Question:
If \(A + B + C = 180^\circ\), then what is \(\sin 2A – \sin 2B – \sin 2C\) equal to?
यदि \(A + B + C = 180^\circ\), तो \(\sin 2A – \sin 2B – \sin 2C\) किसके बराबर है?
Options:
- \(-4\sin A \sin B \sin C\)
- \(-4\cos A \sin B \cos C\)
- \(-4\cos A \cos B \sin C\)
- \(-4\sin A \cos B \cos C\)
- \(-4\sin A \sin B \sin C\)
- \(-4\cos A \sin B \cos C\)
- \(-4\cos A \cos B \sin C\)
- \(-4\sin A \cos B \cos C\)
Correct Answer: d)
Explanation:
Using sum-to-product formula for the first two terms:
\[\sin 2A – \sin 2B = 2\cos(A + B)\sin(A – B) = 2\cos(180^\circ – C)\sin(A – B) = -2\cos C \sin(A – B)\]
Now subtracting \(\sin 2C = 2\sin C \cos C\):
\[-2\cos C [\sin(A – B) + \sin C] = -2\cos C [\sin(A – B) + \sin(180^\circ – (A + B))]\]
\[= -2\cos C [\sin(A – B) + \sin(A + B)] = -2\cos C [2\sin A \cos B] = -4\sin A \cos B \cos C\]
Hence, the correct option is (d).
त्रिकोणमितीय सूत्रों का उपयोग करने पर:
\[\sin 2A – \sin 2B = -2\cos C \sin(A – B)\]
\(\sin 2C = 2\sin C \cos C\) घटाने पर:
\[-2\cos C [\sin(A – B) + \sin(A + B)] = -2\cos C [2\sin A \cos B] = -4\sin A \cos B \cos C\]
अतः सही विकल्प (d) है।