| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Solve equation using proven identity |
| Difficulty | Standard +0.3 Part (a) requires algebraic manipulation of trigonometric fractions using standard identities (tan²θ + 1 = sec²θ), which is routine for P1 level. Part (b) applies the proven identity to solve a quadratic equation in tan θ, requiring careful attention to the domain restriction but following a standard 'hence' structure with no novel insight needed. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{\sin\theta(\sin\theta - \cos\theta) + \cos\theta(\sin\theta + \cos\theta)}{(\sin\theta + \cos\theta)(\sin\theta - \cos\theta)} \left[= \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta - \cos^2\theta}\right]\) | *M1 | Sight of a correct common denominator, either in one or two fractions, condone missing brackets if recovered. In the numerator condone \(\pm\) sign errors only. |
| \(\dfrac{\dfrac{\sin^2\theta}{\cos^2\theta} + \dfrac{\cos^2\theta}{\cos^2\theta}}{\dfrac{\sin^2\theta}{\cos^2\theta} - \dfrac{\cos^2\theta}{\cos^2\theta}}\) | DM1 | Divide throughout by \(\cos^2\theta\). |
| \(\dfrac{\tan^2\theta + 1}{\tan^2\theta - 1}\) AG | A1 | |
| Alternative method: | ||
| \(\dfrac{\dfrac{\sin^2\theta}{\cos^2\theta} + 1}{\dfrac{\sin^2\theta}{\cos^2\theta} - 1} \times \dfrac{\cos^2\theta}{\cos^2\theta}\) or equivalent step \(\left[= \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta - \cos^2\theta}\right]\) | *M1 | Replace \(\tan^2\theta\) with \(\frac{\sin^2\theta}{\cos^2\theta}\) and multiply top and bottom by \(\cos^2\theta\). Condone \(\pm\) sign errors. |
| Sight of convincing use of partial fractions | DM1 | |
| \(\dfrac{\sin\theta}{\sin\theta + \cos\theta} + \dfrac{\cos\theta}{\sin\theta - \cos\theta}\) AG | A1 | Note: M1 DM1 A1 for working on both sides at the same time and finishing at the same correct expression. M1 DM1 for starting separately and finishing at the same correct expression and A1 if there is a final conclusion e.g. QED. Do not allow cross multiplication. Condone use of s, c and t and omission of \(\theta\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{\tan^2\theta+1}{\tan^2\theta-1}=2 \Rightarrow \tan^2\theta+1=2(\tan^2\theta-1)\) | *M1 | Equate expression from (a) to 2 and clear fraction |
| \(\tan\theta=[\pm]\sqrt{3}\) | DM1 | Simplify as far as \(\tan\theta=\). May be implied by a correct final answer in degrees or radians |
| Alternative: \(\frac{\sin^2\theta+\cos^2\theta}{\sin^2\theta-\cos^2\theta}=2 \Rightarrow 1=2\sin^2\theta-2(1-\sin^2\theta)\) | *M1 | Equate to 2, clear fraction and use trig identities to form equation in \(\sin\theta\) or \(\cos\theta\) only |
| \(\sin\theta=[\pm]\sqrt{\frac{3}{4}}\) or \(\cos\theta=[\pm]\sqrt{\frac{1}{4}}\) | DM1 | Simplify as far as \(\sin\theta=\) or \(\cos\theta=\) |
| \(\theta=\frac{1}{3}\pi, \frac{2}{3}\pi\) | A1 | A1 for either correct answer then A1FT for second value being \(\pi-\)(their first), no others in range \(0\leq\theta\leq\pi\), both values exact and in radians. SC: B1 for \(\theta=60°,120°\) or \(0.333\pi, 0.667\pi\) AWRT or 1.05, 2.09 AWRT |
| A1FT |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{\sin\theta(\sin\theta - \cos\theta) + \cos\theta(\sin\theta + \cos\theta)}{(\sin\theta + \cos\theta)(\sin\theta - \cos\theta)} \left[= \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta - \cos^2\theta}\right]$ | *M1 | Sight of a correct common denominator, either in one or two fractions, condone missing brackets if recovered. In the numerator condone $\pm$ sign errors only. |
| $\dfrac{\dfrac{\sin^2\theta}{\cos^2\theta} + \dfrac{\cos^2\theta}{\cos^2\theta}}{\dfrac{\sin^2\theta}{\cos^2\theta} - \dfrac{\cos^2\theta}{\cos^2\theta}}$ | DM1 | Divide throughout by $\cos^2\theta$. |
| $\dfrac{\tan^2\theta + 1}{\tan^2\theta - 1}$ AG | A1 | |
| **Alternative method:** | | |
| $\dfrac{\dfrac{\sin^2\theta}{\cos^2\theta} + 1}{\dfrac{\sin^2\theta}{\cos^2\theta} - 1} \times \dfrac{\cos^2\theta}{\cos^2\theta}$ or equivalent step $\left[= \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta - \cos^2\theta}\right]$ | *M1 | Replace $\tan^2\theta$ with $\frac{\sin^2\theta}{\cos^2\theta}$ and multiply top and bottom by $\cos^2\theta$. Condone $\pm$ sign errors. |
| Sight of convincing use of partial fractions | DM1 | |
| $\dfrac{\sin\theta}{\sin\theta + \cos\theta} + \dfrac{\cos\theta}{\sin\theta - \cos\theta}$ AG | A1 | **Note:** M1 DM1 A1 for working on both sides at the same time and finishing at the same correct expression. M1 DM1 for starting separately and finishing at the same correct expression and A1 if there is a final conclusion e.g. QED. Do not allow cross multiplication. Condone use of s, c and t and omission of $\theta$. |
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{\tan^2\theta+1}{\tan^2\theta-1}=2 \Rightarrow \tan^2\theta+1=2(\tan^2\theta-1)$ | *M1 | Equate expression from (a) to 2 and clear fraction |
| $\tan\theta=[\pm]\sqrt{3}$ | DM1 | Simplify as far as $\tan\theta=$. May be implied by a correct final answer in degrees or radians |
| **Alternative:** $\frac{\sin^2\theta+\cos^2\theta}{\sin^2\theta-\cos^2\theta}=2 \Rightarrow 1=2\sin^2\theta-2(1-\sin^2\theta)$ | *M1 | Equate to 2, clear fraction and use trig identities to form equation in $\sin\theta$ or $\cos\theta$ only |
| $\sin\theta=[\pm]\sqrt{\frac{3}{4}}$ or $\cos\theta=[\pm]\sqrt{\frac{1}{4}}$ | DM1 | Simplify as far as $\sin\theta=$ or $\cos\theta=$ |
| $\theta=\frac{1}{3}\pi, \frac{2}{3}\pi$ | A1 | A1 for either correct answer then A1FT for second value being $\pi-$(their first), no others in range $0\leq\theta\leq\pi$, both values exact and in radians. **SC:** B1 for $\theta=60°,120°$ or $0.333\pi, 0.667\pi$ AWRT or 1.05, 2.09 AWRT |
| | A1FT | |
---7
\begin{enumerate}[label=(\alph*)]
\item Prove the identity $\frac { \sin \theta } { \sin \theta + \cos \theta } + \frac { \cos \theta } { \sin \theta - \cos \theta } \equiv \frac { \tan ^ { 2 } \theta + 1 } { \tan ^ { 2 } \theta - 1 }$.
\item Hence find the exact solutions of the equation $\frac { \sin \theta } { \sin \theta + \cos \theta } + \frac { \cos \theta } { \sin \theta - \cos \theta } = 2$ for $0 \leqslant \theta \leqslant \pi$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q7 [7]}}