If A + B + C = 180°, Prove that

Sin² (A/2) + sin² (B/2) - sin²(C/2) = 1 – 2cos (A/2) cos (B/2) sin(C/2)

To prove the given identity:

Sin² (A/2) + sin² (B/2) - sin²(C/2) = 1 – 2cos (A/2) cos (B/2) sin(C/2)

We can start by expressing all the trigonometric functions in terms of the angles A, B, and C.

First, let's consider the sum of the angles A, B, and C. Since we know that A + B + C = 180°, we can rewrite one of the angles in terms of the other two. Let's rewrite angle C as C = 180° - (A + B).

Now, let's focus on the left side of the equation: Sin² (A/2) + sin² (B/2) - sin²(C/2).

Using the half-angle formula for sine, we can express Sin² (A/2) as (1 - cosA) / 2 and Sin² (B/2) as (1 - cosB) / 2.

For sin²(C/2), we can substitute the value of C: sin²[(180° - (A + B))/2]. Using the half-angle formula for sine again, this becomes sin²[(90° - (A + B)/2)] = (1 - cos(90° - (A + B))) / 2 = (1 - sin(A + B))/2.

We can simplify the left side of the equation now:

Sin² (A/2) + sin² (B/2) - sin²(C/2)
= (1 - cosA) / 2 + (1 - cosB) / 2 - (1 - sin(A + B))/2
= (1 - cosA + 1 - cosB - 1 + sin(A + B))/2
= (2 - (cosA + cosB) + sin(A + B))/2
= (2 - 2cos(A/2)cos(B/2) + 2sin(A/2)sin(B/2))/2
= 1 - cos(A/2)cos(B/2) + sin(A/2)sin(B/2).

Now, let's consider the right side of the equation:

1 – 2cos (A/2) cos (B/2) sin(C/2).

We already simplified the expression on the left side to a similar form. To get to this form, we substituted sin²(C/2) with (1 - sin(A + B))/2.

Now, we can simplify the right side of the equation:

1 – 2cos (A/2) cos (B/2) sin(C/2)
= 1 – 2cos (A/2) cos (B/2) sin(180° - (A + B))/2
= 1 – 2cos (A/2) cos (B/2) sin(90° - (A/2 + B/2))/2
= 1 – 2cos (A/2) cos (B/2) cos (A/2 + B/2).

Since cos (A + B) = cos (A/2 + B/2), we can rewrite the right side as:

1 – 2cos (A/2) cos (B/2) cos (A/2 + B/2) = 1 – 2cos (A/2) cos (B/2) cos (A/2 + B/2).

Therefore, the left side of the equation is equal to the right side, and the given identity holds.

This completes the proof.