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 $4 \cos A \sin B \cos C$.