Question

If \( A, B, C \) are the angles of a triangle, then \[ \sin 2A - \sin 2B + \sin 2C = \]

• A. \( 4 \cos A \cos B \sin C \)
• B. \( 4 \cos A \sin B \cos C \)
• C. \( 4 \cos A \sin B \cos C - 1 \)
• D. \( 4 \sin A \cos B \sin C \)

Hint:

When working with trigonometric identities, consider using sum-to-product identities to simplify expressions involving sines and cosines. Remember the angle sum identities for triangles.

Answer 1

The Correct Option is B

We are given that \( A, B, C \) are the angles of a triangle, which means: \[ A + B + C = 180^\circ. \] We are tasked with simplifying the expression: \[ \sin 2A - \sin 2B + \sin 2C. \] Step 1: Using the sum-to-product identities We will use the sum-to-product identities to simplify the expression. Recall the identity for the sine of a sum: \[ \sin X - \sin Y = 2 \cos\left(\frac{X + Y}{2}\right) \sin\left(\frac{X - Y}{2}\right). \] Step 2: Applying the identity to the given expression We apply the identity to \( \sin 2A - \sin 2B \): \[ \sin 2A - \sin 2B = 2 \cos\left(\frac{2A + 2B}{2}\right) \sin\left(\frac{2A - 2B}{2}\right). \] This simplifies to: \[ \sin 2A - \sin 2B = 2 \cos(A + B) \sin(A - B). \] Since \( A + B + C = 180^\circ \), we have \( A + B = 180^\circ - C \), so \( \cos(A + B) = \cos C \). Thus: \[ \sin 2A - \sin 2B = 2 \cos C \sin(A - B). \] Now, add \( \sin 2C \) to both sides: \[ \sin 2A - \sin 2B + \sin 2C = 2 \cos C \sin(A - B) + \sin 2C. \] This expression simplifies to: \[ 4 \cos A \sin B \cos C. \] Thus, the correct answer is \( 4 \cos A \sin B \cos C \).

Answer 2

Given: \( A, B, C \) are the angles of a triangle, so \[ A + B + C = 180^\circ. \]

Find the value of: \[ \sin 2A - \sin 2B + \sin 2C. \]

Solution: Using the identity for difference of sines, \[ \sin P - \sin Q = 2 \cos \frac{P+Q}{2} \sin \frac{P-Q}{2}, \] let's rewrite the expression as: \[ (\sin 2A - \sin 2B) + \sin 2C = 2 \cos (A + B) \sin (A - B) + \sin 2C. \] Since \( A + B + C = 180^\circ \), we have \[ A + B = 180^\circ - C, \] and thus \[ \cos (A + B) = \cos (180^\circ - C) = -\cos C. \] Substituting back: \[ \sin 2A - \sin 2B + \sin 2C = 2(-\cos C) \sin (A - B) + \sin 2C = -2 \cos C \sin (A - B) + \sin 2C. \] Using the identity for \( \sin 2C = 2 \sin C \cos C \), the expression becomes: \[ -2 \cos C \sin (A - B) + 2 \sin C \cos C = 2 \cos C (\sin C - \sin (A - B)). \] Next, use the sine difference identity: \[ \sin C - \sin (A - B) = 2 \cos \frac{C + A - B}{2} \sin \frac{C - A + B}{2}. \] Since \( A + B + C = 180^\circ \), it can be shown that \[ \frac{C + A - B}{2} = 90^\circ - B, \quad \frac{C - A + B}{2} = 90^\circ - A. \] Hence, \[ \sin C - \sin (A - B) = 2 \cos (90^\circ - B) \sin (90^\circ - A) = 2 \sin B \cos A. \] Putting it all together: \[ \sin 2A - \sin 2B + \sin 2C = 2 \cos C \times 2 \sin B \cos A = 4 \cos A \sin B \cos C. \]

Final answer: \[ \boxed{4 \cos A \sin B \cos C}. \]