| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2006 |
| Session | November |
| Marks | 4 |
| 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 the Pythagorean identity sin²x + cos²x = 1 and the definition tan x = sin x/cos x. Given sin x = 2/5, students simply calculate cos²x = 1 - (2/5)² = 21/25, then tan²x = sin²x/cos²x = 4/21. It requires only basic recall and arithmetic, making it easier than average with no problem-solving or novel insight needed. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\cos^2 x = 1 - \sin^2 x = \frac{21}{25}\) | M1 A1 [2] | Formula only-use of 90° triangle ok co - loses if decimals blatantly used |
| (ii) \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{4}{21}\) | M1 A1 [2] | Formula only – or use of triangle ok. Correct from his answer to (i) |
$x = \sin^{-1}\frac{2}{5} \rightarrow \sin x = \frac{2}{5}$
(i) $\cos^2 x = 1 - \sin^2 x = \frac{21}{25}$ | M1 A1 [2] | Formula only-use of 90° triangle ok co - loses if decimals blatantly used
(ii) $\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{4}{21}$ | M1 A1 [2] | Formula only – or use of triangle ok. Correct from his answer to (i)
---2 Given that $x = \sin ^ { - 1 } \left( \frac { 2 } { 5 } \right)$, find the exact value of\\
(i) $\cos ^ { 2 } x$,\\
(ii) $\tan ^ { 2 } x$.
\hfill \mbox{\textit{CAIE P1 2006 Q2 [4]}}