Trigonometry & Inverse Trigonometry
Trigonometry
star_batch_jee_advanced_2025
Grade 11
Question:
If in a $\triangle ABC$, $\sum\sin 3A = 0$, then at least one angle of $\triangle ABC$ is:
Step-by-Step Solution
Key Concept: Sum-to-product identities convert the trigonometric equation into a product form that reveals when at least one angle equals $60°$.
Given $\sum \sin 3A = 0$, we expand to $\sin 3A + \sin 3B + \sin 3C = 0$. Using sum-to-product formulas and simplifying, we obtain $\cos \frac{3A}{2} \cos \frac{3B}{2} \cos \frac{3C}{2} = 0$. This gives $\frac{3A}{2} = \frac{\pi}{2}$ or $\frac{3B}{2} = \frac{\pi}{2}$ or $\frac{3C}{2} = \frac{\pi}{2}$, leading to $A = \frac{\pi}{3}$ or $B = \frac{\pi}{3}$ or $C = \frac{\pi}{3}$. Therefore, at least one angle of triangle $ABC$ is $60°$.
Correct Answer: 1