Question

在锐角三角形ABC中,4sin^2[(B+C)/2]-cos2A=7/2,求角A的大小

Answer 1

4sin^2[(π-A)/2]-1+2sin^2(A)=7/2

4cos^2(A/2)+8sin^2(A/2)cos^2(A/2)=9/2

令cos^2(A/2)=x

4x+8(1-x)x=9/2

8x+16x-16x^2=9

16x^2-24x+9=0

(4x-3)^2=0

x=3/4

cos(A/2)=根号3/2

A/2=π/6

A=π/3