SIN2X怎样化成f（x）=Asin（wx+φ）
2cos^2(x π/6) √3sin2x=1 cos(2x π/3) √3sin2x=1 cos2xcosπ/3-sin2xsinπ/3 √3sin2x
=1 (1/2)cos2x (√3/2)sin2x
=sin(2x π/6) 1
绛旓細锛1锛夌敱鍑芥暟鐨勫浘璞″彲寰桝=2锛屽嚱鏁f锛坸锛=2sin2x锛屼护2k蟺+蟺2鈮2x鈮2k蟺+3蟺2锛宬鈭坺锛屽彲寰梜蟺+蟺4鈮2x鈮蟺+3蟺4锛宬鈭坺锛屾晠鍑芥暟鐨勫噺鍖洪棿涓篬k蟺+蟺4锛宬蟺+3蟺4]锛宬鈭坺锛屾晠鍑芥暟y=f锛坸锛夊湪鍖洪棿[蟺4锛3蟺4]涓婃槸鍑忓嚱鏁帮紝鍑芥暟鐨勬渶澶у间负2锛庯紙2锛夌敱锛1锛夊彲寰楀嚱鏁扮殑鍛ㄦ湡T=2蟺2...
绛旓細1 f(x)=a路b=1/2*sin2x-鈭3/2*cos2x=sin(2x-蟺/3)鈭-1锛渇(x)锛0鈭-1<sin(2x-蟺/3)<0 鈭祒鈭圼娲/6锛7娲/6]鈭 0<2x-蟺/3<2蟺 鈭聪<2x-蟺/3<2蟺 鈭2蟺/3<x<7蟺/6 2 鈭垫暟鍒楋經Sn+1锝濇槸鍏瘮涓2鐨勭瓑姣旀暟鍒椼傗埓Sn+1=(a1+1)*2^(n-1)=2^n 鈭碨n=2^n-1 ...
绛旓細绁濅綘寮蹇冿紒甯屾湜鑳藉府鍒颁綘锛屽鏋滀笉鎳傦紝璇疯拷闂紝绁濆涔犺繘姝ワ紒O(鈭鈭)O
绛旓細(1)f(x)=a*b =(鏍瑰彿3,-1)*(sin2x,cos2x)=鏍瑰彿3*sin2x-cos2x =2*sin(2x-鍏/6) (杈呭姪瑙)鍙坒(x)=0 鍒2*sin(2x-鍏/6)=0 sin(2x-鍏/6)=0 鍒2x-鍏/6=K*鍏 x=鍏/12+K/2*鍏 (K涓烘暣鏁)鍙0<x<鍏 鍒檟=鍏/12 OR =7鍏/12 (2)鐢眆(x)=2sin(2x-鍏/6)姹傚嚱鏁癴...
绛旓細鍙栨渶澶у兼椂sin2x+cos2x搴斾负锛堚垰2/4+鈭2/4锛塻in2x=sin蟺/4锛寈=蟺/8鍑芥暟鏈夊懆鏈2k蟺 x=蟺/8+2k蟺鍚岀悊鏈灏忓約in2x+cos2x搴斾负锛-鈭2/4-鈭2/4锛夐泦鍚堜负鏈澶x|x=蟺/8+2k蟺,x鈭圧}鏈灏弡x|x=蟺/8+2k蟺,x鈭圧} 闄愬瓧鐣ュ啓 ...
绛旓細鍚戦噺a=(鈭3sin2x,cos2x)锛屽悜閲廱=(sin2x,sin2x)锛屽嚱鏁f(x)=a*b+t 锛屾晠 f(x)=(鈭3sin2x,cos2x)*(sin2x,sin2x)+t =鈭3sin^2 2x+cos2xsin2x+t=[鈭3(1-cos4x)]/2+(sin4x)/2+t =(1/2)(sin4x-鈭3cos4x)+鈭3/2+t=sin(4x-蟺/3)+鈭3/2+t 褰-蟺/12鈮鈮は/6...
绛旓細=(sin2xsin胃+cos2xcos胃)/cos胃+1=cos(2x-胃)/cos胃+1 褰2x=胃锛屽嵆x=胃/2鏃跺緱maxf(x)=f(胃/2)=1/cos胃+1=3锛1/cos胃=2锛宑os胃=1/2锛屾晠胃=60掳锛屸埓a=sin胃/cos胃=tan胃=tan60掳=鈭3 f(x)=(鈭3)sin2x+2cos²x=2cos(2x-60掳)+1 鍗曞鍖洪棿锛氱敱-180掳+k...
绛旓細y^2+4(sin2x)^2-4y*sin2x=1-(sin2x)^2 5(sin2x)^2-4y*sin2x+y^2-1=0 涓婃柟绋嬫湭鐭ユ暟涓(sin2x)鐨勫垽鍒紡鈻斥墺0,鍗 (4y)^2-4*5*(y^2-1)鈮0 y^2鈮5 -鈭5鈮鈮も垰5 鍗筹細 a=2,f(x)鏈澶у=鈭5 2銆f(x)=2sin2x+cos2x f(x)=鈭5sin锛2x+arcsin鈭5/5锛夎浣縡(x)...
绛旓細f(x)=a.b=(2cosx)^2+鈭3sin2x =1+cos2x+鈭3sin2x =1+2[1/2*cos2x+(鈭3/2)*sin2x]=1+2sin(2x+30)浠(x)=1-鈭3锛屽嵆锛1+2sin(2x+30搴)=1-鈭3 鍗2sin(2x+30)=-鈭3 sin(2x+30)=-鈭3/2,鍦ㄦ墍杩拌寖鍥村唴锛屾湁鍞竴瑙ｏ細2x+30搴=-60搴 x=-45搴 鎴杧=-蟺/4(寮у害...
绛旓細f(x)浠i涓哄懆鏈燂紝鎵浠(pi+2)=f(2)銆傚張f(x)鏄竴涓鍑芥暟锛屾墍浠(2)=-f(-2)=-4銆傛墍浠(pi+2)=-4銆
