# math

if A+B+C=180degrees, prove that 2sin(B+C)sinA=1-cos2A

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1. 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

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2. in part 1, how does sin(180-A) equal -sinA from line 4-5

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3. 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

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2. 👎

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