| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Solve 2sinθ = tanθ type equation |
| Difficulty | Standard +0.3 Part (a) requires standard manipulation of a trig equation using tan θ = sin θ/cos θ, leading to a factorised form with straightforward solutions. Part (b) is a routine identity proof using sin²θ + cos²θ = 1. Both parts are slightly above average due to the algebraic manipulation required, but follow standard A-level techniques without requiring novel insight. |
| Spec | 1.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4\sin^2\theta = \frac{\sin^2\theta}{\cos^2\theta}\) | B1 | Not incorrect notation, e.g. \(\left(\frac{\sin}{\cos}\right)^2\theta\) |
| \(\cos^2\theta = \frac{1}{4}\) | M1 | Attempt \(\div\) bs by \(\sin^2\theta\) & \(\sqrt{\text{bs}}\), rearrange to this form. Allow errors |
| \(\cos\theta = \pm\frac{1}{2}\) or \(2\cos\theta = \pm 1\) | ||
| \(\theta = 60°\) | A1 | |
| or \(120°\) | A1 | Allow \(240°\) and/or \(300°\) but no other extras |
| or \(\sin\theta = 0\), \(\theta = 0°\) or \(180°\) | B1 | Allow \(360°\) but no other extras |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4\sin^2\theta\cos^2\theta = \sin^2\theta\) | ||
| \(4\sin^4\theta - 3\sin^2\theta = 0\) | ||
| \(\sin^2\theta = \frac{3}{4}\) | M1 | Attempt use \(s^2+c^2=1\), rearrange to quartic in \(\sin\theta\) & obtain \(\sin^2\theta = \ldots\) or \(\sin\theta = \ldots\) Allow errors |
| \(\sin\theta = \pm\frac{\sqrt{3}}{2}\) | Allow \(\sin\theta = \frac{\sqrt{3}}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1-\cos^2\theta - 1 + \cos\theta}{1-\cos\theta}\) \(\left(\equiv \frac{\cos\theta - \cos^2\theta}{1-\cos\theta}\right)\) | M1 | Use of \(\sin^2\theta + \cos^2\theta = 1\) to obtain correct fraction in \(\cos\) only |
| \(\equiv \frac{\cos\theta(1-\cos\theta)}{1-\cos\theta}\) | A1 | Correct factorised numerator |
| \(\equiv \cos\theta\) | A1 | Must see previous line and result. Allow \(=\) instead of \(\equiv\) throughout. Allow no mention that \(\cos\theta \neq 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1-\cos^2\theta - 1 + \cos\theta}{1-\cos\theta} = \cos\theta\) | M1 | |
| \(-\cos^2\theta + \cos\theta = \cos\theta(1-\cos\theta)\) | A1 | Any correct manipulation of the original identity that finishes with a statement that is correct |
| \(-\cos^2\theta + \cos\theta = \cos\theta - \cos^2\theta\) | A1 |
## Question 3:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4\sin^2\theta = \frac{\sin^2\theta}{\cos^2\theta}$ | B1 | Not incorrect notation, e.g. $\left(\frac{\sin}{\cos}\right)^2\theta$ |
| $\cos^2\theta = \frac{1}{4}$ | M1 | Attempt $\div$ bs by $\sin^2\theta$ & $\sqrt{\text{bs}}$, rearrange to this form. Allow errors |
| $\cos\theta = \pm\frac{1}{2}$ or $2\cos\theta = \pm 1$ | | |
| $\theta = 60°$ | A1 | |
| or $120°$ | A1 | Allow $240°$ and/or $300°$ but no other extras |
| or $\sin\theta = 0$, $\theta = 0°$ or $180°$ | B1 | Allow $360°$ but no other extras |
**Alternative for M1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4\sin^2\theta\cos^2\theta = \sin^2\theta$ | | |
| $4\sin^4\theta - 3\sin^2\theta = 0$ | | |
| $\sin^2\theta = \frac{3}{4}$ | M1 | Attempt use $s^2+c^2=1$, rearrange to quartic in $\sin\theta$ & obtain $\sin^2\theta = \ldots$ or $\sin\theta = \ldots$ Allow errors |
| $\sin\theta = \pm\frac{\sqrt{3}}{2}$ | | Allow $\sin\theta = \frac{\sqrt{3}}{2}$ |
**Summary:** Any largely correct method obtaining $\cos^2\theta = \ldots$ or $\sin^2\theta = \ldots$ B1M1; $60°$ and $120°$ A1A1; $0°$ and $180°$ A1
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1-\cos^2\theta - 1 + \cos\theta}{1-\cos\theta}$ $\left(\equiv \frac{\cos\theta - \cos^2\theta}{1-\cos\theta}\right)$ | M1 | Use of $\sin^2\theta + \cos^2\theta = 1$ to obtain correct fraction in $\cos$ only |
| $\equiv \frac{\cos\theta(1-\cos\theta)}{1-\cos\theta}$ | A1 | Correct factorised numerator |
| $\equiv \cos\theta$ | A1 | Must see previous line and result. Allow $=$ instead of $\equiv$ throughout. Allow no mention that $\cos\theta \neq 1$ |
**Alternative:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1-\cos^2\theta - 1 + \cos\theta}{1-\cos\theta} = \cos\theta$ | M1 | |
| $-\cos^2\theta + \cos\theta = \cos\theta(1-\cos\theta)$ | A1 | Any correct manipulation of the original identity that finishes with a statement that is correct |
| $-\cos^2\theta + \cos\theta = \cos\theta - \cos^2\theta$ | A1 | |
---3 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Solve the equation $4 \sin ^ { 2 } \theta = \tan ^ { 2 } \theta$ for $0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }$.
\item Prove that $\frac { \sin ^ { 2 } \theta - 1 + \cos \theta } { 1 - \cos \theta } \equiv \cos \theta \quad ( \cos \theta \neq 1 )$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q3 [8]}}