在△ABC中,若sin(2派-A)

将第一个式子化成：sinA=√2sinB,第二个式子化为：√3cosA=√2cosB,将一二两式相乘得：√(3/2)sin(2A)=sin(2B),再因为A+B

(1)因为是锐角三角形所以C派/2-B根据正弦函数的单调性可知sinA>sin(派/2-B)=cosB(2)f(x)=sin(2x+2分之3派)=-cos(2x)故f(x)为偶函数（3）选A

sin(2∏-A)=-√2sin(∏-B)-sinA=-√2sinB①√3cosA=-√2cos(∏-B)√3cosA=√2cosB②①²＋②²,得2cos²A=1cos

解析：1）∵(cosA+cosB)sinC=sinA+sinB∴cosA+cosB=(sinA+sinB)/sinC即（b^2+c^2-a^2)/2bc+（a^2+c^2-b^2)/2ac=(a+b)

sin²A=sin²B+sin²C,a/sinA=b/sinB=c/sinC=2R(a/2R)^2=(b/2R)^2+(c/2R)^2a^2=b^2+c^2,ABC是直角

sin²a+cos²a=1cos2a=2cos²a-1sin(B+C)/2=sin(π-A)/2=cos(A/2),

f(x)=√3*cosx+sinx+√3=2sin(x+π/3)+√3f(17π/12)=√3-√22sin(C+π/3)+√3=√3+1C=π/2a/sinA=b/sinB=c/sinC=csinA

由和差化积公式：sinA=sin(B+C)=sinBcosC+cosBsinC=2sinBcosC,所以cosBsinC-sinBcosC=0,即sin(B-C)=0.从而B=C,因此三角形ABC是等

∵在△ABC中,sin(A+B)=sinC∴sinC·sin(A-B)=sin²Csin(A-B)=sinC又∵sinC=sin(A+B)∴sin(A-B)=sin(A+B)sinAcosB

由题可得,A=3分之派

锐角三角形,高中数学题做过.

由正弦定理和已知可以得到:a^2=b^2+c^2.所以三角形为直角三角形.

cosA=（-4/5）所以A为钝角.则sinA=3/5；由正弦定理得：sinB=AC*sinA/BC；则sinB=2/5；cos=√21/5sin(2B+π/6)=sin2Bcos(π/6)+cos(

sin^2A+sin^2B=sin^2C=sin^2(A+B)=(sinAcosB+sinBcosA)^2=sin^2Acos^2B+sin^2Bcos^2A+2sinAcosAsinBcosB左边减

a²≤b²+c²-bcbc≤b²+c²-a²1/2≤(b²+c²-a²)/2bccosa≥1/2a≤60°

根据正弦定理：a/sinA=b/sinB=c/sinC=2R,R为该三角形外接圆半径,则：a/2R=sinAb/2R=sinBc/2R=sinC因此：sinA:sinB:sinC=a:b:c=3:2:

sin^2A+sin^2B+sin^2C=(1-cosA)/2+(1-cosB)/2+(1-cos^2C)=2-cos(A+B)cos(A-B)-cos^2C=2+cosCsoc(A-B)-cos^2

根据正弦定理,原式可化为sin^2Bsin^2C+sin^2Csin^2B=2sinBsinCcosBcosC2sin^2Csin^2B=2sinBsinCcosBcosCsinBsinC=cosBc

/c=sinB/sinC&bsinB=csinC=>sinB/sinC=c/b=>b/c=c/b=>b^2=c^2i.e.b=c=>B=C=>A=180度-2B=>sinA=sin(2B)=>sin^

改了结果相同由正弦定理a/sinA=b/sinB=c/sinC(sinA)^2=(sinB)^2+(sinC)^2等价于a^2=b^2+c^2可知△ABC直角三角形A=π/2sinA=2sinBcos