| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Quadratic in sin²/cos²/tan² |
| Difficulty | Moderate -0.8 This is a routine C2 trigonometric equation requiring the standard identity cos²θ = 1 - sin²θ to convert to a single trigonometric function, then basic algebraic manipulation and calculator work. The algebraic transformation is guided in part (a), making part (b) straightforward application with four solutions to find in the given range. Easier than average due to the scaffolding and standard technique. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(3\sin^2\theta - 2\cos^2\theta = 1\) | ||
| \(3\sin^2\theta - 2(1 - \sin^2\theta) = 1\) | M1 | (Use of \(\sin^2\theta + \cos^2\theta = 1\)) |
| \(3\sin^2\theta - 2 + 2\sin^2\theta = 1\) | ||
| \(5\sin^2\theta = 3\) (cso) | AG; A1 (2) | |
| (b) \(\sin^2\theta = \frac{3}{5}\), so \(\sin\theta = (\pm)\sqrt{0.6}\) | M1 | |
| Attempt to solve both \(\sin\theta = +...\) and \(\sin\theta = -\) (may be implied by later work) | M1 | |
| \(\theta = 50.7685°\) (awrt \(\theta = 50.8°\)) (dependent on first M1 only) | A1 | |
| \(\theta (= 180° - 50.7685°) = 129.23...°\) (awrt \(129.2°\)) | M1; A1 √ |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin\theta = -\sqrt{0.6}\) | ||
| \(\theta = 230.7859°\) and \(309.23152°\) | (awrt \(230.8°, 309.2°\)) (both) | M1A1 (7) [9] |
**(a)** $3\sin^2\theta - 2\cos^2\theta = 1$ | |
$3\sin^2\theta - 2(1 - \sin^2\theta) = 1$ | M1 | (Use of $\sin^2\theta + \cos^2\theta = 1$)
$3\sin^2\theta - 2 + 2\sin^2\theta = 1$ | |
$5\sin^2\theta = 3$ (cso) | AG; A1 (2) |
**(b)** $\sin^2\theta = \frac{3}{5}$, so $\sin\theta = (\pm)\sqrt{0.6}$ | M1 |
Attempt to solve both $\sin\theta = +...$ and $\sin\theta = -$ (may be implied by later work) | M1 |
$\theta = 50.7685°$ (awrt $\theta = 50.8°$) (dependent on first M1 only) | A1 |
$\theta (= 180° - 50.7685°) = 129.23...°$ (awrt $129.2°$) | M1; A1 √ |
[f.t. dependent on first M and 3rd M]
$\sin\theta = -\sqrt{0.6}$ | |
$\theta = 230.7859°$ and $309.23152°$ | (awrt $230.8°, 309.2°$) (both) | M1A1 (7) [9] |
**Notes:**
- (a) N.B: AG; need to see at least one line of working after substituting $\cos^2\theta$.
- (b) First M1: Using $5\sin^2\theta = 3$ to find value for $\sin\theta$ or $\theta$
- Second M1: Considering the $-$ value for $\sin\theta$. (usually later)
- First A1: Given for awrt 50.8°. Not dependent on second M.
- Third M1: For $(180 - 50.8°)°$, need not see written down
- Final M1: Dependent on second M (but may be implied by answers)
- Final A1: Requires both values. (no follow through)
- [Finds $\cos^2\theta = k$ $(k = 2/5)$ and so $\cos\theta = (±)...$M1, then mark equivalently]
---\begin{enumerate}[label=(\alph*)]
\item Show that the equation
$$3 \sin^2 \theta - 2 \cos^2 \theta = 1$$
can be written as
$$5 \sin^2 \theta = 3.$$
[2]
\item Hence solve, for $0° \leq \theta < 360°$, the equation
$$3 \sin^2 \theta - 2 \cos^2 \theta = 1,$$
giving your answers to 1 decimal place. [7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2008 Q4 [9]}}