| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Standard +0.3 Part (i) is a straightforward identity proof using cos²x = 1 - sin²x and expanding. Part (ii) requires substituting the identity, collecting terms to form a quadratic in sin²x, then solving—a standard multi-step procedure with no novel insight required. Slightly easier than average due to the guided structure and routine algebraic manipulation. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos^4 x = (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x\) AG | B1 [1] | Could be LHS to RHS or vice versa |
| Answer | Marks | Guidance |
|---|---|---|
| \(8\sin^4 x + 1 - 2\sin^2 x + \sin^4 x = 2(1 - \sin^2 x)\) | M1 | Substitute for \(\cos^4 x\) and \(\cos^2 x\); OR sub for \(\sin^4 x \rightarrow 3\cos^2 x = 2 \rightarrow \cos x = (\pm)\sqrt{2/3}\) |
| \(9\sin^4 x = 1\) | A1 | |
| \(x = 35.3°\) (or any correct solution) | A1 | Allow first 2 A1 marks for radians \((0.616, 2.53, 3.76, 5.67)\) |
| Any correct second solution from \(144.7°, 215.3°, 324.7°\) | A1\(\checkmark\) | |
| The remaining 2 solutions | A1 [5] |
## Question 6:
### Part (i):
$\cos^4 x = (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x$ AG | **B1** [1] | Could be LHS to RHS or vice versa
### Part (ii):
$8\sin^4 x + 1 - 2\sin^2 x + \sin^4 x = 2(1 - \sin^2 x)$ | **M1** | Substitute for $\cos^4 x$ and $\cos^2 x$; OR sub for $\sin^4 x \rightarrow 3\cos^2 x = 2 \rightarrow \cos x = (\pm)\sqrt{2/3}$
$9\sin^4 x = 1$ | **A1** |
$x = 35.3°$ (or any correct solution) | **A1** | Allow first 2 **A1** marks for radians $(0.616, 2.53, 3.76, 5.67)$
Any correct second solution from $144.7°, 215.3°, 324.7°$ | **A1**$\checkmark$ |
The remaining 2 solutions | **A1** [5] |
---6 (i) Show that $\cos ^ { 4 } x \equiv 1 - 2 \sin ^ { 2 } x + \sin ^ { 4 } x$.\\
(ii) Hence, or otherwise, solve the equation $8 \sin ^ { 4 } x + \cos ^ { 4 } x = 2 \cos ^ { 2 } x$ for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.
\hfill \mbox{\textit{CAIE P1 2016 Q6 [6]}}