sin a sin b sin c formula

PTsin a/2 + sin b/2 + sin c/2 -1 = 4sin((π-a)/4) .sin( (π-b)/4) .sin ((π-c)/4 ) - 3142841. Sin a/2 + sin b/2 + sin c/2 -1 {apply formula on first two terms sinC Itshould be: sin(a) sin(b-c) + sin(b) sin(c-a) + sin(c) sin(a-b) = 0 Just use the addition formula for sine, i.e. sin(x+y) = sin(x)cos(y) + cos(x)sin(y) and the fact that sin(-x) = -sin(x) and cos(-x) = cos(x), and everything cancels out. Howdo I derive the formula: cos(A+B)=cosAcosB-sinAsinB. from the formula: cos(A-B)=cosAcosB+sinAsinB? The only difference that I noticed is the negative and positive sign. I was thinking that first, I replace B with (-B), but then after that how does cos(-B) turn to cos(B), and sin(-B) turn to -sin(B)? Thank you, can someone please explain to me. cosx) = sin(x) = 1 = sin²(q) + cos²(q) sin(-q) = -sin(q) sin(q + p) = -sin(q) sin(p -q) = sin(q) cos(-q) = cos(q) cos(q + p) = -cos(q) cos(p -q) = -cos(q) tan(-q Vay Tien Nhanh Ggads.

sin a sin b sin c formula