在三角形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

sin^2A+sin^2B=sin^2C利用三角形正弦定理sinA/a=sinB/b=sinC/c显然a^2+b^2=c^2所以边c所对的角C为直角.

设三角形的顶点为A、B、C,对应的边长为a、b、c.过顶点B做AC边上的垂线,设垂线长度为h,则有h=asinC.SΔABC＝h*b/2=absinC/2正弦定理a/sinA=b/sinB可得b=as

sin²A+sin²B=2sin²C由正弦定理a^2+b^2=2c^2代入余弦定理：cosC=(a^2+b^2-c^2)/(2ab)=c^2/(2ab)>0所以：cosC

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

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

不妨设C为钝角sin²C

sinA/a=sinB/b=sinC/c=rsin^2A=sin^2B+sin^2C得出a^2*r^2=（b^2+c^2）*r^2即a^2=b^2+c^2所以ABC是直角三角形

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,sinA=2sinBsinC所以sin^2A-sinA=sin^2B+sin^2C-sinA=sin^2B+sin^2C-2sinBsinC即sinA(sin

sin(A/2)=cos((A+B)/2),得sin(A/2)=cos(90度-(C/2))=sin(C/2)就有A/2=C/2或A/2=180度-C/2,故A=C（A+C=360度舍去）,因此三角形

【1】sin方a+sin方b+sin方c=sin方a+sin方b+sin方(180-(a+b))=sin方a+sin方b+sin方(a+b)=sin方a+sin方b+(sina*cosb+cosa*s

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

2sinAcosB=sin(A+B)+sin(A-B)=sinC+sin(A-B)=sinC所以sin(A-B)=0所以A=B所以,△ABC是等腰三角形.完毕.