三角函数几道题,50分!答得好的100分!
1.tan(3X/2)-tan(X/2)=sin(3X/2)÷cos(3X/2)-sin(X/2)÷cos(X/2)={sin(3X/2)cos(X/2)-cos(3X/2)sin(X/2)}÷{cos(3X/2)cos(X/2)}=sin(3X/2-X/2)÷{cos(X/2+X)cos(X/2)}=sinX÷{cos²(X/2)cosX-sin(X/2)cos(X/2)sinX}=2sinX÷{2cos²(X/2)cosX-2sin(X/2)cos(X/2)sinX}=2sinX÷{2cos²(X/2)cosX-cosX+sin²X+cosX}=2sinX÷{cos²X-sin²X+cosX}=2sinX÷{cos2X+cosX}2.tan2A
tan(30-A)
+
tan2A
tan(60-A)+
tan(30-A)tan(60-A)=[tan2A
{tan(30-A)
+
tan(60-A)}]÷{1-tan(30-A)tan(60-A)}×{1-tan(30-A)tan(60-A)}+
tan(30-A)tan(60-A)=(tan2Acot2A)×{1-tan(30-A)tan(60-A)}+
tan(30-A)tan(60-A)=1-tan(30-A)tan(60-A)+
tan(30-A)tan(60-A)=1
3.sinB=Msin(2A+B)可得sin(A+B-A)=Msin(A+A+B)可得sin(A+B)cosA-cos(A+B)sinA=M{sin(A+B)cosA+cos(A+B)sinA}可得sin(A+B)cosA(1-M)=cos(A+B)sinA(1+M)可得tan(A+B)=(1+M)tanA÷(1-M)
5.sinXsin(60+X)=sinX{sin60cosX+cos60sinX}=(
√3/2)sinXcosX+(1/2)
sin²X=(
√3/4)sin2X-(1/4)(1-2
sin²X)+1/4=(
√3/4)sin2X-(1/4)cos2X
+1/4=(1/2){(
√3/2)
sin2X
-(1/2)cos2X}
=(1/2)sin(2X-30°)
所以当2X-30°=90°时,(1/2)sin(2X-30°)最大值为1/2,所以sinXsin(60+X)最大值为1/2.
绛旓細1.tan(3X/2)-tan(X/2)=sin(3X/2)梅cos(3X/2)-sin(X/2)梅cos(X/2)={sin(3X/2)cos(X/2)-cos(3X/2)sin(X/2)}梅{cos(3X/2)cos(X/2)}=sin(3X/2-X/2)梅{cos(X/2+X)cos(X/2)}=sinX梅{cos²(X/2)cosX-sin(X/2)cos(X/2)sinX}=2sinX梅{2cos²(X...
绛旓細1.鍑芥暟y锛漵in(2x+鈭/4)鐨勫浘鍍忓悜鍙冲钩绉烩垙/8涓崟浣嶉暱搴︼紝鍐嶆妸鍚勭偣鐨勭旱鍧愭爣鎵╁ぇ鍒板師鏉ョ殑2鍊嶏紝鎵寰楀浘鍍忕殑瑙f瀽寮忔槸( )绛旀锛歽=1/2sinx 2.2,宸茬煡鍑芥暟y=2sinwx(w>0)鐨勫浘鍍忎笌鐩寸嚎y+2=0鐨勭浉閭讳袱涓叕鍏辩偣涔嬮棿鐨勮窛绂讳负2鈭/3锛屽垯w 鐨勫间负锛 锛堿. 3 B.3/2 C.2/3 D.1/3 绛旀 A 锛...
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绛旓細b
绛旓細1鍕捐偂瀹氱悊a^2+b^2=c^2 鈶燽^2=c^2-a^2=(35鈭2)^2-35^2=35^2鎵浠=35.a=b杩欐槸绛夎叞涓夎褰.鈭燗=鈭燘=45搴 鈶^2=a^2+b^2=12+4=16 鎵浠=4.鍙坰inB=b/c=1/2.鎵浠モ垹B=30 鈭燗=60搴 鈶inA=a/c=2/3.鍙坈=6鎵浠=4 b^2=c^2-a^2=36-16=20鎵浠=2鈭5 鈶...
绛旓細1銆佸仛杈呭姪绾匡細寤堕暱CD锛岃繃B鍋欰C鐨勫钩琛岀嚎锛屽垯涔熷瀭鐩翠簬LD锛屽瀭瓒充负H銆傜敱CD鏄腑绾匡紝鎵浠モ柍ACD鍏ㄧ瓑浜庘柍BHD锛岃AD=x锛屽垯BD=x锛孉D=BD=3銆倀an胃=1/3锛屾墍浠鈭(x^2-9)]/6=1/3銆倄=鈭13锛孉B=2鈭13锛孉C=BH=2锛孊C=2鈭10 2銆
绛旓細瑙佸浘
绛旓細=1/(-(tanx-(1/2))^2+(1/4))>=4 f(x)鐨勬渶灏忓=4 2.Y=cos2x+k(cosx-1)=2(cosx)^2+kcosx-(k+1)=2(cosx+(k/4))^2-((k^2/16)+k+1)>=-((k^2/16)+k+1)鏈灏忓)-((k^2/16)+k+1)3.f锛坸锛=(1+cos2x+8sin²x)/sin2x =(2(cosx)^2+8(sinx)^...
绛旓細3梅cos30掳=2鈭3鈮3.464 (2)鍦ㄧ煩褰BCD涓,O鏄袱鏉″瑙掔嚎鐨勭劍鐐,AE鍨傜洿BD浜庣偣E,鑻E:OD=1:2,AE=鏍瑰彿3,鍒橠E鏄灏慍M 鈭礝E:OD=1:2 鈭碠E=ED 鈭碅D=AO=OD 鈭粹垹ADO=60掳 DE=AE•cot鈭燗DO=1 鍙傝冭祫鏂欙細濡傛灉鎮ㄧ殑鍥炵瓟鏄粠鍏朵粬鍦版柟寮曠敤锛岃琛ㄦ槑鍑哄 ...
绛旓細璁続B=x锛孊E=y.鍥犱负CDA=45掳锛屾墍浠D=AE=x+y 鍥犱负鈭燛CB=30掳锛屾墍浠E=y/tan30掳 鍥犱负鈭燛CA=60掳锛屾墍浠E/CE=tan60掳,鍗筹細锛坸+y锛/(y/tan30掳)=tan60掳 鎵浠=2y,鍗硑=x/2 鍙堝洜涓篊D=CE+ED=a锛屾墍浠/tan30掳 +x+y=a鍗硏/(2tan30掳)+x+x/2=a 鎵浠=2a/(3+鈭3)...