| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Solve equation using proven identity |
| Difficulty | Moderate -0.3 Part (i) is a straightforward algebraic manipulation using the Pythagorean identity (sin²θ + cos²θ = 1) to factor and simplify—a standard textbook exercise. Part (ii) requires substituting the proven identity and solving a quadratic equation in sin²θ, then finding angles in the given range. While multi-step, both parts follow routine procedures with no novel insight required, making this slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(s^2 - c^2\right)\left(s^2 + c^2\right)\) OR \(s^2(1 - c^2) - c^2(1 - s^2)\) | M1 | OR \(\sin^2\theta - (1 - \sin^2\theta)^2\), \(\sin^2\theta - (1 - 2\sin^2\theta + \sin^4\theta) = 2\sin^2\theta - 1\) AG |
| \(\sin^2\theta - \cos^2\theta\), \(2\sin^2\theta - 1\) www | A1 | |
| \(2\sin^2\theta - 1\) www AG | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(2\sin^2\theta - 1 = \frac{1}{2} \Rightarrow \sin\theta = (\pm)\frac{\sqrt{3}}{2}\) or \((\pm)0.866\) | B1 | OR \(\cos 2\theta = -\frac{1}{2} \Rightarrow 2\theta = 120, 240\) etc. |
| \(\theta = 60°\), \(\theta = 120°\) | B1 | Ft for \(180 - \text{their } 60\). Ft for \(180 + \text{their } 60, 360 - \text{their } 60\) |
| \(\theta = 240°, 300°\) | B1 | Allow \(\frac{\pi}{3}, \frac{2\pi}{3}\) etc. Extra sols in range \(–1\) [4] |
### (i)
$\left(s^2 - c^2\right)\left(s^2 + c^2\right)$ OR $s^2(1 - c^2) - c^2(1 - s^2)$ | **M1** | OR $\sin^2\theta - (1 - \sin^2\theta)^2$, $\sin^2\theta - (1 - 2\sin^2\theta + \sin^4\theta) = 2\sin^2\theta - 1$ AG
$\sin^2\theta - \cos^2\theta$, $2\sin^2\theta - 1$ www | **A1** |
$2\sin^2\theta - 1$ www AG | **A1** | [3]
### (ii)
$2\sin^2\theta - 1 = \frac{1}{2} \Rightarrow \sin\theta = (\pm)\frac{\sqrt{3}}{2}$ or $(\pm)0.866$ | **B1** | OR $\cos 2\theta = -\frac{1}{2} \Rightarrow 2\theta = 120, 240$ etc.
$\theta = 60°$, $\theta = 120°$ | **B1** | Ft for $180 - \text{their } 60$. Ft for $180 + \text{their } 60, 360 - \text{their } 60$
$\theta = 240°, 300°$ | **B1** | Allow $\frac{\pi}{3}, \frac{2\pi}{3}$ etc. Extra sols in range $–1$ [4]
---\begin{enumerate}[label=(\roman*)]
\item Show that $\sin^2 \theta - \cos^4 \theta = 2 \sin^2 \theta - 1$. [3]
\item Hence solve the equation $\sin^2 \theta - \cos^4 \theta = \frac{1}{2}$ for $0° \leq \theta \leq 360°$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2014 Q5 [7]}}