| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Complete or critique given proof |
| Difficulty | Standard +0.8 This requires students to complete a proof by contradiction involving trigonometric identities, including expanding brackets, applying Pythagorean identity, and recognizing that the contradiction arises from the obtuse angle constraint. It demands multi-step algebraic manipulation and understanding of proof structure, going beyond routine exercises. |
| Spec | 1.01d Proof by contradiction1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((\sin x - \cos x)^2 < 1 \Rightarrow \sin^2 x - 2\sin x\cos x + \cos^2 x\ ({<}1)\) o.e. | M1 | AO 1.1b — expands \((\sin x-\cos x)^2\) to obtain \(\sin^2 x \pm k\sin x\cos x + \cos^2 x\), \(k=1\) or \(2\) |
| \(1 - 2\sin x\cos x < 1\), i.e. \(-\sin 2x < 0\) o.e. | A1 | AO 2.2a — uses \(\sin^2 x+\cos^2 x=1\) to obtain correct inequality not including \(\sin^2 x\) and \(\cos^2 x\) terms |
| As \(x\) is obtuse: \(-2\sin x\cos x\) is positive since \(\sin x>0\) and \(\cos x<0\), so we have a contradiction. Therefore \(\sin x - \cos x \not\leq 1\) | A1* | AO 2.4 — fully correct work with: convincing argument explaining contradiction; statement of contradiction; conclusion \(\sin x - \cos x \ldots 1\); no contradictory statements; no mixed variables |
## Question 15:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\sin x - \cos x)^2 < 1 \Rightarrow \sin^2 x - 2\sin x\cos x + \cos^2 x\ ({<}1)$ o.e. | M1 | AO 1.1b — expands $(\sin x-\cos x)^2$ to obtain $\sin^2 x \pm k\sin x\cos x + \cos^2 x$, $k=1$ or $2$ |
| $1 - 2\sin x\cos x < 1$, i.e. $-\sin 2x < 0$ o.e. | A1 | AO 2.2a — uses $\sin^2 x+\cos^2 x=1$ to obtain correct inequality not including $\sin^2 x$ and $\cos^2 x$ terms |
| As $x$ is obtuse: $-2\sin x\cos x$ is positive since $\sin x>0$ and $\cos x<0$, so we have a contradiction. Therefore $\sin x - \cos x \not\leq 1$ | A1* | AO 2.4 — fully correct work with: convincing argument explaining contradiction; statement of contradiction; conclusion $\sin x - \cos x \ldots 1$; no contradictory statements; no mixed variables |
**(3 marks)**\begin{enumerate}
\item A student attempts to answer the following question:
\end{enumerate}
Given that $x$ is an obtuse angle, use algebra to prove by contradiction that
$$\sin x - \cos x \geqslant 1$$
The student starts the proof with:
Assume that $\sin x - \cos x < 1$ when $x$ is an obtuse angle
$$\begin{aligned}
& \Rightarrow ( \sin x - \cos x ) ^ { 2 } < 1 \\
& \Rightarrow \ldots
\end{aligned}$$
The start of the student's proof is reprinted below.\\
Complete the proof.
Assume that $\sin x - \cos x < 1$ when $x$ is an obtuse angle
$$\Rightarrow ( \sin x - \cos x ) ^ { 2 } < 1$$
\hfill \mbox{\textit{Edexcel Paper 2 2023 Q15 [3]}}