| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Standard +0.3 This is a slightly above-average Core 3/4 question requiring knowledge of reciprocal trig identities (cot²x + 1 = cosec²x) and double angle formulas. Part (a) is routine algebraic manipulation. Part (b) requires converting cos2x using the double angle formula to get a quadratic in cosx, then solving—a standard technique. The multi-step nature and combination of identities makes it moderately challenging but still follows predictable patterns for C3/4 material. |
| Spec | 1.05k 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 intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Way 1: \(\frac{\cot^2 x}{1+\cot^2 x} \equiv \frac{\cos^2 x/\sin^2 x}{\csc^2 x} \equiv \frac{\cos^2 x/\sin^2 x}{1/\sin^2 x} \equiv \cos^2 x\) | M1, M1, A1* | Uses one valid identity; uses two valid identities; completes proof |
| Way 2: \(\frac{\cot^2 x}{1+\cot^2 x} \times \frac{\sin^2 x}{\sin^2 x} \equiv \frac{\cos^2 x}{\sin^2 x+\cos^2 x} \equiv \frac{\cos^2 x}{1} \equiv \cos^2 x\) | M1, M1, A1* | |
| Way 3: \(\frac{\cot^2 x}{1+\cot^2 x} \equiv \frac{1/\tan^2 x}{1+1/\tan^2 x} \equiv \frac{1}{1+\tan^2 x} \equiv \frac{1}{\sec^2 x} \equiv \cos^2 x\) | M1, M1, A1* | |
| Way 4: \(\frac{\cot^2 x}{1+\cot^2 x} \equiv \frac{\csc^2 x - 1}{\csc^2 x} \equiv 1-\sin^2 x \equiv \cos^2 x\) | M1, M1, A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\cos^2 x = 8(2\cos^2 x - 1) + 2\cos x\) | M1 | Use part (a) and correct double angle formula \(\cos 2x = 2\cos^2 x - 1\) |
| \(15\cos^2 x + 2\cos x - 8 = 0\) | A1 | Correct three-term quadratic, all terms same side |
| \(\cos x = \frac{2}{3}\) or \(-\frac{4}{5}\) | M1 | Solves quadratic in \(\cos x\) by any method |
| \(x = 48.2°\) or \(143.1°\) or \(216.9°\) or \(311.8°\) | dM1, A1, A1 | dM1 for one correct inverse cos; A1 two correct answers; A1 all four correct answers |
# Question 9(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| **Way 1:** $\frac{\cot^2 x}{1+\cot^2 x} \equiv \frac{\cos^2 x/\sin^2 x}{\csc^2 x} \equiv \frac{\cos^2 x/\sin^2 x}{1/\sin^2 x} \equiv \cos^2 x$ | M1, M1, A1* | Uses one valid identity; uses two valid identities; completes proof |
| **Way 2:** $\frac{\cot^2 x}{1+\cot^2 x} \times \frac{\sin^2 x}{\sin^2 x} \equiv \frac{\cos^2 x}{\sin^2 x+\cos^2 x} \equiv \frac{\cos^2 x}{1} \equiv \cos^2 x$ | M1, M1, A1* | |
| **Way 3:** $\frac{\cot^2 x}{1+\cot^2 x} \equiv \frac{1/\tan^2 x}{1+1/\tan^2 x} \equiv \frac{1}{1+\tan^2 x} \equiv \frac{1}{\sec^2 x} \equiv \cos^2 x$ | M1, M1, A1* | |
| **Way 4:** $\frac{\cot^2 x}{1+\cot^2 x} \equiv \frac{\csc^2 x - 1}{\csc^2 x} \equiv 1-\sin^2 x \equiv \cos^2 x$ | M1, M1, A1* | |
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# Question 9(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos^2 x = 8(2\cos^2 x - 1) + 2\cos x$ | M1 | Use part (a) and correct double angle formula $\cos 2x = 2\cos^2 x - 1$ |
| $15\cos^2 x + 2\cos x - 8 = 0$ | A1 | Correct three-term quadratic, all terms same side |
| $\cos x = \frac{2}{3}$ or $-\frac{4}{5}$ | M1 | Solves quadratic in $\cos x$ by any method |
| $x = 48.2°$ or $143.1°$ or $216.9°$ or $311.8°$ | dM1, A1, A1 | dM1 for one correct inverse cos; A1 two correct answers; A1 all four correct answers |
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\item (a) Show that
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$$\frac { \cot ^ { 2 } x } { 1 + \cot ^ { 2 } x } \equiv \cos ^ { 2 } x$$
(b) Hence solve, for $0 \leqslant x < 360 ^ { \circ }$,
$$\frac { \cot ^ { 2 } x } { 1 + \cot ^ { 2 } x } = 8 \cos 2 x + 2 \cos x$$
Give each solution in degrees to one decimal place.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
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\hfill \mbox{\textit{Edexcel C34 2018 Q9 [9]}}