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

Question

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

Answer 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

Answer 2

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

Answer 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

sorry about that

Answer 4

To prove the equation 2sin(B+C)sinA = 1 - cos(2A), we will use the trigonometric identities and the given condition A + B + C = 180 degrees.

Let's start with the left-hand side (LHS) of the equation:

LHS = 2sin(B+C)sinA

Using the double angle formula for sine, we can rewrite sin(B+C) as:

sin(B+C) = sinBcosC + cosBsinC

Substituting this into the equation, we have:

LHS = 2(sinBcosC + cosBsinC)sinA

Now, let's use the product-to-sum identity for cosine:

cosCsinA = (1/2)[sin(C + A) - sin(C - A)]

cosCsinA = (1/2)(sin(C + A) - sin(C - A))

Substituting this back into the equation, we get:

LHS = 2(sinBcosC + (1/2)(sin(C + A) - sin(C - A))sinA

Simplifying further, we have:

LHS = 2sinBcosC + sinAsin(C+A) - sinAsin(C-A)

Using the sum-to-product identity for sine:

sinAsin(C + A) = (1/2)(cos(A - C) - cos(A + C))

sinAsin(C - A) = (1/2)(cos(A + C) - cos(A - C))

Substituting these back into the equation, we have:

LHS = 2sinBcosC + (1/2)(cos(A - C) - cos(A + C)) - (1/2)(cos(A + C) - cos(A - C))

Simplifying, we get:

LHS = 2sinBcosC + (1/2)(cos(A - C) - cos(A + C)) - (1/2)(cos(A + C) - cos(A - C))

The cosines cancel out, leading to:

LHS = 2sinBcosC

Now, let's simplify the right-hand side (RHS) of the equation:

RHS = 1 - cos(2A)

Using the double angle formula for cosine, we have:

RHS = 1 - (cos^2(A) - sin^2(A))

RHS = 1 - cos^2(A) + sin^2(A)

RHS = 1 - (1 - sin^2(A)) + sin^2(A)

RHS = 1 - 1 + sin^2(A) + sin^2(A)

RHS = 2sin^2(A)

We can rewrite 2sin^2(A) as 2(1 - cos^2(A)) using the identity sin^2(A) = 1 - cos^2(A), so:

RHS = 2(1 - cos^2(A))

RHS = 2 - 2cos^2(A)

Now, comparing the LHS and RHS, we see that they are both equal to 2sinBcosC. Therefore, we have successfully proved that:

2sin(B+C)sinA = 1 - cos(2A)