If $A = 60^\circ$ and $B = 30^\circ$, verify that : $\sin(A + B) = \sin A \cos B + \cos A \sin B$
Generated by claude-sonnet-4-6 · 2026-06-15 10:27 · grounding rag
Model Answer
Given: A = 60°, B = 30°
LHS: $\sin(A + B) = \sin(60° + 30°) = \sin 90° = 1$
RHS: $\sin A \cos B + \cos A \sin B$
$= \sin 60° \cos 30° + \cos 60° \sin 30°$
$= \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} \times \dfrac{1}{2}$
$= \dfrac{3}{4} + \dfrac{1}{4} = 1$
Since LHS = RHS = 1, the identity is verified. ✓
Source: Chapter 8, Section 8.3 (Trigonometric Ratios of Specific Angles)
---
Explanation
- Always compute LHS and RHS separately, then show they are equal — examiners award marks for this structure.
- Substitute standard values from Table 8.1: sin 60° = √3/2, cos 30° = √3/2, cos 60° = 1/2, sin 30° = 1/2.
- The key result sin 90° = 1 comes directly from the table.
- Do not skip the arithmetic steps — each step can carry partial marks.