三角形ABC为锐角三角形,a,b,c为A,B,C对边,且(sinA)^2=sin(π/3+B)sin

三角形ABC为锐角三角形,a,b,c为A,B,C对边,且(sinA)^2=sin(π/3+B)sin

∵（sinA)ˆ2=sin(π/3+B)sin(π/3-B)+sin(B)ˆ2∴ (sinA)ˆ2=(sinπ/3cosB+cosπ/3sinB)(sinπ/3cosB-cosπ/3sinB)+sin(B)ˆ2∴ (sinA)ˆ2=[(3ˆ1/2)/2cosB+1/2sinB][3ˆ1/2)/2cosB-1/2sinB]+sin(B)ˆ2∴ (sinA)ˆ2=3/4(cosB)ˆ2-1/4(sinB)ˆ2+sin(B)ˆ2=3/4[(cosB)ˆ2+sin(B)ˆ2]=3/4∴ sinA=(3ˆ1/2)/2∵ A为锐角∴ A=π/3======以下答案可供参考======供参考答案1:忘记公式呢 有公式 那个公式 我记不到呢供参考答案2:忘公式了供参考答案3:sin(π/3+B) = sin(π/3)cos(B)+cos(π/3)sin(B) sin(π/3-B) = sin(π/3)cos(B)-cos(π/3)sin(B) sin(π/3+B) sin(π/3-B) = sin(π/3)cos(B)sin(π/3)cos(B) - cos(π/3)sin(B) cos(π/3)sin(B) = 3/4 * (cos(B))^2 - 1/4 *(sinB)^2(sinA)^2=sin(π/3+B)sin(π/3-B)+(sinB)^2= 3/4 (cos(B))^2 -1/4 (sinB)^2 +(sin(B))^2 = 3/4 [(sinB)^2 + (cosB)^2] = 3/4A=π/3供参考答案4:sinA^2=(sin60°cosB+cos60°sinB)(sin60°cosB-cos60°sin)+sinB^2 =(根号3/2cosB)^2-(1/2sinB)^2+sinB^2 =3/4(cosB^2+sinB^2) =3/4cosA=根号1-sinA^2=1/2所以A=60°

我好好复习下

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