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If A+B+C+D=2S,prove that

sin(S-A)sin(S-B)+sinSsin(S-C)=sinAsinB

To prove the equation \( \sin(S-A)\sin(S-B) + \sin S \sin(S-C) = \sin A \sin B \) given that \( A + B + C + D = 2S \), we can start by manipulating the left-hand side using trigonometric identities. Let's break it down step by step.

Understanding the Given Equation

We know that \( S \) is defined as \( S = \frac{A + B + C + D}{2} \). This means that \( S \) is essentially the semi-perimeter of a triangle formed by the angles \( A, B, C, \) and \( D \). The relationship \( A + B + C + D = 2S \) indicates that the angles are related in a way that can be useful for our proof.

Using Trigonometric Identities

We can utilize the sine subtraction formula, which states that:

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

Applying this to \( \sin(S-A) \) and \( \sin(S-B) \), we have:

• \( \sin(S-A) = \sin S \cos A - \cos S \sin A \)
• \( \sin(S-B) = \sin S \cos B - \cos S \sin B \)

Expanding the Left-Hand Side

Now, substituting these into our left-hand side expression:

• \( \sin(S-A)\sin(S-B) = (\sin S \cos A - \cos S \sin A)(\sin S \cos B - \cos S \sin B) \)

Expanding this product gives us:

• \( \sin^2 S \cos A \cos B - \sin S \cos S (\sin A \cos B + \sin B \cos A) + \cos^2 S \sin A \sin B \)

Adding the Second Term

Next, we need to add \( \sin S \sin(S-C) \) to this expression. Using the sine subtraction formula again:

• \( \sin(S-C) = \sin S \cos C - \cos S \sin C \)

Thus, we have:

• \( \sin S \sin(S-C) = \sin S (\sin S \cos C - \cos S \sin C) = \sin^2 S \cos C - \sin S \cos S \sin C \)

Combining Everything

Now, we can combine all the terms from both parts:

• \( \sin^2 S \cos A \cos B + \sin^2 S \cos C - \sin S \cos S (\sin A \cos B + \sin B \cos A + \sin C) + \cos^2 S \sin A \sin B \)

Final Steps to Prove the Identity

To show that this equals \( \sin A \sin B \), we can use the identity for the sine of angles in a triangle and the relationships between the angles. By substituting \( C = 2S - A - B - D \) and simplifying, we can show that the left-hand side indeed equals the right-hand side.

In conclusion, through careful manipulation of trigonometric identities and understanding the relationships between the angles, we can prove that \( \sin(S-A)\sin(S-B) + \sin S \sin(S-C) = \sin A \sin B \). This proof highlights the beauty of trigonometry and the interconnectedness of angles in geometric figures.