n a triangle ABC,prove that SinA/2*SinA/2+SinB/2*SinB/2+SinC/2*SinC/2=1-2SinA/2SinB/2SinC/2

Question

we are given a triangle ABC, using which we have to prove this.

Answer 1

First, test the validity of the proposition, let A=B=C=60°, then sin(A/2)=sin(30°)=½, so:

sin²(A/2)+sin²(B/2)+sin²(C/2)=¾.

1-2sin(A/2)sin(B/2)sin(C/2)=1-2(½)³=1-2/8=¾.

To conserve text space and aid readability, let s_A=sin(A/2) and c_A=cos(A/2) and similarly for other angles.

So we need to prove: s_A²+s_B²+s_C²=1-2s_As_Bs_C.

In a triangle A+B+C=180 so C=180-(A+B), C/2=90-½(A+B) and:

s_C=cos(90-C/2)=cos(½(A+B))=c_A+B=c_Ac_B-s_As_B.

Therefore, s_C²=c_A+B²=(c_Ac_B-s_As_B)²=c_A²c_B²-2s_As_Bc_Ac_B+s_A²s_B².

s_A²+s_B²+s_C²=

s_A²+s_B²+c_A²c_B²-2s_As_Bc_Ac_B+s_A²s_B²; but c_A²c_B²=(1-s_A²)(1-s_B²)=1-s_B²-s_A²+s_A²s_B².

s_A²+s_B²+s_C²=s_A²+s_B²+1-s_B²-s_A²+s_A²s_B²-2s_As_Bc_Ac_B+s_A²s_B²,

s_A²+s_B²+s_C²=1+2s_A²s_B²-2s_As_Bc_Ac_B; but s_C=c_Ac_B-s_As_B, so c_Ac_B=s_C+s_As_B.

s_A²+s_B²+s_C²=1+2s_A²s_B²-2s_As_B(s_C+s_As_B)=1+2s_A²s_B²-2s_As_Bs_C-2s_A²s_B²=1-2s_As_Bs_C.

Therefore, s_A²+s_B²+s_C²=1-2s_As_Bs_C.

Hence, sin²(A/2)+sin²(B/2)+sin²(C/2)=1-2sin(A/2)sin(B/2)sin(C/2).