| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Given one function find others |
| Difficulty | Moderate -0.8 This is a straightforward application of standard trigonometric identities with minimal problem-solving required. Part (a) is direct recall of the reciprocal definition, part (b) uses a compound angle formula, and part (c) requires the double angle formula with Pythagorean identity. All are routine textbook exercises for C3/C4 level, making this easier than average but not trivial since it requires multiple identity applications. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05l Double angle formulae: and compound angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sec\theta° = \frac{1}{\cos\theta°} = \frac{1}{p}\) | B1 | Accept \(\sec\theta° = p^{-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sin(\theta-90)° = \sin\theta°\cos 90° - \cos\theta°\sin 90° = -p\) | M1 | Attempts \(\sin(\theta-90)° = \sin\theta°\cos 90° \pm \cos\theta°\sin 90°\) with \(\cos\theta° = p\) used |
| \(\sin(\theta - 90)° = -p\) | A1 | Allow \(-p\) for both marks as long as no incorrect work used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sin 2\theta° = 2\sin\theta°\cos\theta°\) | B1 | States identity \(\sin 2\theta° = 2\sin\theta°\cos\theta°\) |
| Uses \(\sin\theta° = \sqrt{1-\cos^2\theta°} = \sqrt{1-p^2}\) | M1 | Attempts \(\sin^2\theta + \cos^2\theta = 1\) with \(\cos\theta° = p\) to get \(\sin\theta°\) in terms of \(p\). Allow \(\sin^2\theta + \cos^2\theta = 1 \Rightarrow \sin\theta = \sqrt{1-p}\) as slip |
| \(\Rightarrow \sin 2\theta° = 2p\sqrt{1-p^2}\) | A1 | But NOT \(\sin 2\theta° = \pm 2p\sqrt{1-p^2}\). Final answer; do not isw. Note \(\sin 2\theta° = 2p\sqrt{1-p^2} = 2p(1-p)\) scores B1 M1 A0 |
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sec\theta° = \frac{1}{\cos\theta°} = \frac{1}{p}$ | B1 | Accept $\sec\theta° = p^{-1}$ |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sin(\theta-90)° = \sin\theta°\cos 90° - \cos\theta°\sin 90° = -p$ | M1 | Attempts $\sin(\theta-90)° = \sin\theta°\cos 90° \pm \cos\theta°\sin 90°$ with $\cos\theta° = p$ used |
| $\sin(\theta - 90)° = -p$ | A1 | Allow $-p$ for both marks as long as no incorrect work used |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sin 2\theta° = 2\sin\theta°\cos\theta°$ | B1 | States identity $\sin 2\theta° = 2\sin\theta°\cos\theta°$ |
| Uses $\sin\theta° = \sqrt{1-\cos^2\theta°} = \sqrt{1-p^2}$ | M1 | Attempts $\sin^2\theta + \cos^2\theta = 1$ with $\cos\theta° = p$ to get $\sin\theta°$ in terms of $p$. Allow $\sin^2\theta + \cos^2\theta = 1 \Rightarrow \sin\theta = \sqrt{1-p}$ as slip |
| $\Rightarrow \sin 2\theta° = 2p\sqrt{1-p^2}$ | A1 | But NOT $\sin 2\theta° = \pm 2p\sqrt{1-p^2}$. Final answer; do not isw. Note $\sin 2\theta° = 2p\sqrt{1-p^2} = 2p(1-p)$ scores B1 M1 A0 |
---3. Given\\
$\cos \theta ^ { \circ } = p$, where $p$ is a constant and $\theta ^ { \circ }$ is acute use standard trigonometric identities to find, in terms of $p$,
\begin{enumerate}[label=(\alph*)]
\item $\sec \theta ^ { \circ }$
\item $\sin ( \theta - 90 ) ^ { \circ }$
\item $\sin 2 \theta ^ { \circ }$
Write each answer in its simplest form.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2018 Q3 [6]}}