼笅鐝簡
绛旓細鎵浠²+b²+1/2-a-b鈮0 鍗砤²+b²+1/2鈮+b
绛旓細(sin 胃cos胃) ²=2/9 鍥犱负胃涓虹涓夎薄闄愯锛屾墍浠in 胃cos胃锛濃垰 2/3
绛旓細绛夊樊鏁板垪an=a1+(n-1)d a3=a1+2d=3 a6=a1+5d=-3 瑙ｅ嚭a1= 7 , d=-2 a15=a1+14d=7-14x2=-21
绛旓細1.杩欎袱绉嶅懡棰橈細鍘熷懡棰橈細鑻鍒橯锛涢嗗惁鍛介锛氳嫢銆孮 鍒欍孭锛涒滃師鍛介锛氳嫢P鍒橯鈥 鈥滈嗗惁鍛介锛氳嫢銆孮 鍒欍孭鈥濈湡锛1锛 鐪燂紙1锛夊亣锛0锛 鍋囷紙0锛夆滃鏋淧,鍒橯鈥濆拰鈥滃鏋溿孮锛屽垯銆孭鈥濅袱涓懡棰樻槸鍚岀湡鍚屽亣 锛屼緥濡傦細P: a=0 锛決锛歛b=0 锛涘垯鍘熷懡棰樷滆嫢a=0锛屽垯ab=0鈥濇槸鐪熷懡...
绛旓細姹傚悜閲忕殑棰樼洰锛灏辨槸鐢ㄦ湯鐐瑰潗鏍囧噺鍘诲鐐圭殑 鍧愭爣銆傝瘉鏄庡悜閲忓拰涓0锛屽彧瑕佹妸鍑犱釜鍚戦噺鐨 鍧愭爣鐩稿姞鍗冲彲銆傝В锛氬悜閲廇B=[4-锛-2锛锛1-锛-1锛塢=锛6锛2锛夊悜閲廈C=[锛-1锛-4锛3-1]=锛-5锛2锛夊悜閲廋A=[-2-锛-1锛夛紝-1-3]=锛-1锛-4锛夊悜閲廇B+鍚戦噺BC+鍚戦噺CA= 锛 6+锛-5锛+锛-1锛夛紝2+2+...
绛旓細濡傚浘
绛旓細闄嶅箓鍏紡锛歴in²a=(1-cos2a)/2锛宑os²a=(1+cos2a)/2 鍘熷紡=(1+cos30掳)/2+(1+cos150掳)/2+cos15掳sin15掳 =(2+鈭3)/4+(2-鈭3)/4+1/4 =5/4
绛旓細瑙ｏ細杞﹁締杞儙鐨勫懆闀縇=2蟺r=2蟺*0.33鈮2绫 涓鍒嗛挓鍐呰浆杩96鍦堬紝閭ｄ箞涓涓皬鏃惰浆杩囩殑鍦堟暟A=60*96=5760鍦 鈭存昏矾绋婼=A*L=11520绫
绛旓細鎵浠1x2+y1y1.鍐嶇畻浜屾鏂圭▼鐨勫垽鍒紡锛屼护鍏堕潪璐熴傚嵆鍙 2.鍚岀涓棰橈紝璁炬枩鐜囷紝浣嗗厛瑕佽璁烘枩鐜囦笉瀛樺湪鏃剁殑鎯呭喌锛屽嵆MN鍨傜洿X杞达紝鍦嗚兘杩囧師鐐广傝嫢鏂滅巼瀛樺湪锛屾柟娉曞悓绗竴棰橈紝璁剧洿绾挎柟绋嬶紝杩炵珛锛岀敤闊﹁揪瀹氱悊锛屽彧瑕佸悜閲廜M涓嶰N鍨傜洿锛屽渾灏辫兘杩囧渾鐐广傛墍浠ョ畻绠楀悜閲廜M涓嶰N鏄惁鍨傜洿鍗冲彲銆
