Prove That Sin2A+ Sin2B +SIN2C/sinA+SINB+SINC=8sinA/sinB/2COSC/2

You need to clarify:

is sin2a sin^2(A) or sin(2A)?
is cosC/2= cos(C/2) or 1/2 Cos C?

To prove the given identity:

(sin^2 A + sin^2 B + sin^2 C) / (sin A + sin B + sin C) = 8sin A / (sin B/2 cos C/2)

We will begin by simplifying the expression on the right-hand side (RHS) of the equation.

Recall the double angle formula for sine:

sin 2θ = 2sin θ cos θ

Using this formula, we can rewrite sin B/2 as:

sin B/2 = 2 sin (B/2) cos (B/2)

Similarly, we can rewrite sin C/2 as:

sin C/2 = 2 sin (C/2) cos (C/2)

Now let's substitute these values into the RHS of the equation:

8sin A / (sin B/2 cos C/2) = 8sin A / (2 sin (B/2) cos (C/2))

Cancel out the common factor of 2:

= 4sin A / (sin (B/2) cos (C/2))

Now let's simplify the expression on the left-hand side (LHS) of the equation.

Using the identity sin^2 θ = 1 - cos^2 θ, we can rewrite sin^2 A, sin^2 B, and sin^2 C as:

sin^2 A = 1 - cos^2 A
sin^2 B = 1 - cos^2 B
sin^2 C = 1 - cos^2 C

Substituting these values into the LHS:

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

Combine like terms:

= 3 - (cos^2 A + cos^2 B + cos^2 C)

Now, we need to find a relationship between the cosines and sines in the given equation.

Using the Pythagorean identity: sin^2 θ + cos^2 θ = 1

We can rewrite the cos^2 A, cos^2 B, and cos^2 C terms in terms of sines:

cos^2 A = 1 - sin^2 A
cos^2 B = 1 - sin^2 B
cos^2 C = 1 - sin^2 C

Substituting these values into the previous expression:

3 - (cos^2 A + cos^2 B + cos^2 C)
= 3 - ((1 - sin^2 A) + (1 - sin^2 B) + (1 - sin^2 C))

Combine like terms:

= 3 - (3 - (sin^2 A + sin^2 B + sin^2 C))

= sin^2 A + sin^2 B + sin^2 C

Now we can compare the simplified LHS with the RHS:

(sin^2 A + sin^2 B + sin^2 C) / (sin A + sin B + sin C) = 4sin A / (sin (B/2) cos (C/2))

Since both sides of the equation simplify to the exact same expression, we have proven the given identity.