| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Given one function find others |
| Difficulty | Standard +0.3 This is a multi-part question requiring the identity sec²θ = 1 + tan²θ and algebraic manipulation to form a quadratic in m, followed by straightforward application of right-angled triangle methods. While it has several steps (5 marks total), each technique is standard for C3/C4 level with no novel insight required—slightly easier than average due to the guided structure. |
| 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^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sec^2 x = 1 + m^2\) or \(\sec^2 y = 1 + \left(\frac{8m+5}{4}\right)^2\) or \(16\sec^2 y = 16 + 16(8m+5)^2\) | M1 | Attempts to express \(\sec^2 x\) or \(\sec^2 y\) in terms of \(m\) using a correct identity |
| \(16(\sec^2 x + \sec^2 y) = 16\left(1 + m^2 + 1 + \left(\frac{8m+5}{4}\right)^2\right) = 537\) | M1 | Uses expressions in \(m\) and 537 to obtain quadratic in \(m\) (may be unsimplified) |
| \(m^2 + m - 6 = 0 \Rightarrow m = 2, -3\) | M1 | Solves their 3TQ as far as \(m = \ldots\) |
| \(m = 2, -3\) | A1 | Correct values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan x = 2 \Rightarrow \sin x = \frac{2}{\sqrt{1^2 + 2^2}} = \ldots\) | M1 | Correct method for value of \(\sin x\). Must be from appropriate identity or exact work. e.g. \(\sin(\tan^{-1} 2) = 0.8944\ldots\) scores M0. Can be for either value of \(m\) |
| \(= \frac{2}{\sqrt{5}}\) | A1 | cao (oe) and no other values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan y = \frac{21}{4} \Rightarrow \cot y = \frac{4}{21}\) | M1 | Uses \(4\tan y = 8m+5\) and their \(m\) to find value of \(\tan y\) and finds reciprocal. Can be for either value of \(m\) |
| \(\cot y = \frac{4}{21}\) | A1 | cao (oe) and no other values |
## Question 5:
Given: $\tan x = m$ and $4\tan y = 8m + 5$
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sec^2 x = 1 + m^2$ or $\sec^2 y = 1 + \left(\frac{8m+5}{4}\right)^2$ or $16\sec^2 y = 16 + 16(8m+5)^2$ | M1 | Attempts to express $\sec^2 x$ or $\sec^2 y$ in terms of $m$ using a correct identity |
| $16(\sec^2 x + \sec^2 y) = 16\left(1 + m^2 + 1 + \left(\frac{8m+5}{4}\right)^2\right) = 537$ | M1 | Uses expressions in $m$ and 537 to obtain quadratic in $m$ (may be unsimplified) |
| $m^2 + m - 6 = 0 \Rightarrow m = 2, -3$ | M1 | Solves their 3TQ as far as $m = \ldots$ |
| $m = 2, -3$ | A1 | Correct values |
**(4 marks)**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan x = 2 \Rightarrow \sin x = \frac{2}{\sqrt{1^2 + 2^2}} = \ldots$ | M1 | Correct method for value of $\sin x$. Must be from appropriate identity or exact work. e.g. $\sin(\tan^{-1} 2) = 0.8944\ldots$ scores M0. Can be for either value of $m$ |
| $= \frac{2}{\sqrt{5}}$ | A1 | cao (oe) **and no other values** |
**(2 marks)**
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan y = \frac{21}{4} \Rightarrow \cot y = \frac{4}{21}$ | M1 | Uses $4\tan y = 8m+5$ and their $m$ to find value of $\tan y$ and finds reciprocal. Can be for either value of $m$ |
| $\cot y = \frac{4}{21}$ | A1 | cao (oe) **and no other values** |
**(2 marks) — Total: 8**
---5. The angle $x$ and the angle $y$ are such that
$$\tan x = m \text { and } 4 \tan y = 8 m + 5$$
where $m$ is a constant.\\
Given that $16 \sec ^ { 2 } x + 16 \sec ^ { 2 } y = 537$
\begin{enumerate}[label=(\alph*)]
\item find the two possible values of $m$.
Given that the angle $x$ and the angle $y$ are acute, find the exact value of
\item $\sin x$
\item $\cot y$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2019 Q5 [8]}}