已知3sin²a 2sin²b=2sina

sin^2a+sin^b=sina-sin(2a)/2=sina-sinacosa假设f(a)=sina-sinacosa求导得f'(a)=cosa-((cosa)^2-(sina)^2)=-2(co

要证3sinB=sin(2A+B)即证3sin(A+B-A)=sin(A+B+A)即证3sin(A+B)cosA-3cos(A+B)sinA=sin(A+B)cosA+cos(A+B)sinA即证2s

5sinB=sin(2A+B)=sin(A+B+A)=sin(A+B)cosA+cos(A+B)sinA,sinB=sin(A+B-A)=sin(A+B)cosA-cos(A+B)sinA5sin(A

由正弦定理a/sinA=b/cosB=c/sinC令a/sinA=b/cosB=c/sinC=1/k则sinA=aksinB=bksinC=cksin^2A=sin^C+sin^B+根号3sinCsi

sin(a+b)=0.5,sin(a-b)=1/3,两式相加得sinacosb=5/12sinbcosa=1/12tana/tanb=(sinacosb)/(sinbcosa)=5

90°

3(sinA)^2+2(sinB)^2=5sinA(sinA)^2+(sinB)^2=5sinA/2-(sinA)^2/25sinA/2-(sinA)^2/2=-(1/2)(sinA-5/2)^2+2

解题思路：第一问利用正弦定理求解，第二问先证明三角形是直角三角形，然后求出外接圆面积解题过程：

sin(A+B)=3/5,sin(A-B)=1/5则：sin(A+B)=3sin(A-B)sinAcosB+cosAsinB=3sinAcosB-3cosAsinB2sinAcosB=4cosAsin

先利用正弦定理将角都转为边再用余弦定理即b^2-a^2-c^2=√3acb^2=a^2+c^2+√3ac=a^2+c^2-2accosBcosB=-√3/2又B∈(0,π)B=120°

由题意得:COSθ=-3Sinθ,将其带入所求式中则(sinθ-2cosθ)/(3sinθ+5cosθ)=7sinθ/(-12sinθ)=-7/12很高兴为您解答

cosC/cosB=（3a-c）/b用余弦定理：【（a^2+b^2-c^2）/2ab】/【（a^2+c^2-b^2）/2ac】=（3a-c）/b化简后得：2ac=3a^2+3c^2-3b^2（a^2+

由sin(a+b)=sinacosb+cosasinbsin(a+b)=sinacosb+cosasinb可知sinacosb=[sin(a+b)+sin(a-b)]/2=5/24cosasinb=[

1sin(a+b)=2/3sina*cosb+cosa*sinb=2/3-----------(1)sin(a-b)=1/5sina*cosb-cosa*sinb=1/5-----------(2)联

解sin(π/2-b)*cos(a+b)-sin(π+b)*sin(a+b)=3/5即cosbcos(a+b)+sinbsin(a+b)=cos[b-(a+b)]=cos(-a)=cosa∴cosa=

3sinB=sin(2A+B)3sin(A+B-A)=sin(A+b+A)3sin(A+B)cosA-3cos(A+b)sinA=sin(A+B)cosA+sinAcos(a+b)2sin(A+b)c

证明：5sin[(A+B)-A]=sin[A+(A+B)]5sin(A+B)cosA-5cos(A+B)sinA=sin(A+B)cosA+cos(A+B)sinA4sin(A+B)cosA=6cos

sin(a+b)=sinacosb+cosasinb=1/2sin(a-b)=sinacosb-cosasinb=1/3所以sinacosb=(1/2+1/3)/2=5/12cosasinb=(1/2

cos[(a＋b)＋a]=3sin[(a＋b)－a]cos(a＋b)cosa－sin(a＋b)sina=3sin(a＋b)cosa－3cos(a＋b)sinacos(a＋b)[cosa＋3sina]=

sin(a-b)cosa-cos(b-a)sina=-[sin(b-a)cosa+cos(b-a)sina]=-sin(b-a+a)=-sinb=>sinb=-3/5=>cosb=-4/5sin(b+