Q1. [1] § 8.3 Trigonometric Ratios of Some Specific Angles
Given that $\sin 2\alpha = \frac{\sqrt{3}}{2}$, the value of $\sin 3\alpha$ is :
- A $\frac{3\sqrt{3}}{4}$
- B $\frac{1}{2}$
- C $1$
- D $\frac{\sqrt{3}}{4}$
Previously asked in CBSE board exam
2026 30/3/1 Q15
Generated by claude-sonnet-4-6 · 2026-06-15 10:25 · grounding rag
Model Answer
$\sin 2\alpha = \dfrac{\sqrt{3}}{2} \Rightarrow 2\alpha = 60° \Rightarrow \alpha = 30°$. Therefore, $\sin 3\alpha = \sin 90° = \mathbf{1}$. Answer: (C)
Explanation
From Table 8.1, $\sin 60° = \dfrac{\sqrt{3}}{2}$, so $2\alpha = 60°$, giving $\alpha = 30°$. Then $3\alpha = 90°$ and $\sin 90° = 1$. Students must recall standard angle values from the table and work step-by-step.
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