如图这道题是如何把cos变sin的 特别是后面为什么要加兀/2？
sin\u56fe\u50cf\u548ccos\u56fe\u50cf\u628a\u6a2a\u5750\u6807\u6269\u5927\u4e3a2\u500d\uff0c\u5219x\u7684\u7cfb\u6570\u4e581/2
sin(x/2+\u03c0/4)
\u5047\u8bbe\u662f\u5411\u5de6\u79fb\u03c0/3
\u5219x\u6362\u6210x+\u03c0/3
sin[(x+\u03c0/3)/2+\u03c0/4]=sin(x/2+5\u03c0/12)
\u5982\u679c\u662f\u5411\u53f3\u79fb\u5219x\u6362\u6210x-\u03c0/3
\u5148\u5de6\u79fb\u03c0/3
\u5219x\u6362\u6210x+\u03c0/3
=sin[(x+\u03c0/3)+\u03c0/4]=sin(x+7\u03c0/12)
\u5728\u6a2a\u5750\u6807\u6269\u5927\u4e3a2\u500d\uff0c\u5219x\u7684\u7cfb\u6570\u4e581/2
\u662fsin(x/2+7\u03c0/12)
\u89e3\uff1asin2x\u7684\u56fe\u50cf\u53d8\u6210sinx\u7684\u56fe\u50cf
sin2x\u7684\u5468\u671f\u4e3a2\u03c0/2=\u03c0
sinx\u7684\u5468\u671f\u4e3a2\u03c0
\u9664\u4e86\u5468\u671f\u53d8\u4e3a\u4e00\u534a\uff0c\u5176\u4ed6\u7684\u4e0d\u53d8
sin2x\u7684\u4e00\u4e2a\u5b8c\u6574\u7684\u5468\u671f\uff1a[0,\u03c0]
sinx\u7684\u4e00\u4e2a\u5b8c\u6574\u7684\u5468\u671f\uff1a[0,2\u03c0]
绛旓細X1=0.04cos(2蟺t+(1/2)蟺)(SI),X2=0.03cos(2蟺t+蟺)(SI),鍒欒鐗╀綋鐨勫悎鎸姩鏂圭▼涓 鐢ㄦ棆杞煝閲忔硶 浜岀煝閲忓瀭鐩 鍚堟尟骞呬负鏍瑰彿锛0.06*0.06+0.04*0.04锛夊垵鐩镐綅涓篴rctg锛3/2锛-蟺/4 鍚堟尟鍔ㄨ〃杈惧紡涓篨=鏍瑰彿锛0.06*0.06+0.04*0.04锛塩os锛2蟺t/T+arctg锛3/2锛- (蟺/4) 锛...
绛旓細鏍规嵁蹇冩硶锛屽鍙橈紝鎵浠in瑕鍙樻垚cos锛堝搴旂殑濡傛灉鏄痶an灏卞彉cot锛宑os灏鍙榮in锛宑ot灏卞彉tan锛夆滅鍙风湅璞￠檺鈥濈敱浜庢垜浠皢a褰撳仛閿愯锛堢1璞￠檺锛夛紝閭ｄ箞a+蟺/2灞炰簬绗2璞￠檺锛屽湪杩欎釜璞￠檺閲屻俿in鐨勫兼槸姝ｇ殑銆傛墍浠ユ垜浠渶鍚庣殑鍊煎氨鏄鐨刢os鐨勫笺傚啀鏉ヤ緥棰2锛歴in锛坅-蟺锛=-sina 棣栧厛瑙掑害鏄痑-2鍊嶇殑蟺/2鎵浠ユ槸...
绛旓細sinA+sinB=sinC锛坈osA+cosB锛夆埖A+B+C=180掳鈭磗inC=sin[蟺锛嶏箼A锛婤锕歖=sin锕橝锛婤锕氾紳sinAcosB锛媍osAsinBsinC锛坈osA+cosB锕氾紳锕檚inAcosB锛媍osAsinB锕氾箼cosA+cosB锕氾紳sinAcosAcosB锛cos??AsinB锛sinAcos??B锛媍osAsinBcosB锛濓箼sinAcosAcosB锛媍osAsinBcosB锕氾紜锕檆os??AsinB锛媠inAcos??B锕氾紳cosAco...
绛旓細涓寮у害绛変簬180搴﹂櫎浠ユ淳
绛旓細绗竴棰橈紝璇卞鍏紡锛岃浣忎竴鍙ヨ瘽锛屽鍙樺伓涓嶅彉锛岀鍙风湅璞￠檺锛屽厛鐪嬪垎鏁扮嚎涓婇潰鐨勶紝锛埾鎵撹捣鏉ュお楹荤儲锛屾崲鎴恜i锛 cos(a+pi锛=-cosa sin(a+3pi)=-sina 鎵浠ヤ笂闈㈠氨=-cosa *sin^2 a 涓嬮潰锛宑os(-pi-a)=cos(a+pi) tan(pi+a)*cos^3(a+pi)=sin(a+i)cos^2(a+pi)=-sinacos^2a...
绛旓細杩欓噷鏈変袱涓叕寮 cos(蟺-a)=-cosa sin(蟺+a)=-sina 棣栧厛鍋杩欎釜棰樼洰涔嬪墠浣犲緱鐭ラ亾cosa鍜宻ina閮芥槸鍛ㄦ湡涓2蟺鐨勫懆鏈熷嚱鏁 鍥犱负k涓哄伓鏁帮紝鍒 sin[(k+1)蟺+a]=sin(k蟺+蟺+a)=sin(蟺+a)=-sina cos[(k+1)蟺-a]=cos(k蟺+蟺-a)=cos(蟺-a)=-cosa ...
绛旓細(1)鐩存帴鎶妜鍊间唬鍏ュ嵆鍙倄=10澶勶細y1=0.25cos(15t-0.2*10)=0.25cos(15t-2)x=20澶勶細y2=0.25cos(15t-0.2*20)=0.25cos(15t-4)(2)t=4鏃讹細鍥犱负鏄疭I鍒讹紝鎵浠ユ墍鏈夎搴﹀潎浠ュ姬搴﹁〃绀恒倅1=0.25cos(15*4-2)=0.25cos58=0.0298 y1=0.25cos(15*4-4)=0.25cos56=0.2133 ...
绛旓細鍥炵瓟锛氱湅涓嶆竻銆併併併
绛旓細1 OA+OB+OC=0鎺ㄥ嚭o鏄噸蹇冿紝鐢讳釜鍥 灏辫兘鏄庣櫧 OA*OB=OB*OC 鎺ㄥ嚭OB鍨傜洿浜嶤A 缁ц屽彲鐭ヨ涓夎褰㈡槸绛夎叞涓夎褰 鐢讳釜鍥 灏辫 涔熷彲浠ョЩ椤逛笉杩囧お澶嶆潅 浣犲彲浠ュ弬鑰冧笅http://zhidao.baidu.com/question/94325490.html?si=6 2 sinAcosA+sinBcosB=sinCcosC sin2A+sin2B=2sinCcosC鍏抽敭鏄繖姝 澶櫄浜 ...
绛旓細tan伪^2-tan伪-2=0 tan伪=2鎴 tan伪=-1 鐢遍tan伪=2 sin伪=鏍瑰彿5,cos伪=浜斿垎涔嬩簩鏍瑰彿浜 sin(伪 -蟺鈭3锛=鏍瑰彿5*浜屽垎涔嬫牴鍙3-0.5*浜斿垎涔嬩簩鏍瑰彿浜 =鍗佸垎涔嬩簲鏍瑰彿鍗佷簲鍑忎簩鏍瑰彿浜,5,宸茬煡sin伪 涓洪攼瑙,涓攕in伪 ^2-sin伪 cos伪 -2cos伪 ^2=0.(1)姹倀an伪 鐨勫.锛2锛夋眰sin...
