| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: tan/sin/cos identity manipulation |
| Difficulty | Moderate -0.3 This is a standard C2 trigonometric equation requiring routine manipulation (converting tan to sin/cos, using sin²θ + cos²θ = 1) to reach a given quadratic form, then solving by factorization. The proof is guided by providing the target form, and the quadratic factors easily. Slightly easier than average due to the structured guidance and straightforward algebraic steps. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin\theta\tan\theta = \sin\theta \times \frac{\sin\theta}{\cos\theta} = \frac{1-\cos^2\theta}{\cos\theta}\) | M1 | For use of \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) |
| M1 | For use of \(\cos^2\theta + \sin^2\theta = 1\) | |
| Hence \(1-\cos^2\theta = \cos\theta(\cos\theta+1)\), i.e. \(2\cos^2\theta + \cos\theta - 1 = 0\), or equiv | A1 | 3 For showing given equation correctly |
| Answer | Marks | Guidance |
|---|---|---|
| \((2\cos\theta-1)(\cos\theta+1)=0\) | M1 | For solution of quadratic equation in \(\cos\theta\) |
| Hence \(\cos\theta = \frac{1}{2}\) or \(-1\) | A1 | For both values of \(\cos\theta\) correct |
| So \(\theta = 60°\) | A1 | For correct answer \(60°\) |
| \(300°, 180°\) | A1 | For correct answer \(180°\) |
| A1\(\checkmark\) | 5 For a correct non-principal-value answer, following their value of \(\cos\theta\) (excluding \(\cos\theta = -1, 0, 1\)) and no other values for \(\theta\) |
# Question 5:
## Part (i):
| $\sin\theta\tan\theta = \sin\theta \times \frac{\sin\theta}{\cos\theta} = \frac{1-\cos^2\theta}{\cos\theta}$ | M1 | For use of $\tan\theta = \frac{\sin\theta}{\cos\theta}$ |
|---|---|---|
| | M1 | For use of $\cos^2\theta + \sin^2\theta = 1$ |
| Hence $1-\cos^2\theta = \cos\theta(\cos\theta+1)$, i.e. $2\cos^2\theta + \cos\theta - 1 = 0$, or equiv | A1 | **3** For showing given equation correctly |
## Part (ii):
| $(2\cos\theta-1)(\cos\theta+1)=0$ | M1 | For solution of quadratic equation in $\cos\theta$ |
|---|---|---|
| Hence $\cos\theta = \frac{1}{2}$ or $-1$ | A1 | For both values of $\cos\theta$ correct |
| So $\theta = 60°$ | A1 | For correct answer $60°$ |
| $300°, 180°$ | A1 | For correct answer $180°$ |
| | A1$\checkmark$ | **5** For a correct non-principal-value answer, following their value of $\cos\theta$ (excluding $\cos\theta = -1, 0, 1$) **and** no other values for $\theta$ |
---5 (i) Prove that the equation
$$\sin \theta \tan \theta = \cos \theta + 1$$
can be expressed in the form
$$2 \cos ^ { 2 } \theta + \cos \theta - 1 = 0$$
(ii) Hence solve the equation
$$\sin \theta \tan \theta = \cos \theta + 1$$
giving all values of $\theta$ between $0 ^ { \circ }$ and $360 ^ { \circ }$.
\hfill \mbox{\textit{OCR C2 2005 Q5 [8]}}