Question

如图，证明三角函数恒等式

Answer 1

sinA+sinC-sin(A+C)
=4sin(A/2)sin(C/2)sin(A/2+C/2)
∵A+C=180°
∴sinA+sinC=4sin(A/2)sin(C/2)
同理
sinB+sinD=4sin(B/2)sin(D/2)
代入即可得证。

Answer 2

证：
sinx+siny=sin(x+y)+4sin(x/2)sin(y/2)sin[(x+y)/2]
(sinA+sinB+sinC+sinD)/4
=[(sinA+sinC)+(sinB+sinD)]/4
=[sin(A+C)+4sin(A/2)sin(C/2)sin[(A+C)/2]+sin(B+D)+4sin(B/2)sin(D/2)sin[(B+D)/2]]/4
=[sin180°+4sin(A/2)sin(C/2)sin90°+sin180°+4sin(B/2)sin(D/2)sin90°]/4
=[0+4sin(A/2)sin(C/2)+0+4sin(B/2)sin(D/2)]/4
=sin(A/2)sin(C/2)+sin(B/2)sin(D/2)

Answer 3

追答