| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| 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 Pythagorean identity and double angle formula. Part (i) requires basic manipulation ($\sin\theta = -\sqrt{1-k^2}$ for reflex angle, then $\tan\theta = \sin\theta/\cos\theta$), and part (ii) tests understanding that reflex angles place $\theta$ in quadrant III or IV where $\sin 2\theta = 2\sin\theta\cos\theta$ must be negative. Easier than average due to routine algebraic manipulation and standard reasoning about quadrants. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (i) (a) \(\sin\theta = -\sqrt{1-k^2}\) | B1 B1 [2] | (−) B1 rest B1 |
| (b) Uses \(t=s/c \to \frac{-\sqrt{1-k^2}}{k}\) | B1✓ [1] | ✓ for (i) ÷ k. |
| (ii) \(\theta\) is in 4th quadrant. \(2\theta\) lies between 540° and 720° \(\sin2\theta\) is negative in both these quadrants. | B1 B1 [2] | co co |
reflex angle $\theta$ is such that $\cos\theta = k,$
**(i) (a)** $\sin\theta = -\sqrt{1-k^2}$ | B1 B1 [2] | (−) B1 rest B1
**(b)** Uses $t=s/c \to \frac{-\sqrt{1-k^2}}{k}$ | B1✓ [1] | ✓ for (i) ÷ k.
**(ii)** $\theta$ is in 4th quadrant. $2\theta$ lies between 540° and 720° $\sin2\theta$ is negative in both these quadrants. | B1 B1 [2] | co co3 The reflex angle $\theta$ is such that $\cos \theta = k$, where $0 < k < 1$.\\
(i) Find an expression, in terms of $k$, for
\begin{enumerate}[label=(\alph*)]
\item $\sin \theta$,
\item $\tan \theta$.\\
(ii) Explain why $\sin 2 \theta$ is negative for $0 < k < 1$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2014 Q3 [5]}}