Question

In a triangle \(ABC\), \[ \sin 2A+\sin 2B+\sin 2C= \]

• A. \(4\sin A\sin B\sin C\)
• B. \(2\sin A\sin B\sin C\)
• C. \(4\cos A\cos B\cos C\)
• D. \(2\sin A\cos B\cos C\)

Hint:

In triangle identities, always use \(A+B+C=180^\circ\) to convert sums of angles and simplify trigonometric expressions.

Answer 1

The Correct Option is A

Step 1: Use the angle sum property of a triangle.
In a triangle \(ABC\), \[ A+B+C=180^\circ. \] Therefore, \[ A+B=180^\circ-C. \]

Step 2: Combine \(\sin 2A+\sin 2B\).
Using the identity \[ \sin X+\sin Y=2\sin\frac{X+Y}{2}\cos\frac{X-Y}{2}, \] we get \[ \sin 2A+\sin 2B = 2\sin(A+B)\cos(A-B). \] Since \[ A+B=180^\circ-C, \] we have \[ \sin(A+B)=\sin C. \] Thus, \[ \sin 2A+\sin 2B=2\sin C\cos(A-B). \]

Step 3: Add \(\sin 2C\).
Now, \[ \sin 2A+\sin 2B+\sin 2C = 2\sin C\cos(A-B)+2\sin C\cos C. \] Taking \(2\sin C\) common, \[ =2\sin C[\cos(A-B)+\cos C]. \] Since \[ C=180^\circ-(A+B), \] we get \[ \cos C=-\cos(A+B). \] Therefore, \[ \cos(A-B)+\cos C = \cos(A-B)-\cos(A+B). \] Using the identity \[ \cos X-\cos Y=-2\sin\frac{X+Y}{2}\sin\frac{X-Y}{2}, \] we get \[ \cos(A-B)-\cos(A+B)=2\sin A\sin B. \]

Step 4: Substitute the value.
Hence, \[ \sin 2A+\sin 2B+\sin 2C = 2\sin C(2\sin A\sin B). \] \[ =4\sin A\sin B\sin C. \]

Step 5: Final conclusion.
Therefore, \[ \boxed{4\sin A\sin B\sin C} \]