| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Prove trig identity then solve |
| Difficulty | Moderate -0.3 This is a straightforward two-part question requiring basic trigonometric manipulation (tan = sin/cos and the Pythagorean identity) followed by solving a simple quadratic equation in sin θ. The algebraic steps are routine and the question explicitly guides students through the method, making it slightly easier than average for A-level. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks |
|---|---|
| \(\sin\theta = \cos^2\theta\) and completion to given result | M1 |
| A1 |
| Answer | Marks |
|---|---|
| \(\sin^2\theta + \sin\theta - 1 = 0\) | M1 |
| \([\sin\theta =] \frac{-1 \pm \sqrt{5}}{2}\) oe, may be implied by correct answers | A1 |
| \([\theta =] 38.17...,\) or \(38.2\) and \(141.83...,\) or \(141.8\) or \(142\) | A1 |
**Question 2:**
**(i)**
$\sin\theta = \cos^2\theta$ and completion to given result | M1
| A1
[2]
**(ii)**
$\sin^2\theta + \sin\theta - 1 = 0$ | M1
$[\sin\theta =] \frac{-1 \pm \sqrt{5}}{2}$ oe, may be implied by correct answers | A1
$[\theta =] 38.17...,$ or $38.2$ and $141.83...,$ or $141.8$ or $142$ | A1
[3]
**Guidance notes:**
Allow $1$ on RHS if attempt to complete square
May be implied by correct answers
Ignore extra values outside range; $A0$ if extra values in range or in radians
NB $0.6662$ and $2.4754$ if working in radian mode earns M1A1A0
Condone $y^2 + y - 1 = 0$, mark to benefit of candidate
Ignore any work with negative root and condone omission of negative root with no comment, e.g. M1 for $0.618...$
If unsupported, B1 for one of these, B2 for both. If both values correct with extra values in range, then B1.
NB $0.6662$ and $2.4754$ to $3$sf or more2 (i) Show that the equation $\frac { \tan \theta } { \cos \theta } = 1$ may be rewritten as $\sin \theta = 1 - \sin ^ { 2 } \theta$.\\
(ii) Hence solve the equation $\frac { \tan \theta } { \cos \theta } = 1$ for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.
\hfill \mbox{\textit{OCR MEI C2 Q2 [5]}}