| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Given sin/cos/tan, find other expressions |
| Difficulty | Moderate -0.3 Part (i) is a straightforward simplification using the double angle formula sin 2α = 2sin α cos α and sec α = 1/cos α, yielding 2sin α. Part (ii) requires substituting cos 2β = 2cos²β - 1 to form a quadratic in cos β, then solving—a standard technique but requires multiple steps and careful algebraic manipulation. Overall slightly easier than average due to being routine application of formulas with no novel insight required. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State \(\sin2\alpha = 2\sin\alpha\cos\alpha\) and \(\sec\alpha = 1/\cos\alpha\) | B1 | |
| Obtain \(2\sin\alpha\) | B1 | [2] |
| (ii) Use \(\cos2\beta = 2\cos^2\beta - 1\) or equivalent to produce correct equation in \(\cos\beta\) | B1 | |
| Solve three-term quadratic equation for \(\cos\beta\) | M1 | |
| Obtain \(\cos\beta = -\frac{1}{3}\) only | A1 | [3] |
(i) State $\sin2\alpha = 2\sin\alpha\cos\alpha$ and $\sec\alpha = 1/\cos\alpha$ | B1 |
Obtain $2\sin\alpha$ | B1 | [2]
(ii) Use $\cos2\beta = 2\cos^2\beta - 1$ or equivalent to produce correct equation in $\cos\beta$ | B1 |
Solve three-term quadratic equation for $\cos\beta$ | M1 |
Obtain $\cos\beta = -\frac{1}{3}$ only | A1 | [3]1 (i) Simplify $\sin 2 \alpha \sec \alpha$.\\
(ii) Given that $3 \cos 2 \beta + 7 \cos \beta = 0$, find the exact value of $\cos \beta$.
\hfill \mbox{\textit{CAIE P3 2014 Q1 [5]}}