| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Sketch reciprocal function graphs |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard knowledge of reciprocal trig functions and double angle identities. Part (a) requires sketching cosec x (routine for C3) and using symmetry to relate two angles. Part (b) involves recalling the tan 2θ identity and algebraic manipulation with cot/tan relationships. All techniques are standard C3 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.05f Trigonometric function graphs: symmetries and periodicities1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05l Double angle formulae: and compound angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) Draw at least two correctly shaped branches, one for \(y > 0\), one for \(y < 0\) | M1 | otherwise located anywhere including \(x < 0\) now |
| Draw four correct branches | M1 | (more or less) correctly located; |
| Draw (more or less) correct graph | A1 | 3 with some indication of horiz scale (perhaps only 4\(x\) indicated); with asymptotic behaviour shown (but not too fussy about branch drifting slightly away from asymptotic value nor about branch touching asymptote) but branches must not obviously cross asymptotic value; with −1 and 1 shown (or implied by presence of sine curve or by presence of only one of them on a reasonably accurate sketch); no need for vertical (dotted) lines drawn to indicate asymptotic values |
| (a)(ii) State expression of form \(k\pi + \alpha\) or \(k\pi - \alpha\) or \(\alpha = k\pi + \beta\) or \(\alpha = k\pi - \beta\) | M1 | any non-zero numerical value of \(k\); M0 if degrees used |
| State \(3\pi - \alpha\) | A1 | 2 or unsimplified equiv |
| (b)(i) State \(\frac{2\tan\theta}{1 - \tan^2\theta}\) | B1 | 1 or equiv such as \(\frac{t + t}{1 - tx t}\) or \(\frac{2\tan A}{1 - \tan^2 A}\) |
| (b)(ii) State or imply \(\tan\phi = \frac{1}{4}\) | B1 | or equiv such as \(\frac{1}{\tan\phi} = 4\) |
| Attempt to evaluate \(\tan 2\phi\) or cot \(2\phi\) | M1 | perhaps within attempt at complete expression but using correct identity |
| Obtain \(\tan 2\phi = \frac{8}{15}\) or cot \(2\phi = \frac{15}{8}\) | A1 | or (unsimplified) equiv; may be implied |
| Attempt to evaluate value of \(\tan 4\phi\) | M1 | perhaps within attempt at complete expression; condone only minor slip(s) in use of relevant identity |
| Obtain \(\frac{240}{161}\) | A1 | or (unsimplified) exact equiv; may be implied |
| Obtain final answer \(\frac{225}{322}\) | A1 | 6 or exact equiv |
**(a)(i)** Draw at least two correctly shaped branches, one for $y > 0$, one for $y < 0$ | M1 | otherwise located anywhere including $x < 0$ now
Draw four correct branches | M1 | (more or less) correctly located;
Draw (more or less) correct graph | A1 | 3 with some indication of horiz scale (perhaps only 4$x$ indicated); with asymptotic behaviour shown (but not too fussy about branch drifting slightly away from asymptotic value nor about branch touching asymptote) but branches must not obviously cross asymptotic value; with −1 and 1 shown (or implied by presence of sine curve or by presence of only one of them on a reasonably accurate sketch); no need for vertical (dotted) lines drawn to indicate asymptotic values
**(a)(ii)** State expression of form $k\pi + \alpha$ or $k\pi - \alpha$ or $\alpha = k\pi + \beta$ or $\alpha = k\pi - \beta$ | M1 | any non-zero numerical value of $k$; M0 if degrees used
State $3\pi - \alpha$ | A1 | 2 or unsimplified equiv
**(b)(i)** State $\frac{2\tan\theta}{1 - \tan^2\theta}$ | B1 | 1 or equiv such as $\frac{t + t}{1 - tx t}$ or $\frac{2\tan A}{1 - \tan^2 A}$
**(b)(ii)** State or imply $\tan\phi = \frac{1}{4}$ | B1 | or equiv such as $\frac{1}{\tan\phi} = 4$
Attempt to evaluate $\tan 2\phi$ or cot $2\phi$ | M1 | perhaps within attempt at complete expression but using correct identity
Obtain $\tan 2\phi = \frac{8}{15}$ or cot $2\phi = \frac{15}{8}$ | A1 | or (unsimplified) equiv; may be implied
Attempt to evaluate value of $\tan 4\phi$ | M1 | perhaps within attempt at complete expression; condone only minor slip(s) in use of relevant identity
Obtain $\frac{240}{161}$ | A1 | or (unsimplified) exact equiv; may be implied
Obtain final answer $\frac{225}{322}$ | A1 | 6 or exact equiv
[SC – (use of calculator and little or no working) State or imply $\tan\phi = \frac{1}{4}$ B1; Obtain $\tan 2\phi = \frac{8}{15}$ B1; State or imply $\tan\phi = \frac{1}{4}$ B1; Obtain $\frac{225}{322}$ B2 (max 3/6)]
---8
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Sketch the graph of $y = \operatorname { cosec } x$ for $0 < x < 4 \pi$.
\item It is given that $\operatorname { cosec } \alpha = \operatorname { cosec } \beta$, where $\frac { 1 } { 2 } \pi < \alpha < \pi$ and $2 \pi < \beta < \frac { 5 } { 2 } \pi$. By using your sketch, or otherwise, express $\beta$ in terms of $\alpha$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Write down the identity giving $\tan 2 \theta$ in terms of $\tan \theta$.
\item Given that $\cot \phi = 4$, find the exact value of $\tan \phi \cot 2 \phi \tan 4 \phi$, showing all your working.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR C3 2011 Q8 [12]}}