| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Express in terms of one function |
| Difficulty | Moderate -0.3 This is a slightly below-average A-level question. Part (a)(i) requires routine manipulation using tan²θ = sin²θ/cos²θ and the Pythagorean identity, which is standard bookwork. Part (a)(ii) is a straightforward solve following from (i). Part (b) involves basic intersection of sin and cos graphs using tanx = 1/2, which is standard technique. All parts are multi-step but use well-practiced methods with no novel insight required. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\tan^2\theta - 1}{\tan^2\theta + 1} = \frac{\frac{\sin^2\theta}{\cos^2\theta}-1}{\frac{\sin^2\theta}{\cos^2\theta}+1}\) | M1 | |
| \(= \frac{\sin^2\theta - \cos^2\theta}{\sin^2\theta + \cos^2\theta}\) | A1 | Multiplying by \(\cos^2\theta\). Intermediate stage can be omitted by multiplying directly by \(\cos^2\theta\) |
| \(= \sin^2\theta - \cos^2\theta = \sin^2\theta - (1-\sin^2\theta) = 2\sin^2\theta - 1\) | A1 | Using \(\sin^2\theta + \cos^2\theta = 1\) twice. Accept \(a=2,\ b=-1\) |
| ALT 1: \(\frac{\sec^2\theta - 2}{\sec^2\theta}\) | M1 | ALT 2: \(\frac{\tan^2\theta - 1}{\sec^2\theta}\) |
| \(1 - \frac{2}{\sec^2\theta} = 1 - 2\cos^2\theta\) | A1 | \((\tan^2\theta - 1)\cos^2\theta\) |
| \(1 - 2(1-\sin^2\theta) = 2\sin^2\theta - 1\) | A1 | \(\sin^2\theta - \cos^2\theta = \sin^2\theta - (1-\sin^2\theta) = 2\sin^2\theta - 1\) |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2\sin^2\theta - 1 = \frac{1}{4} \to \sin\theta = (\pm)\sqrt{\frac{5}{8}}\) or \((\pm)0.7906\) | M1 | OR \(\frac{t^2-1}{t^2+1} = \frac{1}{4} \to 3t^2 = 5 \to t = (\pm)\sqrt{\frac{5}{3}}\) or \(t = (\pm)1.2910\) |
| \(\theta = -52.2°\) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin x = 2\cos x \rightarrow \tan x = 2\) | M1 | Or \(\sin x = \sqrt{\frac{4}{5}}\) or \(\cos x = \sqrt{\frac{1}{5}}\) |
| \(x = 1.11\) with no additional solutions | A1 | Accept \(0.352\pi\) or \(0.353\pi\). Accept in co-ord form ignoring \(y\) co-ord |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Negative answer in range \(-1 < y < -0.8\) | B1 | |
| \(-0.894\) or \(-0.895\) or \(-0.896\) | B1 | |
| 2 |
## Question 7(a)(i):
| $\frac{\tan^2\theta - 1}{\tan^2\theta + 1} = \frac{\frac{\sin^2\theta}{\cos^2\theta}-1}{\frac{\sin^2\theta}{\cos^2\theta}+1}$ | M1 | |
|---|---|---|
| $= \frac{\sin^2\theta - \cos^2\theta}{\sin^2\theta + \cos^2\theta}$ | A1 | Multiplying by $\cos^2\theta$. Intermediate stage can be omitted by multiplying directly by $\cos^2\theta$ |
| $= \sin^2\theta - \cos^2\theta = \sin^2\theta - (1-\sin^2\theta) = 2\sin^2\theta - 1$ | A1 | Using $\sin^2\theta + \cos^2\theta = 1$ twice. Accept $a=2,\ b=-1$ |
| ALT 1: $\frac{\sec^2\theta - 2}{\sec^2\theta}$ | M1 | ALT 2: $\frac{\tan^2\theta - 1}{\sec^2\theta}$ |
| $1 - \frac{2}{\sec^2\theta} = 1 - 2\cos^2\theta$ | A1 | $(\tan^2\theta - 1)\cos^2\theta$ |
| $1 - 2(1-\sin^2\theta) = 2\sin^2\theta - 1$ | A1 | $\sin^2\theta - \cos^2\theta = \sin^2\theta - (1-\sin^2\theta) = 2\sin^2\theta - 1$ |
| **Total: 3** | | |
## Question 7(a)(ii):
| $2\sin^2\theta - 1 = \frac{1}{4} \to \sin\theta = (\pm)\sqrt{\frac{5}{8}}$ or $(\pm)0.7906$ | M1 | OR $\frac{t^2-1}{t^2+1} = \frac{1}{4} \to 3t^2 = 5 \to t = (\pm)\sqrt{\frac{5}{3}}$ or $t = (\pm)1.2910$ |
|---|---|---|
| $\theta = -52.2°$ | A1 | |
| **Total: 2** | | |
## Question 7(b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin x = 2\cos x \rightarrow \tan x = 2$ | M1 | Or $\sin x = \sqrt{\frac{4}{5}}$ or $\cos x = \sqrt{\frac{1}{5}}$ |
| $x = 1.11$ with no additional solutions | A1 | Accept $0.352\pi$ or $0.353\pi$. Accept in co-ord form ignoring $y$ co-ord |
| | **2** | |
## Question 7(b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Negative answer in range $-1 < y < -0.8$ | B1 | |
| $-0.894$ or $-0.895$ or $-0.896$ | B1 | |
| | **2** | |7
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Express $\frac { \tan ^ { 2 } \theta - 1 } { \tan ^ { 2 } \theta + 1 }$ in the form $a \sin ^ { 2 } \theta + b$, where $a$ and $b$ are constants to be found. [3]
\item Hence, or otherwise, and showing all necessary working, solve the equation
$$\frac { \tan ^ { 2 } \theta - 1 } { \tan ^ { 2 } \theta + 1 } = \frac { 1 } { 4 }$$
for $- 90 ^ { \circ } \leqslant \theta \leqslant 0 ^ { \circ }$.
\end{enumerate}\item \\
\includegraphics[max width=\textwidth, alt={}, center]{ea402a1d-3632-4637-9198-2365715b5246-11_549_796_267_717}
The diagram shows the graphs of $y = \sin x$ and $y = 2 \cos x$ for $- \pi \leqslant x \leqslant \pi$. The graphs intersect at the points $A$ and $B$.
\begin{enumerate}[label=(\roman*)]
\item Find the $x$-coordinate of $A$.
\item Find the $y$-coordinate of $B$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2018 Q7 [9]}}