问题补充：
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(π/3-B)+(sinB)^2,求A
答案：
∵（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] = 3/4
A=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°
