三角形内角ABC的对边为abc,且a-c/b-c=sinB/(sinA+sinC),(1)求A(2)若y=2sinB2+cos(π/3-2B),求函数值域 在△ABC中,内角A,B,C的对边分别为a,b,c,已知si...
\u4e09\u89d2\u5f62ABC\u4e2d\u5df2\u77e5\u89d2ABC\u5bf9\u8fb9\u5206\u522b\u4e3aabc\u4e14a-c\uff0fb-c=sinB\uff0fsinA+sinC\u3002\u3002\uff081\uff09\u6c42A\uff082...(a-c)/(b-c)=sinB/(sinA+sinC)
(a-c)(sinA+sinC)=(b-c)sinB
\u6839\u636e\u6b63\u5f26\u5b9a\u7406\uff0c
sinA=a/2R\uff0csinB=b/2R
\u56e0\u6b64(a-c)(a+c)=b(b-c)
\u5373a^2-c^2=b^2-bc
\u79fb\u9879\uff1abc=b^2+c^2-a^2
\u6545cosA=(b^2+c^2-a^2)/2bc=1/2
\u56e0\u6b64A=\u03c0/3
2.f(x)\uff1dcos\u5e73\u65b9\uff08x\uff0bA\uff09\uff0dsin\u5e73\u65b9\uff08x\uff0dA\uff09
=(Cos2(x+A)+1)/2-[1-cos2(x-A)]/2
=1/2[cos2(X+A)-cos2(x-A)]
=cos(2x+2A+2x-2A)/2cos(2x+2A-2x+2A)/2
=cos2xcos2A
=cos2xcos2\u03c0/3
=-1/2Cos2x.
\u9012\u589e\u533a\u95f4\u662f\uff1a0<=2x<=2k\u03c0+\u03c0
\u5373\uff1a\u30100\uff0cK\u03c0+\u03c0/2\u3011
sinB/(sinA+sinC)=1-sinC/(sinA\uff0bsinB)\uff0c
b/(a+c)=1-[c/(a+b)]
\u6574\u7406\u5f97
b^2+c^2-a^2=bc
cosA=(b^2+c^2-a^2)/2bc=bc/2bc=1/2
cosA=1/2
A=\u03c0/3
2)acosC\uff1d-1.
a^2[1-(sinC)^2]=1
(sinC)^2]=(a^2-1)/a^2
a/sinA=c/sinC
a^2/(sinA)^2=c^2/(sinC0^2
4a^2/3=c^2a^2/(a^2-1)
3\u4ee5^2=4a^2-4...................(1)
c^2=b^2+a^2-2abcosC=25+a^2-2*5*(acosC)=a^2+25+10
c^2=a^2+35......................(1)
\u89e3(1)(2)\u5f97
a^2=109
c^2=a^2+35=109+35=144
c=12
S=1/2bcsinA=1/2*12*5*\u221a3/2=15\u221a3
S=15\u221a3
根据正弦定理：sinB=b/(2R)，sinA=a/(2R)，sinC=c/(2R)，代入上式得：
(a-c)/(b-c)=b/(a+c)
b(b-c)=(a-c)(a+c)
b²-bc = a²-c²
b²+c²-a² = bc
余弦定理：cosA = (b²+c²-a²)/(2bc) = bc/(2bc) = 1/2
A=π/3
B+C+A=π
B+C=2π/3
0＜B＜2π/3
y = 2sin²B + cos(π/3-2B)
= （1-cos2B) + cosπ/3cos2B+sinπ/3sin2B
= 1-cos2B +1/2cos2B+√3/2sin2B
= 1 - cos2Bsinπ/6+sin2Bcosπ/6
= sin(2B-π/6) + 1
0＜B＜2π/3
0＜2B＜4π/3
-π/6＜2B-π/6＜7π/6
2B-π/6 = π/2时，有最大值 = 1+1 = 2
2B-π/6 趋近于 -π/6或7π/6时，y趋近于最小值=-1/2+1=1/2
故y值域（1/2，2】
绛旓細2B=A+C,A+B+C=180 B=60搴 a+c=鈭2b sinA+sinC=鈭2sinB sin(120-C)+sinC=鈭2*鈭3/2 鈭3/2cosC+1/2sinC+sinC=鈭2*鈭3/2 鈭3/2cosC+3/2sinC=鈭2*鈭3/2 1/2cosC+鈭3/2sinC=鈭2/2 cos(C-60)=鈭2/2,(A>C)C-60=45 C=105 ...
绛旓細鐢盿锛漛cosC锛嬧垰3csinB鍜屾寮﹀畾鐞嗗緱锛歴inA锛漵inBcosC锛嬧垰3sinCsinB.鏁咃細sin(B+C)=sinBcosC锛嬧垰3sinCsinB 鍗筹細sinBcosC+cosBsinC=sinBcosC锛嬧垰3sinCsinB 鎵浠osBsinC=鈭3sinCsinB 鍥犱负sinC鈮0锛屾墍浠osB=鈭3sinB 鎵浠anB=鈭3/3 鎵浠=30掳 ...
绛旓細瑙ｆ瀽锛氱敱姝ｅ鸡瀹氱悊鏈夛細b/sinB=c/sinC 宸茬煡c=鏍瑰彿3,b=1,B=蟺/6,閭ｄ箞锛歴inC=c*sinB/b=鏍瑰彿3*sin(蟺/6) 梅 1=鏍瑰彿3/2 瑙ｅ緱锛氣垹C=60掳鎴120掳 褰撯垹C=60掳鏃,鈭燗=180掳-60掳-30掳=90掳,姝ゆ椂S鈻ABC=(1/2)*b*c=鏍瑰彿3/2锛涘綋鈭燙=120掳鏃,鈭燗=180掳-120掳-30掳=30掳,姝ゆ椂S...
绛旓細锛1锛夋牴鎹綑寮﹀叕寮忥細c^2=a^2 b^2-2abcosA 鏁咃細a^2 b^2-2abcos(蟺/3)=2^2=4 a^2 b^2-ab=4鈥︹︼紙1锛夊張闈㈢НS=absinC/2=鈭3 absin(蟺/3)=2鈭3 ab=4鈥︹︼紙2锛夌敱锛1锛夊拰锛2锛夎В寰楋細a=2锛宐=2
绛旓細鈭粹柍ABC鐨鍛ㄩ暱=a(sinA+sinB+sinC)/sinA=鈭3(鈭3+鈭11)=3+鈭33.甯歌鐨涓夎褰鎸夎竟鍒嗘湁鏅氫笁瑙掑舰锛堜笁鏉¤竟閮戒笉鐩哥瓑锛夛紝绛夎叞涓夎锛堣叞涓庡簳涓嶇瓑鐨勭瓑鑵颁笁瑙掑舰銆佽叞涓庡簳鐩哥瓑鐨勭瓑鑵颁笁瑙掑舰鍗崇瓑杈逛笁瑙掑舰锛夛紱鎸夎鍒嗘湁鐩磋涓夎褰侀攼瑙掍笁瑙掑舰銆侀挐瑙掍笁瑙掑舰绛夛紝鍏朵腑閿愯涓夎褰㈠拰閽濊涓夎褰㈢粺绉版枩涓夎褰
绛旓細A=蟺/6 瑙ｆ瀽濡備笅锛歴inC =2鈭3sinB 鐢辨寮﹀畾鐞嗗彲鐭ワ細c=2鈭3b 浠ｅ叆锛歛^2-b^2=鈭3bc 鍗筹細a^2=7b^2 鐢变綑寮﹀畾鐞嗭細cosA=(b^2+c^2-a^2)/2bc =(b^2+12b^2-7b^2)/2b*2鈭3b = 鈭3/2 鎵浠=蟺/6
绛旓細鍏充簬涓夎褰abc鐨鍐呰abc鐨勫杈鍒嗗埆涓篴bc锛屽湪涓夎褰bc涓鍐呰abc鐨勫杈鍒嗗埆涓篴bc杩欎釜寰堝浜鸿繕涓嶇煡閬擄紝浠婂ぉ鏉ヤ负澶у瑙ｇ瓟浠ヤ笂鐨勯棶棰橈紝鐜板湪璁╂垜浠竴璧锋潵鐪嬬湅鍚э紒1銆(1)鍥犱负锛坈osA-2cosC锛壝穋osB=锛2c-a锛壝穊 鏍规嵁姝ｅ鸡瀹氱悊锛坈osA-2cosC锛壝穋osB=锛坰inA-2sinC锛壝穝inB鍥犱负cosB=-cos锛圓+C锛塻inB=sin...
绛旓細(1)2sinA=sinC=sin(A+B)2sinA=sinAcosB+sinBcosA=1/2*sinA+鈭3/2*cosA 3sinA=鈭3cosA,鈭磘anA=鈭3/3 鈭礎鏄笁瑙掑舰鍐呰,鈭碅=蟺/6 (2)BA鈫捖稡C鈫=accosB=3,ac=3/cosB=6 sinAsinC=a/2R*c/2R=ac/4R²=1/2,R=鈭3 鈭碽=2RsinB=3 ...
绛旓細宸茬煡涓夎褰ABC鐨勪笁涓鍐呰ABC鐨勫杈逛负abc锛屼笖瑙2B=A+C锛屼笁瑙掑舰ABC鐨勯潰绉负鏍瑰彿3锛岋紙1锛夎嫢cosA=3/5锛屾眰cosC鐨勫 B=60搴 cos(A+B)=cosAcosB-sinAsinB=0.3-0.4鈭3 cosC=-0.3+0.4鈭3 锛2锛夎嫢sinA=4sinC锛屾眰b 涓夎褰BC鐨勯潰绉=0.5acsinB=鈭3 ac=4 a/sinA=c/sinC a=1,c=4 b...
绛旓細a/sinA=b sinA=asinB/b=(1*鈭3/2)/2=鈭3/4 a<b,A<B cosA=鈭(1-sin²A)=鈭13/4 cosC=cos[蟺-(A+B)]=-cos(A+B)=-(cosAcosB-sinAsinB)=(3-鈭13)/8 cos2C=2cos²C-1 =(-5-3鈭13)/16
