To prove the given equation, we will use the following trigonometric identities:

1. Cosine double angle formula: cos(2θ) = 1 - 2sin²(θ)
2. Sine half angle formula: sin(θ/2) = ±√((1 - cos(θ))/2)

Now, let's begin the proof:

We know that cos(A) + cos(B) + cos(C) = 1 + 4sin(A/2)sin(B/2)sin(C/2)

Using the cosine double angle formula, we can rewrite sin(A/2)sin(B/2)sin(C/2) as:

sin(A/2)sin(B/2)sin(C/2) = √((1 - cos(A))/2) * √((1 - cos(B))/2) * √((1 - cos(C))/2)

Multiplying these three terms together, we get:

sin(A/2)sin(B/2)sin(C/2) = √((1 - cos(A))/2 * (1 - cos(B))/2 * (1 - cos(C))/2)

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

1 + 4sin(A/2)sin(B/2)sin(C/2) = 1 + 4√((1 - cos(A))/2 * (1 - cos(B))/2 * (1 - cos(C))/2)

To simplify further, we can rewrite the expression inside the square root as:

(1 - cos(A))/2 * (1 - cos(B))/2 * (1 - cos(C))/2 = (1 - cos(A))(1 - cos(B))(1 - cos(C))/8

Now, let's substitute this back into the equation:

1 + 4√((1 - cos(A))/2 * (1 - cos(B))/2 * (1 - cos(C))/2) = 1 + 4√((1 - cos(A))(1 - cos(B))(1 - cos(C))/8)

To simplify the square root, we can rewrite it as:

√((1 - cos(A))(1 - cos(B))(1 - cos(C))/8) = √((1 - cos(A))(1 - cos(B))(1 - cos(C))/2²)

Now, let's substitute this back into the equation:

1 + 4√((1 - cos(A))(1 - cos(B))(1 - cos(C))/8) = 1 + 4√((1 - cos(A))(1 - cos(B))(1 - cos(C))/2²)

Simplifying further, we get:

1 + 4√((1 - cos(A))(1 - cos(B))(1 - cos(C))/2²) = 1 + 2√((1 - cos(A))(1 - cos(B))(1 - cos(C)))

Now, let's substitute this back into the equation:

1 + 2√((1 - cos(A))(1 - cos(B))(1 - cos(C))) = 1 + 2√((1 - cos(A))(1 - cos(B))(1 - cos(C)))

Therefore, we have proved that cos(A) + cos(B) + cos(C) = 1 + 4sin(A/2)sin(B/2)sin(C/2).