| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Prove trigonometric identity |
| Difficulty | Standard +0.3 Part (i) is a straightforward algebraic manipulation using the sum of cubes factorization (a³+b³=(a+b)(a²-ab+b²)) with sin²x+cos²x=1, requiring 3-4 routine steps. Part (ii) involves standard manipulation (converting tan to sin/cos, factoring) leading to cos θ = 1/3, then calculator work. Both parts are mechanical applications of standard techniques with no novel insight required, making this slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((\sin x + \cos x)(1 - \sin x \cos x) \equiv \sin x + \cos x - \sin^2 x \cos x - \cos^2 x \sin x\) | 1st M1 | Expands bracket to 4 terms; condone sign slips but terms must be correct |
| \(\equiv \sin x + \cos x - (1-\cos^2 x)\cos x - (1-\sin^2 x)\sin x\) | 2nd M1 | Replaces \(\sin^2 x\) by \((1-\cos^2 x)\) and \(\cos^2 x\) by \((1-\sin^2 x)\) |
| \(\equiv \sin^3 x + \cos^3 x\) \(\quad *\) | A1* | Completes proof with no errors; withhold if poor notation or mixed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use LHS \(= (\sin x + \cos x)(\sin^2 x + \cos^2 x - \sin x \cos x)\) | 2nd M1 | |
| \(\equiv \sin^3 x + \sin x\cos^2 x - \sin^2 x\cos x + \sin^2 x\cos x - \cos^2 x\sin x + \cos^3 x\) | 1st M1 | |
| \(\equiv \sin^3 x + \cos^3 x\) \(\quad *\) | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use RHS \(\equiv \sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x + \cos^2 x - \sin x\cos x)\) | M1 | |
| \(= (\sin x + \cos x)(1 - \sin x\cos x)\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) to give \(3\sin\theta = \frac{\sin\theta}{\cos\theta}\) | M1 | Or equivalent to give \(3\cos\theta = 1\) |
| \(\cos\theta = \frac{1}{3}\) | A1 | |
| \(\sin\theta = 0\) | A1 | |
| \(\arccos\!\left(\frac{1}{3}\right)\) leading to one value of \(\theta\) to nearest degree | M1 | |
| \(\theta = 70.5°, 289.5°\) | A1 | Withheld for extra solutions of \(\arccos\!\left(\frac{1}{3}\right)\) in range; ignore extra solutions outside range |
| \(\theta = 0\) and \(180°\) | B1 | Withheld for extra solutions arising from \(\sin\theta = 0\); ignore extra solutions outside range |
## Question 13:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(\sin x + \cos x)(1 - \sin x \cos x) \equiv \sin x + \cos x - \sin^2 x \cos x - \cos^2 x \sin x$ | 1st M1 | Expands bracket to 4 terms; condone sign slips but terms must be correct |
| $\equiv \sin x + \cos x - (1-\cos^2 x)\cos x - (1-\sin^2 x)\sin x$ | 2nd M1 | Replaces $\sin^2 x$ by $(1-\cos^2 x)$ **and** $\cos^2 x$ by $(1-\sin^2 x)$ |
| $\equiv \sin^3 x + \cos^3 x$ $\quad *$ | A1* | Completes proof with no errors; withhold if poor notation or mixed variables |
**Alt I(i):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use LHS $= (\sin x + \cos x)(\sin^2 x + \cos^2 x - \sin x \cos x)$ | 2nd M1 | |
| $\equiv \sin^3 x + \sin x\cos^2 x - \sin^2 x\cos x + \sin^2 x\cos x - \cos^2 x\sin x + \cos^3 x$ | 1st M1 | |
| $\equiv \sin^3 x + \cos^3 x$ $\quad *$ | A1* | |
**Alt II(i):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use RHS $\equiv \sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x + \cos^2 x - \sin x\cos x)$ | M1 | |
| $= (\sin x + \cos x)(1 - \sin x\cos x)$ | M1 A1 | |
### Part (ii): Solve $3\sin\theta = \tan\theta$, $0 \leq \theta \leq 360°$
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use $\tan\theta = \frac{\sin\theta}{\cos\theta}$ to give $3\sin\theta = \frac{\sin\theta}{\cos\theta}$ | M1 | Or equivalent to give $3\cos\theta = 1$ |
| $\cos\theta = \frac{1}{3}$ | A1 | |
| $\sin\theta = 0$ | A1 | |
| $\arccos\!\left(\frac{1}{3}\right)$ leading to one value of $\theta$ to nearest degree | M1 | |
| $\theta = 70.5°, 289.5°$ | A1 | Withheld for extra solutions of $\arccos\!\left(\frac{1}{3}\right)$ in range; ignore extra solutions outside range |
| $\theta = 0$ and $180°$ | B1 | Withheld for extra solutions arising from $\sin\theta = 0$; ignore extra solutions outside range |
Note: Students proceeding from $\frac{\sin\theta}{\tan\theta} = \cos\theta$ can score M1A1A0M1A1B0 (4 out of 6). Radian solutions: withhold only the final A1; solutions are awrt 2dp: $1.23, 5.05, 0, 3.14$.
---\begin{enumerate}
\item (i) Showing each step in your reasoning, prove that
\end{enumerate}
$$( \sin x + \cos x ) ( 1 - \sin x \cos x ) \equiv \sin ^ { 3 } x + \cos ^ { 3 } x$$
(ii) Solve, for $0 \leqslant \theta < 360 ^ { \circ }$,
$$3 \sin \theta = \tan \theta$$
giving your answers in degrees to 1 decimal place, as appropriate.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\hfill \mbox{\textit{Edexcel C12 2015 Q13 [9]}}