Question

高中三角函数问题

在三角形ABC中 求证
(sin(A/2))^2+(sin(B/2))^2+(sin(c/2))^2=1-2sin(A/2)*sin(B/2)*sin(C/2)
详细过程

Answer 1

sin(A/2)^2+sin(B/2)^2+（sinC/2)^2
=[sin(A/2)]^2+(1-cosB)/2+(1-cosC)/2
=1+[sin(A/2)]^2-(cosB+cosC)/2
=1+{cos[(B+C)/2]}^2-cos[(B+C)/2]cos[(B-C)/2] 和差化积
=1+cos[(B+C)/2]{cos[(B+C)/2]-cos[(B-C)/2]} 和差化积
=1+sin(A/2)*(-2)sin(B/2)sin(C/2)
=1-2sin(A/2)sin(B/2)sin(C/2)

Answer 2

这个貌似是常用的 要记住的
sin(A/2)^2+sin(B/2)^2+（sinC/2)^2
=[sin(A/2)]^2+(1-cosB)/2+(1-cosC)/2
=1+[sin(A/2)]^2-(cosB+cosC)/2
=1+{cos[(B+C)/2]}^2-cos[(B+C)/2]cos[(B-C)/2]
=1+cos[(B+C)/2]{cos[(B+C)/2]-cos[(B-C)/2]}
=1+sin(A/2)*(-2)sin(B/2)sin(C/2)
=1-2sin(A/2)sin(B/2)sin(C/2)

Answer 3

看不懂啊