| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: sin²/cos² substitution |
| Difficulty | Moderate -0.3 This is a standard C2 trigonometric equation requiring the identity cos²θ = 1 - sin²θ to convert to quadratic form, then factorising or using the quadratic formula, followed by routine inverse trig. The 'show that' part (i) is straightforward algebraic manipulation, and part (ii) is a textbook exercise with no novel problem-solving required. Slightly easier than average due to its routine nature and clear structure. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(3\sin^2 \theta + \sin \theta - 2 = 0\) | M1, A1 2 | Use \(\cos^2 \theta = 1 - \sin^2 \theta\). Show given equation correctly. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin\theta = \frac{2}{3}\) or \(-1\) | M1, A1, A1 | Attempt to solve quadratic equation in \(\sin\theta\). Both values of \(\sin\theta\) correct. Correct answer of \(270°\). |
| \(\theta = 42°, 138°, 270°\) | A1, A1, A1∨ 5 | Correct answer of \(42°\). Correct answer of \(270°\) (any extra values for \(\theta\) in required range is max 4/5) (radians is max 4/5). SR: answer only (or no supporting method) is B1 for \(42°\), B1∨ for \(138°\), B1 for \(270°\) |
(i) $3(\cos^2 \theta - \sin^2 \theta) = \sin \theta + 1$
$3 - 3\sin^2 \theta = \sin \theta + 1$
$3\sin^2 \theta + \sin \theta - 2 = 0$ | M1, A1 2 | Use $\cos^2 \theta = 1 - \sin^2 \theta$. Show given equation correctly.
(ii) $(3\sin\theta - 2)(\sin\theta + 1) = 0$
$\sin\theta = \frac{2}{3}$ or $-1$ | M1, A1, A1 | Attempt to solve quadratic equation in $\sin\theta$. Both values of $\sin\theta$ correct. Correct answer of $270°$.
$\theta = 42°, 138°, 270°$ | A1, A1, A1∨ 5 | Correct answer of $42°$. Correct answer of $270°$ (any extra values for $\theta$ in required range is max 4/5) (radians is max 4/5). SR: answer only (or no supporting method) is B1 for $42°$, B1∨ for $138°$, B1 for $270°$5 (i) Show that the equation
$$3 \cos ^ { 2 } \theta = \sin \theta + 1$$
can be expressed in the form
$$3 \sin ^ { 2 } \theta + \sin \theta - 2 = 0$$
(ii) Hence solve the equation
$$3 \cos ^ { 2 } \theta = \sin \theta + 1 ,$$
giving all values of $\theta$ between $0 ^ { \circ }$ and $360 ^ { \circ }$.
\hfill \mbox{\textit{OCR C2 2007 Q5 [7]}}