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In triangle ABC, if 2b² = a² + c², then the value of sin 3B divided by sin B is

1. (c² − a²)/(2ca)
2. (c² − a²)/ca
3. ((c² − a²)/ca)²
4. ((c² − a²)/(2ca))²

Correct answer: ((c² − a²)/(2ca))²

Solution

sin3B/sinB=3-4sin^2 B=4cos^2 B-1. With 2b^2=a^2+c^2, cosB=(a^2+c^2-b^2)/(2ac)=(a^2+c^2)/(4ac). Then 4cos^2 B-1 simplifies to ((c^2-a^2)/(2ca))^2.

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