三角形ABC是锐角三角形,sin²A=sin(π/3+B)sin(π/3-B)+sin²B,求角A_数学解答

三角形ABC是锐角三角形,sin²A=sin(π/3+B)sin(π/3-B)+sin²B,求角A

人气：131 ℃　时间：2019-10-23 04:44:47

解答

sin(π/3+B)sin(π/3-B)
=(√3/2*cosB+1/2sinB)(√3/2*cosB-1/2sinB)
=3/4*(cosB)^2-1/4(sinB)^2
sin²A=sin(π/3+B)sin(π/3-B)+sin²B
=3/4*(cosB)^2-1/4(sinB)^2+(sinB)^2
=3/4[(cosB)^2+(sinB)^2]
=3/4 A