| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Solve 2sinθ = tanθ type equation |
| Difficulty | Standard +0.3 Part (a) is a standard trig equation requiring double angle formula and algebraic manipulation (sin 2θ = 2sin θ cos θ, then factoring). Part (b) tests understanding of transformations via simple substitution. This is routine A-level content with straightforward steps, slightly easier than average due to the direct deduction in part (b). |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5\sin 2\theta = 9\tan\theta \Rightarrow 10\sin\theta\cos\theta = 9\times\dfrac{\sin\theta}{\cos\theta}\) leading to \(A\cos^2\theta=B\) or \(C\sin^2\theta=D\) | M1 | Must use \(\sin 2\theta = 2\sin\theta\cos\theta\) and \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\); allow \(P\cos^2\theta\sin\theta = Q\sin\theta\) |
| Correct simplified equation in one trig function e.g. \(10\cos^2\theta = 9\) or \(10\sin^2\theta = 1\) | A1 | Also accept \(10 = 9\sec^2\theta\) |
| Correct order of operations: \(10\cos^2\theta = 9 \Rightarrow \theta = \arccos\left(\pm\sqrt{\dfrac{9}{10}}\right)\) | dM1 | Dependent on previous M; square root before inverse trig |
| Any one value from \(\theta = \pm 18.4°, \pm 161.6°\) | A1 | Allow awrt \(\pm 0.32\) rad or \(\pm 2.82\) rad |
| All four values: \(\theta =\) awrt \(\pm 18.4°, \pm 161.6°\) and no others apart from \(0°, \pm 180°\) | A1 | |
| \(\theta = 0°, \pm 180°\) | B1 | Scored independently of method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to solve \(x - 25° = -18.4°\) | M1 | Where \(\theta\) is a solution from their part (a) |
| \(x = 6.6°\) | A1ft | Follow through on their \(\theta + 25°\) where \(-25 < \theta < 0\); if multiple answers given, correct value must be chosen |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5\sin 2\theta = 9\tan\theta \Rightarrow 10\sin\theta\cos\theta = 9\times\dfrac{\sin\theta}{\cos\theta}$ leading to $A\cos^2\theta=B$ or $C\sin^2\theta=D$ | M1 | Must use $\sin 2\theta = 2\sin\theta\cos\theta$ and $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$; allow $P\cos^2\theta\sin\theta = Q\sin\theta$ |
| Correct simplified equation in one trig function e.g. $10\cos^2\theta = 9$ or $10\sin^2\theta = 1$ | A1 | Also accept $10 = 9\sec^2\theta$ |
| Correct order of operations: $10\cos^2\theta = 9 \Rightarrow \theta = \arccos\left(\pm\sqrt{\dfrac{9}{10}}\right)$ | dM1 | Dependent on previous M; square root before inverse trig |
| Any one value from $\theta = \pm 18.4°, \pm 161.6°$ | A1 | Allow awrt $\pm 0.32$ rad or $\pm 2.82$ rad |
| All four values: $\theta =$ awrt $\pm 18.4°, \pm 161.6°$ and no others apart from $0°, \pm 180°$ | A1 | |
| $\theta = 0°, \pm 180°$ | B1 | Scored independently of method |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to solve $x - 25° = -18.4°$ | M1 | Where $\theta$ is a solution from their part (a) |
| $x = 6.6°$ | A1ft | Follow through on their $\theta + 25°$ where $-25 < \theta < 0$; if multiple answers given, correct value must be chosen |
---\begin{enumerate}
\item (a) Solve, for $- 180 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }$, the equation
\end{enumerate}
$$5 \sin 2 \theta = 9 \tan \theta$$
giving your answers, where necessary, to one decimal place.\\[0pt]
[Solutions based entirely on graphical or numerical methods are not acceptable.]\\
(b) Deduce the smallest positive solution to the equation
$$5 \sin \left( 2 x - 50 ^ { \circ } \right) = 9 \tan \left( x - 25 ^ { \circ } \right)$$
\hfill \mbox{\textit{Edexcel Paper 1 2019 Q6 [8]}}