| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 This is a structured three-part question with clear scaffolding. Part (i) requires algebraic manipulation of standard trig identities (routine for C4), part (ii) is a direct substitution, and part (iii) involves solving a quadratic in tan θ with given bounds. While it requires multiple techniques, each step follows naturally from the previous one with no novel insight needed, making it slightly easier than average. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of \(\cos = 1/\sec\) (or \(\sin = 1/\cosec\)) | B1 | Must be used |
| From RHS: \(\frac{1 - \tan\alpha\tan\beta}{\sec\alpha\sec\beta} = \frac{1 - \sin\alpha/\cos\alpha \cdot \sin\beta/\cos\beta}{1/\cos\alpha \cdot 1/\cos\beta}\) | M1 | Substituting and simplifying as far as having no fractions within a fraction. Need more than \(\frac{1-tt}{\sec\sec} = cc - ss\) ie an intermediate step that can lead to \(cc-ss\) |
| \(= \cos\alpha\cos\beta\left(1 - \frac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}\right) = \cos\alpha\cos\beta - \sin\alpha\sin\beta\) | ||
| \(= \cos(\alpha + \beta)\) | A1 | Convincing simplification and correct use of \(\cos(\alpha+\beta)\). Answer given. |
| OR From LHS: \(\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta\) | B1 | \(\cos = 1/\sec\) or \(\sin = 1/\cosec\) used |
| \(= \frac{1}{\sec\alpha\sec\beta} - \sin\alpha\sin\beta = \frac{1 - \sec\alpha\sin\alpha\sec\beta\sin\beta}{\sec\alpha\sec\beta}\) | M1 | Correct angle formula and substitution and simplification to one term. OR e.g. \(\cos\alpha\cos\beta - \sin\alpha\sin\beta = \cos\alpha\cos\beta(1-\tan\alpha\tan\beta)\) |
| \(= \frac{1 - \tan\alpha\tan\beta}{\sec\alpha\sec\beta}\) | A1 | Simplifying to final answer www. Answer given. Or any equivalent work but must have more than \(cc-ss =\) answer. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\beta = \alpha\): \(\cos 2\alpha = \frac{1-\tan^2\alpha}{\sec^2\alpha}\) | M1 | \(\beta = \alpha\) used. Need to see \(\sec^2\alpha\) |
| \(= \frac{1-\tan^2\alpha}{1+\tan^2\alpha}\) | A1 | Use of \(\sec^2\alpha = 1 + \tan^2\alpha\) to give required result. Answer given. |
| OR without Hence: \(\cos 2\alpha = \cos^2\alpha\left(1 - \frac{\sin^2\alpha}{\cos^2\alpha}\right) = \frac{1}{\sec^2\alpha}(1-\tan^2\alpha) = \frac{1-\tan^2\alpha}{1+\tan^2\alpha}\) | M1 A1 | Use of \(\cos 2\alpha = \cos^2\alpha - \sin^2\alpha\). Simplifying and using \(\sec^2\alpha = 1+\tan^2\alpha\) to final answer. Accept working in reverse to show RHS=LHS. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos 2\theta = \frac{1}{2}\) | M1 | Soi or from \(\tan^2\theta = 1/3\) oe from \(\sin^2\theta\) or \(\cos^2\theta\) |
| \(2\theta = 60°, 300°\) | ||
| \(\theta = 30°, 150°\) | A1 | First correct solution |
| A1 | Second correct solution and no others in the range. SC B1 for \(\pi/6\) and \(5\pi/6\) and no others in the range. |
## Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of $\cos = 1/\sec$ (or $\sin = 1/\cosec$) | B1 | Must be **used** |
| From RHS: $\frac{1 - \tan\alpha\tan\beta}{\sec\alpha\sec\beta} = \frac{1 - \sin\alpha/\cos\alpha \cdot \sin\beta/\cos\beta}{1/\cos\alpha \cdot 1/\cos\beta}$ | M1 | Substituting and simplifying as far as having no fractions within a fraction. Need more than $\frac{1-tt}{\sec\sec} = cc - ss$ ie an intermediate step that can lead to $cc-ss$ |
| $= \cos\alpha\cos\beta\left(1 - \frac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}\right) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$ | | |
| $= \cos(\alpha + \beta)$ | A1 | Convincing simplification and correct use of $\cos(\alpha+\beta)$. Answer given. |
| **OR** From LHS: $\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$ | B1 | $\cos = 1/\sec$ or $\sin = 1/\cosec$ **used** |
| $= \frac{1}{\sec\alpha\sec\beta} - \sin\alpha\sin\beta = \frac{1 - \sec\alpha\sin\alpha\sec\beta\sin\beta}{\sec\alpha\sec\beta}$ | M1 | Correct angle formula and substitution and simplification to one term. OR e.g. $\cos\alpha\cos\beta - \sin\alpha\sin\beta = \cos\alpha\cos\beta(1-\tan\alpha\tan\beta)$ |
| $= \frac{1 - \tan\alpha\tan\beta}{\sec\alpha\sec\beta}$ | A1 | Simplifying to final answer www. **Answer given.** Or any equivalent work but must have more than $cc-ss =$ answer. |
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## Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\beta = \alpha$: $\cos 2\alpha = \frac{1-\tan^2\alpha}{\sec^2\alpha}$ | M1 | $\beta = \alpha$ used. Need to see $\sec^2\alpha$ |
| $= \frac{1-\tan^2\alpha}{1+\tan^2\alpha}$ | A1 | Use of $\sec^2\alpha = 1 + \tan^2\alpha$ to give required result. Answer given. |
| **OR** without Hence: $\cos 2\alpha = \cos^2\alpha\left(1 - \frac{\sin^2\alpha}{\cos^2\alpha}\right) = \frac{1}{\sec^2\alpha}(1-\tan^2\alpha) = \frac{1-\tan^2\alpha}{1+\tan^2\alpha}$ | M1 A1 | Use of $\cos 2\alpha = \cos^2\alpha - \sin^2\alpha$. Simplifying and using $\sec^2\alpha = 1+\tan^2\alpha$ to final answer. Accept working in reverse to show RHS=LHS. |
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## Question 2(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos 2\theta = \frac{1}{2}$ | M1 | Soi or from $\tan^2\theta = 1/3$ oe from $\sin^2\theta$ or $\cos^2\theta$ |
| $2\theta = 60°, 300°$ | | |
| $\theta = 30°, 150°$ | A1 | First correct solution |
| | A1 | Second correct solution and no others in the range. SC B1 for $\pi/6$ and $5\pi/6$ and no others in the range. |
---2 (i) Show that $\cos ( \alpha + \beta ) = \frac { 1 - \tan \alpha \tan \beta } { \sec \alpha \sec \beta }$.\\
(ii) Hence show that $\cos 2 \alpha = \frac { 1 - \tan ^ { 2 } \alpha } { 1 + \tan ^ { 2 } \alpha }$.\\
(iii) Hence or otherwise solve the equation $\frac { 1 - \tan ^ { 2 } \theta } { 1 + \tan ^ { 2 } \theta } = \frac { 1 } { 2 }$ for $0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }$.
\hfill \mbox{\textit{OCR MEI C4 Q2 [8]}}