Verified answer

Let's assume that A, B, and C are angles of a triangle, so they sum to 180.

Since the angles sum to 180, we have

B = 180 - (A+C)

The first term of your equation is

sin B = sin (180 - (A+C))

The angle-difference formula for sines is

sin (x-y) = sin x cos y - cos x sin y

So sin B is given by

sin B = 0*cos(A+C) - (-1)*sin(A+C)

= sin(A+C)

The left side of the equation is now given by

sin(A+C) + sin(A-C)

We expand both of these by using the formulas for angle sums and differences:

(sin A cos C + cos A sin C) + (sin A cos C - cos A sin C)

The second and fourth terms cancel out, and the first and third terms are identical, so this simplifies to

2 sin A cos C

So what I've proved is this:

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

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(Sorry -- I made a bunch of sloppy mistakes, but this version is now correct, finally. It's hard to work in the tiny answer box, particularly when I can't see your updated question at the same time.)

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to review the steps:

The left side of your equation is

sin B + sin (A-C)

I showed that sin B is equal to sin(A+C), so the left side of your equation is equal to

sin (A+C) + sin (A-C)

When we expand these terms (using formulae for the sines of angle sums and differences), we get

2 sin A cos C

Q.E.D.