| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: sin²/cos² substitution |
| Difficulty | Standard +0.3 This is a standard C2 trigonometric equation requiring algebraic manipulation to convert to quadratic form (part a is scaffolded as 'show that'), then solving using the quadratic formula and considering the restricted domain. The manipulation uses cos²θ = 1 - sin²θ, which is routine at this level. Slightly above average difficulty due to the fractional form and two-part structure, but well within expected C2 scope with clear guidance. |
| 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/Working | Marks | Guidance |
| \(\frac{5+\sin\theta}{3\cos\theta}=2\cos\theta \Rightarrow 5+\sin\theta=6\cos^2\theta\) | M1 | Attempts to cross multiply to form equation in form \(5+\sin\theta=A\cos^2\theta\) |
| \(\Rightarrow 5+\sin\theta=6(1-\sin^2\theta) \Rightarrow 6\sin^2\theta+\sin\theta-1=0\) | dM1 | Dependent on previous M. Uses \(\cos^2\theta=\pm1\pm\sin^2\theta\) to get equation in just \(\sin\theta\) |
| Given answer confirmed | A1* | All aspects must be correct. Mixed variables lose mark. \(\cos^2\theta\) written as \(\cos^2\) gives M1 dM1 A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(6\sin^2\theta+\sin\theta-1=0 \Rightarrow (3\sin\theta-1)(2\sin\theta+1)=0\) | M1 | Attempts to factorise; accept formula or calculator. Accept answers in terms of \(x\) |
| \((\sin\theta)=+\frac{1}{3}, -\frac{1}{2}\) | A1 | Values do not need to be simplified; can be implied by correct \(\theta\) |
| \(\theta=19.5°, -30°\) | dM1, A1 | dM1: calculates at least one \(\theta\) from their \(\sin\theta\). A1: both values, no additional solutions in range. Condone \(-30.0°\) |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{5+\sin\theta}{3\cos\theta}=2\cos\theta \Rightarrow 5+\sin\theta=6\cos^2\theta$ | M1 | Attempts to cross multiply to form equation in form $5+\sin\theta=A\cos^2\theta$ |
| $\Rightarrow 5+\sin\theta=6(1-\sin^2\theta) \Rightarrow 6\sin^2\theta+\sin\theta-1=0$ | dM1 | Dependent on previous M. Uses $\cos^2\theta=\pm1\pm\sin^2\theta$ to get equation in just $\sin\theta$ |
| Given answer confirmed | A1* | All aspects must be correct. Mixed variables lose mark. $\cos^2\theta$ written as $\cos^2$ gives M1 dM1 A0 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $6\sin^2\theta+\sin\theta-1=0 \Rightarrow (3\sin\theta-1)(2\sin\theta+1)=0$ | M1 | Attempts to factorise; accept formula or calculator. Accept answers in terms of $x$ |
| $(\sin\theta)=+\frac{1}{3}, -\frac{1}{2}$ | A1 | Values do not need to be simplified; can be implied by correct $\theta$ |
| $\theta=19.5°, -30°$ | dM1, A1 | dM1: calculates at least one $\theta$ from their $\sin\theta$. A1: both values, no additional solutions in range. Condone $-30.0°$ |
---8. In this question the angle $\theta$ is measured in degrees throughout.
\begin{enumerate}[label=(\alph*)]
\item Show that the equation
$$\frac { 5 + \sin \theta } { 3 \cos \theta } = 2 \cos \theta , \quad \theta \neq ( 2 n + 1 ) 90 ^ { \circ } , \quad n \in \mathbb { Z }$$
may be rewritten as
$$6 \sin ^ { 2 } \theta + \sin \theta - 1 = 0$$
\item Hence solve, for $- 90 ^ { \circ } < \theta < 90 ^ { \circ }$, the equation
$$\frac { 5 + \sin \theta } { 3 \cos \theta } = 2 \cos \theta$$
Give your answers to one decimal place, where appropriate.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q8 [7]}}