已知sinA 根号2sinB=2sinc当角C最大时,求三角形的面积

sinA+sinB=√2sinCa/sinA=b/sinB=c/sinC有：(a+b+c)/(sinA+sinB+sinC)=(a+b)/(sinA+sinB)=c/sinC所以有：(√2+1-c)/

等于1啦设A,B,C三个角对应的边为La,Lb,LcLa=(SINA*Lc)/SINC(1)Lb=(SINB*Lc)/SINC(2)所以La+Lb+Lc=(1)+(2)+Lc=根号2+1其中SINA+

A-B=(cosa-cosb,sina-sinb)|A-B|^2=(cosa-cosb)^2+(sina-sinb)^2=2-2(cosacosb+sinasinb)=4/5cosacosb+sina

3sin^2a+2sin^b=2sina2sin^2a+2sin^b=2sina-sin^2a2*(sin^2a+sin^b)=-(sina-1)^2+1-1≤sina≤1-4≤-(sina-1)^2

令cosA+cosB=x---(1)sinA+sinB=√2/2---(2)由两式平方相加的2+2cos(A-B)=x^2+1/2得-1≤cos(A-B)=x^2/2-3/4≤1可得x^2≤7/2,从

sina+sinb=-1/2,cosa+cosb=根号3/2所以(sina+sinb)²=1/4(cosa+cosb)²=3/4故(sina+sinb)²+(cosa+c

|(a-b)|=|(cosa-cosb),(sina-sinb)|=2根号5/5所以有cosa^2-2cosacosb+cosb^2+sina^2-2sinasinb+sinb^2=4/5-2(cos

(sina-sinb)^2=(1-√3/2)^2①(cosa-cosb)^2=(-0.5)^2②①+②(sina)^2+(cosa)^2+(sinb)^2+(cosb)^2-2sinsinb-2cos

由正弦定理,a:b:c=√5：√35：2√5=1：√7：2,∴cosB=(1+4-7)/4=-1/2,∴B=120°,为所求.

sina+sinb=根号2……①cosa+cosb=(根号2)/3……②①^2+②^2,得(sina)^2+(sinb)^2+2sinasinb+(cosa)^2+(cosb)^2+2cosacosb

用韦达定理可求得sina+sinb和sinasinb的值(我觉得题目错了吧,方程的根应该是sina和cosa)sin^3a+cos^3a=(sina+cosa)(sin^2a+cos^2a-sinac

由sinA/a=sinB/b=sinC/c(其中a,b,c为角A,B,C对应的三条边)设sinA/a=sinB/b=sinC/c=k则a=sinA/k,b=sinB/k,c=sinC/k带入(sinB

sin^2a=2sin^2b,cos^2a=2/3cos^2b,两式相加1=2sin^2b+2/3cos^2b,1=2(1-cos2b)/2+2/3(1+cos2b)/2cos2b=1/22b=π/3

答：sina+sinb=√2/2两边平方得：sin²a+2sinasinb+sin²b=1/2…………（1）设cosa+cosb=m两边平方得：cos²a+2cosaco

正确答案：=lzsb不用谢,也不用鼓掌,只要闷声点个赞

由题意知a=2b，a2=b2+c2-2bccosA，2b2=b2+c2-2bccosA，又c2=b2+2bc，∴cosA=22，A=45°，sinB=12，B=30°，∴C=105°．故答案为：45°

亲,我写给你,你要给我好评哦～～～再问：恩再问：答案呢再答：等下再答：再问：亲，答案有正和负再答：那是平方，忘了～”再问：亲，那个怎么变形等于1我看不懂再答：用三角函数再问：sina的平方/4+9/4

∵a,B均为锐角∴0

由正弦定理a/sinA=b/sinB=c/sinCsinA+sinB=√2sinC所以a+b=√2ca+b+c=2√2+2所以√2c+c=2√2+2所以AB=c=2a+b=√2c=2√2S=1/2ab