| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Simplify or verify trig identity with acute angle |
| Difficulty | Easy -1.2 This is a straightforward application of basic trigonometric identities requiring only recall of Pythagorean identity and complementary angle relationships. All three parts are direct substitutions with no problem-solving or multi-step reasoning needed, making it easier than average for A-level. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\sin x = \sqrt{1 - p^2}\) | B1 | Allow \(1 - p\) if following \(\sqrt{1 - p^2}\) ± is B0. |
| (ii) \(\tan x = \frac{\sin x}{\cos x} = \frac{\sqrt{1 - p^2}}{p}\) | B1✓ [1] | ✓ for answer to (i) used. |
| (iii) \(\tan(90 - x) = \frac{p}{\sqrt{1 - p^2}}\) | B1✓ [1] | ✓ for reciprocal of (ii) |
(i) $\sin x = \sqrt{1 - p^2}$ | B1 | Allow $1 - p$ if following $\sqrt{1 - p^2}$ ± is B0.
(ii) $\tan x = \frac{\sin x}{\cos x} = \frac{\sqrt{1 - p^2}}{p}$ | B1✓ [1] | ✓ for answer to (i) used.
(iii) $\tan(90 - x) = \frac{p}{\sqrt{1 - p^2}}$ | B1✓ [1] | ✓ for reciprocal of (ii)1 Given that $\cos x = p$, where $x$ is an acute angle in degrees, find, in terms of $p$,\\
(i) $\sin x$,\\
(ii) $\tan x$,\\
(iii) $\tan \left( 90 ^ { \circ } - x \right)$.
\hfill \mbox{\textit{CAIE P1 2013 Q1 [3]}}