OK, B+C = 180-A
so
LS = 2sin(B+C)sinA
= 2sin(180 - A)sinA
= 2(-sinA)sinA
= -2 sin^2 A
RS = 1 - cos 2A
= 1 -(2sin^2 A - 1)
= -2sin^2 A
= LS
if A+B+C=180degrees, prove that 2sin(B+C)sinA=1-cos2A
3 answers
in part 1, how does sin(180-A) equal -sinA from line 4-5
Actually I have 2 errors in my solution
1.
sin(180-A) = sinA , not -sinA
e.g. sin (150) = sin(180-150) = sin 30
2. cos 2A = 1 - 2sin^2 A, I had it backwards
so here is the correct solution
LS = 2sin(B+C)sinA
= 2sin(180 - A)sinA
= 2(sinA)sinA
= 2 sin^2 A
RS = 1 - cos 2A
= 1 -(1 - 2sin^2 A)
= 2sin^2 A
= LS
sorry about that
1.
sin(180-A) = sinA , not -sinA
e.g. sin (150) = sin(180-150) = sin 30
2. cos 2A = 1 - 2sin^2 A, I had it backwards
so here is the correct solution
LS = 2sin(B+C)sinA
= 2sin(180 - A)sinA
= 2(sinA)sinA
= 2 sin^2 A
RS = 1 - cos 2A
= 1 -(1 - 2sin^2 A)
= 2sin^2 A
= LS
sorry about that