高一数学问题
\u9ad8\u4e00\u6570\u5b66\u9898\u89e3\uff1ay=1-2sinx^2-sinx\uff0c\u56e0\u4e3ax\u5c5e\u4e8e(0,\u03c0/2\u3011,\u6240\u4ee5sinx\u5c5e\u4e8e\uff080\uff0c1\u3011\uff0c\u8bbet=sinx\uff0c y=-2t^2-t+1=-2(t+1/2)^2+3/2\uff0c\u629b\u7269\u7ebf\u5f00\u53e3\u5411\u4e0b\uff0c\u5bf9\u79f0\u8f74x=-1/2\uff0cy\u6700\u5927\u503c\u4e3af(0)=1,y\u6700\u5c0f\u4e3af(1)=-2\uff0c\u6240\u4ee5\u51fd\u6570\u7684\u503c\u57df\u4e3a\u3010-2\uff0c1\uff09
1\u3001\u5df2\u77e5\u76f4\u7ebfl\uff1akx-y-3k=0\uff1b\u5706M\uff1ax2+y2-8x-2y+9=0\uff0c\u6c42\u8bc1\uff1a\u76f4\u7ebfl\u4e0e\u5706M\u5fc5\u76f8\u4ea4\uff1b
\u8bc1\u660e\uff1a\u56edM\uff1a(x-4)²+(y-1)²=8\uff0c\u5706\u5fc3M(4\uff0c1\uff09\uff1b\u534a\u5f84R=2\u221a2
\u76f4\u7ebfL\uff1akx-y-3k=0\u8fc7\u5b9a\u70b9P(3\uff0c0\uff09
\u2502MP\u2502=\u221a[(4-3)²+(1-0)²]=\u221a2<R\uff0c\u6545\u4e0d\u8bbak\u4e3a\u4f55\u503c\uff0cL\u5fc5\u4e0e\u56edM\u76f8\u4ea4\u3002
3\u3001\u82e5\u76f4\u7ebfax\uff0bby\uff0d1=0\u4e0e\u5706x ²\uff0by ²=1\u76f8\u4ea4,\u5219\u70b9P(a,b)\u7684\u4f4d\u7f6e\u662f( )
A\u3001\u5728\u5706\u4e0a B\u3001\u5728\u5706\u5916
C\u3001\u5728\u5706\u5185 D\u3001\u4ee5\u4e0a\u7686\u6709\u53ef\u80fd
\u89e3\uff1a\u2235\u76f4\u7ebfax+by-1=0\u4e0e\u56edx²+y²=1\u76f8\u4ea4(\u6ce8\u610f\u662f\u76f8\u4ea4\uff0c\u4e0d\u662f\u76f8\u5207)\uff0c\u6545\u5706\u5fc3(0\uff0c0)\u5230\u8be5\u76f4\u7ebf\u7684\u8ddd\u79bbd=1/\u221a(a²+b²)1\uff1b\u800c\u2502OP\u2502=\u221a(a²+b²)\uff1b\u2234\u5fc5\u6709\u2502OP\u2502>1\uff0c\u5373\u70b9P\u5fc5\u5728\u56ed\u5916\uff0c\u6545
\u5e94\u9009B.
sin150°=sin30°=1/2
cos150°=-cos30°=-√3/2
∴tanα=sin150°/cos150°=-√3/3
2、
终边在直线y=-3x上
∴过点 (1,-3) 或 (-1,3)
∵√（1²+（-3）²）=√10
∴sinα=1/√10，cosα=-3/√10
或sinα=-1/√10，cosα=3/√10
tanα=P纵坐标/P横坐标=cos150°/sin150°=负三分之根号三 那符号懒得找。。。
2、同理了：直线Y=-3X上 所以tanα=-3
由那个sinα和tanα转化公式，你在树上找找，很久没用，不记得了，好像是cos^2(α)=1/【1+tan^2(α)】,就能知道sinα=十分之三倍根号十，cosα =负十分之根号十
或者sinα=负十分之三倍根号十，cosα =十分之根号十
找到公式了。。。刚才 SIN和COS 弄错了
绛旓細绗31棰 钂欐棩闂Monge's Problem 鐢讳竴涓渾,浣垮叾涓庝笁宸茬煡鍦嗘浜ゃ傜32棰 闃挎尝娲涘凹鏂浉鍒囬棶棰楾he Tangency Problem of Apollonius 鐢讳竴涓笌涓変釜宸茬煡鍦嗙浉鍒囩殑鍦嗐傜33棰 椹储鑻ュ凹鍦嗚闂Macheroni's Compass Problem 璇佹槑浠讳綍鍙敤鍦嗚鍜岀洿灏烘墍浣滅殑鍥惧潎鍙彧鐢ㄥ渾瑙勪綔鍑恒傜34棰 鏂潶绾崇洿灏洪棶棰楽teiner'...
绛旓細寰堢畝鍗曠殑涓閬楂樹竴鏁板鍗曡皟鎬ч 鏂规硶1锛 瑙ｏ細 y=k/x 璁撅細x1锛渪2 鏈夛細y(x2)-y(x1)=k(1/x2-1/x1) 鍗筹細y(x2)-y(x1)=k(x1-x2)/[(x1)(x2)] 1銆佸綋k锛0鏃讹紝鏈夛細y(x2)-y(x1)锛0 姝ゆ椂锛寉涓哄噺鍑藉紡锛 2銆佸綋k锛0鏃讹紝鏈夛細y(x2)-y(x1)锛0 姝ゆ椂锛寉涓...
绛旓細1.浣犺浣犵殑宸ヤ汉涓轰綘宸ヤ綔7澶,缁欏伐浜虹殑鎶ラ叕鏄竴鏍归噾鏉,閲戞潯骞冲垎鎴愮浉杩炵殑7娈,浣犲繀椤诲湪姣忓ぉ缁撴潫鏃剁粰浠栦滑涓娈甸噾鏉,濡傛灉鍙浣犱袱娆℃妸閲戞潯寮勬,浣犲浣曠粰宸ヤ汉浠樿垂?鎶婇噾鏉″垎鎴愪笁娈碉紙灏辨槸鍒嗕袱娆★紝鎴栬呭垏涓ゅ垁锛夛紝鍒嗗埆鏄暣鏍归噾鏉＄殑1/7銆2/7 4/7 绗竴澶╋細缁1/7鐨勶紝绗簩澶╋細缁2/7鐨勶紝鏀跺洖1/7鐨 绗笁澶...
绛旓細1.杩囩偣P锛4,2锛変綔鐩寸嚎L鍒嗗埆浜杞淬亂杞翠簬A锛孊涓ょ偣锛屽綋鈯緼OB闈㈢Н鏈灏忔椂锛屾眰鐩寸嚎L鐨勬柟绋嬨2.璁剧洿绾縇鐨勬柟绋嬩负锛坢锛2锛墄锛3y锛漨锛屾牴鎹笅鍒楁潯浠跺垎鍒眰m鐨勫笺傦紙1锛塋鍦▁杞翠笂鐨勬埅璺濅负锛2锛2锛夋枩鐜囦负锛1 3.宸茬煡鐩寸嚎L缁忚繃鐐癸紙锛2,2锛変笖涓庝袱鍧愭爣杞村洿鎴愬崟浣嶉潰绉殑涓夎褰紝姹傝鐩寸嚎鐨勬柟绋嬨4....
绛旓細鐗╀笉鐭ュ叾鏁闂鍑鸿嚜涓鍗冨叚鐧惧勾鍓嶆垜鍥藉彜浠鏁板鍚嶈憲銆婂瓩瀛愮畻缁忋嬨傚師棰樹负锛"浠婃湁鐗╀笉鐭ュ叾鏁帮紝涓変笁鏁颁箣浜岋紝浜斾簲鏁颁箣涓夛紝涓冧竷鏁颁箣浜岋紝闂墿鍑犱綍锛"杩欓亾棰樼殑鎰忔濇槸锛氭湁涓鎵圭墿鍝侊紝涓嶇煡閬撴湁鍑犱欢銆傚鏋滀笁浠朵笁浠跺湴鏁帮紝灏变細鍓╀笅涓や欢锛涘鏋滀簲浠朵簲浠跺湴鏁帮紝灏变細鍓╀笅涓変欢锛涘鏋滀竷浠朵竷浠跺湴鏁帮紝涔熶細鍓╀笅涓...
绛旓細1.f(x)鏄鍑芥暟涓攛鈭(0锛+鈭)涓婁负澧炲嚱鏁帮紝鍒檉(x)鍦(-鈭烇紝0)涔熸槸澧炲嚱鏁般備簬鏄痻路f(x)锛0绛変环浜庯細x>0 锛宖(x)锛0銆鎴杧锛0 锛宖(x)>0銆銆傚洜涓篺(-2)=0锛屾墍浠(2)=0锛屾墍浠>0 锛宖(x)锛0鐨勮В鏄0锛渪锛2锛泋锛0 锛宖(x)>0鐨勮В鏄紞2锛渪锛0銆傚洜姝や笉绛夊紡x路f(x)锛0...
绛旓細楂樹竴鏁板棰樺瀷鍙婅В棰樻妧宸у涓嬶細鏁板瑙ｉ鐨勬濈淮杩囩▼ 鏁板瑙ｉ鐨勬濈淮杩囩▼鏄寚浠庣悊瑙闂寮濮嬶紝缁忚繃鎺㈢储鎬濊矾锛岃浆鎹㈤棶棰樼洿鑷宠В鍐抽棶棰橈紝杩涜鍥為【鐨勫叏杩囩▼鐨勬濈淮娲诲姩銆傚浜庢暟瀛﹁В棰樻濈淮杩囩▼锛孏.娉㈠埄浜氭彁鍑轰簡鍥涗釜闃舵瑙侀檮褰曪紝鍗冲紕娓呴棶棰樸佹嫙瀹氳鍒掋佸疄鐜拌鍒掑拰鍥為【銆傝繖鍥涗釜闃舵鎬濈淮杩囩▼鐨勫疄璐紝鍙互鐢ㄤ笅鍒楀叓涓瓧鍔犱互...
绛旓細tan(arctanx1+arctanx2)=(x1+x2)/(1-x1x2)=sin(蟺/5)/[1-cos(4蟺/5)]=sin(蟺/5)/[1+cos(蟺/5)]=tan(蟺/10)鈭-蟺/2<arctanx1<蟺/2,-蟺/2<arctanx2<蟺/2 鈭-蟺<arctanx1+arctanx2<蟺 鈭碼rctanx1+arctanx2=蟺/10 鈶sin(伪/2)+cos(伪/2)]²=1+...
绛旓細瑕佽浣忎竴涓唴瀹癸紝鏈濂界殑鍔炴硶锛屽氨鏄妸鍐呭鎬荤粨涓洪傚悎浜庤嚜宸辫蹇嗗拰鎺屾彙鐨勭煭鍙ャ備緥濡傦紝鏈涓嶅鏄撴帉鎻＄殑鍏崷鐨勫啓娉曪細涔句笁鑱旓紝鍧ゅ叚鏂紱绂讳腑铏氾紝鍧庝腑婊★紱闇囦话鐩傦紝鑹纰楋紱鍏戜笂缂猴紝宸戒笅鏂備粎渚涘弬鑰冿紝浣犲彲浠ラ夋嫨浣犺嚜宸辩殑鏂瑰紡鏉ユ帉鎻°傚洜涓鏁板澶氫负閫昏緫鎬濈淮锛屽鍋氶鏈夋椂灏辫兘璁颁綇瀹氱悊銆佸叕寮忋佸畾涔夌瓑鍐呭銆
绛旓細x 鍖洪棿 [-1, 2] 涓婂崟璋冮掑锛屾墍浠ュ叾鍊煎煙涓猴細-a+2鈮(x)鈮2a+2 鍙堝洜涓哄湪 x 鍖洪棿 [-1, 2] 涓婏紝f(x₁)=g(x₂)锛屾墍浠 f(x) 鐨勫煎煙 [-1, 3] 搴旀槸 g(x) 鍊煎煙鐨勫瓙闆 鎵浠ユ湁锛-a+2鈮-1锛2a+2鈮3 姹傚緱锛歛鈮3 鍗冲疄鏁 a 鐨勫彇鍊艰寖鍥翠负锛 [3, +鈭)...
