| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 This is a slightly-above-average C3 question. Part (a) requires standard manipulation of double angle formulas (sin 2θ = 2sinθcosθ, cos 2θ = 1-2sin²θ) to prove an identity—routine for C3 students. Part (b)(i) applies the result with θ=15° requiring knowledge that sin 30°=1/2 and cos 30°=√3/2. Part (b)(ii) involves recognizing the given equation matches the proven identity with 2θ=4x, then solving tan 2x=1 for multiple solutions in the given range. While multi-step, all techniques are standard C3 material with no novel insight required. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(\frac{1}{\sin2\theta} - \frac{\cos2\theta}{\sin2\theta} = \frac{1-\cos2\theta}{\sin2\theta}\) | M1 | |
| \(= \frac{2\sin^2\theta}{2\sin\theta\cos\theta}\) | M1A1 | |
| \(= \frac{\sin\theta}{\cos\theta} = \tan\theta\) | A1* | cso, (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(\tan15° = \frac{1}{\sin30°} - \frac{\cos30°}{\sin30°}\) | M1 | |
| \(\tan15° = \frac{1}{\frac{1}{2}} - \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = 2 - \sqrt{3}\) | dM1 A1* | cso, (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(\tan15° = \tan(60°-45°)\) or \(\tan(45°-30°)\) | ||
| \(\tan15° = \frac{\tan60-\tan45}{1+\tan60\tan45}\) or \(\frac{\tan45-\tan30}{1+\tan45\tan30}\) | M1 | |
| \(\tan15° = \frac{\sqrt{3}-1}{1+\sqrt{3}}\) or \(\frac{1-\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}}\) | M1 | |
| Rationalises to produce \(\tan15° = 2-\sqrt{3}\) | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(\tan2x = 1\) | M1 | |
| \(2x = 45°\) | A1 | |
| \(2x = 45° + 180°\) | M1 | |
| \(x = 22.5°, 112.5°, 202.5°, 292.5°\) | A1(any two), A1 | (5 marks) |
## Question 6:
### Part (a):
| Working | Mark | Notes |
|---------|------|-------|
| $\frac{1}{\sin2\theta} - \frac{\cos2\theta}{\sin2\theta} = \frac{1-\cos2\theta}{\sin2\theta}$ | M1 | |
| $= \frac{2\sin^2\theta}{2\sin\theta\cos\theta}$ | M1A1 | |
| $= \frac{\sin\theta}{\cos\theta} = \tan\theta$ | A1* | cso, (4 marks) |
### Part (b)(i):
| Working | Mark | Notes |
|---------|------|-------|
| $\tan15° = \frac{1}{\sin30°} - \frac{\cos30°}{\sin30°}$ | M1 | |
| $\tan15° = \frac{1}{\frac{1}{2}} - \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = 2 - \sqrt{3}$ | dM1 A1* | cso, (3 marks) |
**Alternative for (b)(i):**
| Working | Mark | Notes |
|---------|------|-------|
| $\tan15° = \tan(60°-45°)$ or $\tan(45°-30°)$ | | |
| $\tan15° = \frac{\tan60-\tan45}{1+\tan60\tan45}$ or $\frac{\tan45-\tan30}{1+\tan45\tan30}$ | M1 | |
| $\tan15° = \frac{\sqrt{3}-1}{1+\sqrt{3}}$ or $\frac{1-\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}}$ | M1 | |
| Rationalises to produce $\tan15° = 2-\sqrt{3}$ | A1* | |
### Part (b)(ii):
| Working | Mark | Notes |
|---------|------|-------|
| $\tan2x = 1$ | M1 | |
| $2x = 45°$ | A1 | |
| $2x = 45° + 180°$ | M1 | |
| $x = 22.5°, 112.5°, 202.5°, 292.5°$ | A1(any two), A1 | (5 marks) |
---6. (a) Prove that
$$\frac { 1 } { \sin 2 \theta } - \frac { \cos 2 \theta } { \sin 2 \theta } = \tan \theta , \quad \theta \neq 90 n ^ { \circ } , n \in \mathbb { Z }$$
(b) Hence, or otherwise,
\begin{enumerate}[label=(\roman*)]
\item show that $\tan 15 ^ { \circ } = 2 - \sqrt { 3 }$,
\item solve, for $0 < x < 360 ^ { \circ }$,
$$\operatorname { cosec } 4 x - \cot 4 x = 1$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2011 Q6 [12]}}