Question

高一数学，急急急~

已知△ABC的内角为A、B、C , 求证:sin^2（A/2）+sin^2（B/2）+sin^2（C/2）≥3/4

Answer 1

sin^2（A/2）+sin^2（B/2）+sin^2（C/2）
=(1/2)[1-cosA+1-cosB+1-cosC]
=3/2-(1/2)[2cos(A+B)/2cos(A-B)/2-cos(A+B)]
=3/2-(1/2)[2cos(A+B)/2cos(A-B)/2-2cos^2(A+B)/2+1]
=1-(1/2)*2cos(A+B)/2[cos(A-B)/2-cos(A+B)/2]
=1-2sin(A/2)sin(B/2)sin(C/2)
因sin^2（A/2）+sin^2（B/2）+sin^2（C/2）≥3[sin(A/2)sin(B/2)sin(C/2)]^(2/3)
设sin(A/2)sin(B/2)sin(C/2)=x
则1-2x≥3x^(2/3)
(1-2x)^3≥27x^2
(1+x)^2(1-8x)≥0
∵(1+x)^2≥0
∴1-8x≥0 x≤1/8
1-2sin(A/2)sin(B/2)sin(C/2)≤1-2*(1/8)=3/4
得证

Answer 2

利用和差化积公式一顿推导

Answer 3

用倍角公式 化成单角的就可以了