If $a\cos\theta + b\sin\theta = m$ and $a\sin\theta - b\cos\theta = n$, then prove that $a^2 + b^2 = m^2 + n^2$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:24 · grounding rag
Model Answer
Proof:
Given: $a\cos\theta + b\sin\theta = m$ ... (1)
$a\sin\theta - b\cos\theta = n$ ... (2)
Squaring and adding (1) and (2):
$$m^2 + n^2 = (a\cos\theta + b\sin\theta)^2 + (a\sin\theta - b\cos\theta)^2$$
$$= a^2\cos^2\theta + 2ab\cos\theta\sin\theta + b^2\sin^2\theta + a^2\sin^2\theta - 2ab\sin\theta\cos\theta + b^2\cos^2\theta$$
$$= a^2(\cos^2\theta + \sin^2\theta) + b^2(\sin^2\theta + \cos^2\theta)$$
$$= a^2(1) + b^2(1) = a^2 + b^2$$
$$\therefore\quad a^2 + b^2 = m^2 + n^2 \qquad \textbf{(Proved)}$$
Source: Chapter 8, Section 8.4 Trigonometric Identities
---
Explanation
- Key step: Square both equations and add — the cross terms $(\pm\,2ab\sin\theta\cos\theta)$ cancel each other.
- Identity used: $\sin^2\theta + \cos^2\theta = 1$ (the fundamental Pythagorean identity from Section 8.4).
- Examiners award 1 mark for correct squaring and adding, and 1 mark for applying the identity and concluding. Write "Proved" or "Hence proved" at the end.