| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Multiple independent equations — includes show/prove component |
| Difficulty | Standard +0.3 This is a straightforward two-part trigonometric question slightly easier than average. Part (i) requires dividing to get tan(θ+30°) and solving a basic linear tangent equation. Part (ii) involves routine algebraic manipulation using sin²x + cos²x = 1 to derive tan²x = 4, then solving in all four quadrants. All techniques are standard C2 content with no novel insight required, making it slightly easier than a typical A-level question. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3\sin(\theta+30°)=2\cos(\theta+30°) \Rightarrow \tan(\theta+30°)=\frac{2}{3}\) | M1 | For stating \(\tan(\theta+30°)=k\), \(k\neq 0\); allow even where candidate writes \(\tan(\theta+30°)=\frac{3}{2}\) |
| \(\theta+30°=\arctan\left(\frac{2}{3}\right)=33.69°, 213.69° \Rightarrow \theta=...\) | dM1 | Taking arctan, subtracting 30 and proceeding to \(\theta=...\); do not allow mixed units; scored when they reach \(\theta=26.3°\) |
| \(\theta = 3.69°, 183.69°\) | A1, A1 [4] | A1: \(\theta=3.69°\) or \(183.69°\); A1: \(3.69°\) and \(183.69°\) only in range \(0\to360\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\cos^2 x+2\sin^2 x}{1-\sin^2 x}=5 \Rightarrow \frac{\cos^2 x+2\sin^2 x}{\cos^2 x}=5\) | M1 | Use of \(1-\sin^2 x = \cos^2 x\) or equivalent |
| \(\Rightarrow 1+2\tan^2 x=5\) | M1 | Dividing both terms by \(\cos^2 x\) and using \(\frac{\sin^2 x}{\cos^2 x}=\tan^2 x\) leading to \(\tan^2 x=k\) |
| \(\Rightarrow \tan^2 x=2\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\tan^2 x=2 \Rightarrow \tan x=\pm\sqrt{2}\) | M1 | Taking square root and stating \(\tan x=\sqrt{k}\) or \(\tan x=-\sqrt{k}\); accept decimals |
| \(\Rightarrow x=0.955, 2.186, 4.097, 5.328\) | M1 A1, A1 [7] | M1: taking arctan and finding two of the 4 angles; A1: two of awrt \(x=0.96, 2.19, 4.10, 5.33\); A1: all four angles in radians (no extras in range) awrt \(x=0.955, 2.186, 4.097, 5.328\) |
# Question 12:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3\sin(\theta+30°)=2\cos(\theta+30°) \Rightarrow \tan(\theta+30°)=\frac{2}{3}$ | M1 | For stating $\tan(\theta+30°)=k$, $k\neq 0$; allow even where candidate writes $\tan(\theta+30°)=\frac{3}{2}$ |
| $\theta+30°=\arctan\left(\frac{2}{3}\right)=33.69°, 213.69° \Rightarrow \theta=...$ | dM1 | Taking arctan, subtracting 30 and proceeding to $\theta=...$; do not allow mixed units; scored when they reach $\theta=26.3°$ |
| $\theta = 3.69°, 183.69°$ | A1, A1 [4] | A1: $\theta=3.69°$ or $183.69°$; A1: $3.69°$ and $183.69°$ only in range $0\to360$ |
## Part (ii)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\cos^2 x+2\sin^2 x}{1-\sin^2 x}=5 \Rightarrow \frac{\cos^2 x+2\sin^2 x}{\cos^2 x}=5$ | M1 | Use of $1-\sin^2 x = \cos^2 x$ or equivalent |
| $\Rightarrow 1+2\tan^2 x=5$ | M1 | Dividing both terms by $\cos^2 x$ and using $\frac{\sin^2 x}{\cos^2 x}=\tan^2 x$ leading to $\tan^2 x=k$ |
| $\Rightarrow \tan^2 x=2$ | A1 | cao |
## Part (ii)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tan^2 x=2 \Rightarrow \tan x=\pm\sqrt{2}$ | M1 | Taking square root and stating $\tan x=\sqrt{k}$ or $\tan x=-\sqrt{k}$; accept decimals |
| $\Rightarrow x=0.955, 2.186, 4.097, 5.328$ | M1 A1, A1 [7] | M1: taking arctan and finding two of the 4 angles; A1: two of awrt $x=0.96, 2.19, 4.10, 5.33$; A1: all four angles in radians (no extras in range) awrt $x=0.955, 2.186, 4.097, 5.328$ |
---12. (i) Solve, for $0 < \theta \leqslant 360 ^ { \circ }$,
$$3 \sin \left( \theta + 30 ^ { \circ } \right) = 2 \cos \left( \theta + 30 ^ { \circ } \right)$$
giving your answers, in degrees, to 2 decimal places.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)\\
(ii) (a) Given that
$$\frac { \cos ^ { 2 } x + 2 \sin ^ { 2 } x } { 1 - \sin ^ { 2 } x } = 5$$
show that
$$\tan ^ { 2 } x = k , \quad \text { where } k \text { is a constant. }$$
(b) Hence solve, for $0 < x \leqslant 2 \pi$,
$$\frac { \cos ^ { 2 } x + 2 \sin ^ { 2 } x } { 1 - \sin ^ { 2 } x } = 5$$
giving your answers, in radians, to 3 decimal places.
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel C12 2017 Q12 [11]}}