sinA=0.6,sinB=0.07求角A,B大小

sina(sinb+cosb)-sinc=0sinasinb+sinacosb-sin(a+b)=0sinasinb+sinacosb-(sinacosb+cosasinb)=0sinasinb-co

[-1.5,0.5]

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

设（cosa)^2+(cosb)^2=k乘22（cosa)^2+2(cosb)^2=2k这是1式3(sina)^2+2(sinb)^2-2sina=0这是2式12式相加得(sina)^2-2sina+

cos(B-C)=cosBcosC+sinBsinc又sinB+sinc=-sinAcosB+cosC=-cosA所以同时平方sinB^2+sinc^2+2sinBsinc=sinA^2cosB^2+

证明:要证sin(a+b)sin(a-b)=(sina+sinb)(sina-sinb)只须证(sina*cosb+cosa*sinb)(sina*cosb-cosa*sinb)=(sina+sinb

A、B、C为三角形的三内角，且方程（sinB-sinA）x2+（sinA-sinC）x+（sinC-sinB）=0有等根，故有△=（sinA-sinC）2-4（sinB-sinA）（sinC-sinB

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

证:∵△ABC为锐角三角形,∴A+B>90°得A>90°-B∴sinA>sin(90°-B)=cosB,即sinA>cosB,同理可得sinB>cosC,sinC>cosA上面三式相加:sinA+si

解题思路：考查了正弦定理、余弦定理的应用，以及已知特殊角的三角函数值求角。解题过程：

不好意思,我好长时间没做这样的题了,不过不会的题可以问老师嘛,还有同学呀

cos(B-C)=cosBcosC+sinBsincsinB+sinc=-sinAcosB+cosC=-cosA所以同时平方sinB^2+sinc^2+2sinBsinc=sinA^2cosB^2+c

sinA/sinB=cosB/cosA即sinAcosA=sinBcosBsin2A=sin2B2A=2B2A=180-2B.

sinA+sinB=2[sin(A+B)/2]*[cos(A-B)/2]=2[sin60]*[cos(A-B)/2]=根号3*[cos(A-B)/2]当A=B=60时原式有最大值根号3

原式=sin²a+2sinasinb+sin²b+cos²a+2cosacosb+cos²b=(sin²a+cos²a)+2(cosacos

(1)原式=(sinA-sinC)2-4（sinB-sinA)(sinC-sinB)=sin2A-2sinAsinC+sin2C-4(sinBsinC-sinAsinC-sin2B+sinAsinB)

2(sin²a+sin²b)=-sin²a+2sina=-(sina-1)²+12sin²b=2sina-3sin²a因为0

利用2倍角公式化简等式sinAcosA=sinBcosB2sinAcosA=2sinBcosBsin2A=sin2B因为0

∵acosA+bcosB=ccosC∴sinAcosA+sinBcosB=sinCcosC∴sin2A+sin2B=sin2C=sin(2π-2A-2B)=-sin(2A+2B)∴0=sin2A+si

(1)(sinA-sinC)²-4（sinB-sinA)(sinC-sinB)=sin²A-2sinAsinC+sin²C-4(sinBsinC-sinAsinC-sin