Given that $\sin\theta + \cos\theta = x$, prove that $\sin^4\theta + \cos^4\theta = \frac{2 - (x^2 - 1)^2}{2}$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:26 · grounding rag
Model Answer
Given: $\sin\theta + \cos\theta = x$
Step 1: Square both sides:
$$x^2 = (\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta$$
$$\therefore \sin\theta\cos\theta = \frac{x^2 - 1}{2}$$
Step 2: Find $\sin^4\theta + \cos^4\theta$:
$$\sin^4\theta + \cos^4\theta = (\sin^2\theta + \cos^2\theta)^2 - 2\sin^2\theta\cos^2\theta$$
$$= 1 - 2(\sin\theta\cos\theta)^2$$
$$= 1 - 2\left(\frac{x^2-1}{2}\right)^2$$
$$= 1 - \frac{(x^2-1)^2}{2} = \frac{2-(x^2-1)^2}{2}$$
= RHS $\qquad \square$
Source: Chapter 8, Section 8.4 Trigonometric Identities
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Explanation
- Key identity used: $\sin^2\theta + \cos^2\theta = 1$ (from Section 8.4).
- Algebraic trick: $\sin^4\theta + \cos^4\theta = (\sin^2\theta+\cos^2\theta)^2 - 2\sin^2\theta\cos^2\theta$ — this is the standard $a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2$ identity. Examiners expect you to show this step explicitly.
- Squaring $x = \sin\theta+\cos\theta$ to extract $\sin\theta\cos\theta$ is the essential starting move — missing it loses marks.