典型例题分析1: 在△ABC中,角A,B,C所对的边分别为a,b,c,若sin(A+B)=1/3,a=3,c=4,则sinA=( ) A.2/3 B.1/4 C.3/4 D.1/6 解:∵A+B+C=π, ∴sin(A+B)=sinC=1/3, 又∵a=3,c=4, ∴a/sinA=c/sinC, 即3/sinA=4/(1/3), ∴sinA=1/4, 故选B. 考点分析: 正弦定理. 题干分析: 由内角和定理及诱导公式知sin(A+B)=sinC=1/3,再利用正弦定理求解. ![](http://image109.360doc.com/DownloadImg/2020/09/2117/202774086_1_20200921051328743_wm) 函数f(x)=Asin(ωx+φ)({A>0,ω>0,|φ|<π/2)在某一周期内图象最低点与最高点的坐标分别为(7π/3,-√3)和(13π/3,√3)(Ⅱ)设△ABC的三内角A,B,C的对边分别为a,b,c,且f(A)=√3,a=3,求△ABC周长的取值范围.解:(1)由题意得:A=√3,T/2=13π/3﹣7π/3=2π,T=4π,ω=1/2,∴函数表达式:f(x)=√3sin(x/2+π/3),(Ⅱ)∵f(A)=√3sin(A/2+π/3)=√3,sin(A/2+π/3)=1,A∈(0,π),A/2+π/3∈(π/3,5π/6),可得:A/2+π/3=π/2,解:A=π/3,由正弦定理得:a/sinA=b/sinB=c/sinC=3/(√3/2)=2√3,三角形的周长L=a+b+c=b+c+√3=2√3sinB=2√3sinC+3,=2√3[sinB+sin(2π/3﹣B)]+3,=2√3(sinB+√3/2·cosB+sinB/2)+3,(Ⅰ)由正弦函数的性质,求得A及T的值,ω=2π/T,求得ω,将(13π/3,√3)代入求得φ的值,即可求得函数表达式;(Ⅱ)根据正弦定理求得角A值,b、c关系,L=a+b+c=6sin(B+π/6)+3,根据正弦函数最值,求得L的取值范围.![](http://image109.360doc.com/DownloadImg/2020/09/2117/202774086_2_20200921051328805_wm) 在△ABC中,内角A,B,C所对的边分别为a,b,c.已知a=√3,且b2+c2=3+bc.![](http://image109.360doc.com/DownloadImg/2020/09/2117/202774086_3_20200921051328868_wm) (I)由余弦定理可得:cosA=(b2+c2-a2)/2bc=(3+bc-3)/2bc=1/2,即可得出.(II)由正弦定理可得:可得b=asinB/sinA,可得bsinC=2sinBsin(2π/3-B)=sin(2B-π/6)+1/2,根据B∈(0,2π/3)即可得出.▽▽▽▽▽▽▽▽▽▽▽▽▽▽▽▽▽▽▽▽▽▽▽▽
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