| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Equation with non-equation preliminary part (sketch/proof/identity) |
| Difficulty | Standard +0.3 Part (a) requires using the double angle identity cos 2θ = 1 - 2sin²2θ to convert to a quadratic in cos 2θ, then solving for θ in the given range—a standard multi-step trigonometric equation. Part (b) involves finding parameters of a transformed tan graph using two given points, which is routine substitution. Both parts are typical textbook exercises requiring standard techniques without novel insight, placing this slightly easier than average. |
| Spec | 1.05f Trigonometric function graphs: symmetries and periodicities1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) | ( ) | |
| 3 1−cos22θ +8cos2θ=0 → 3cos22θ−8cos2θ−3 (=0 ) | M1 | Use s2 =1−c2 and simplify to 3-term quadratic in 2θ |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | A1 | Ignore other solution |
| 2θ = 109.(47)º or 250.(53)º | A1 | One solution is sufficient, may be implied by either of |
| Answer | Marks | Guidance |
|---|---|---|
| θ = 54.7º or 125.3º | A1A1ft | Ft for 180º ‒ other solution |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 7(b) | √3=a+tan0 → a=√3 | B1 |
| 0=tan(−bπ/6)+√3 taken as far as tan−1, angle units consistent | M1 | ( ) |
| Answer | Marks |
|---|---|
| b=2 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 7:
--- 7(a) ---
7(a) | ( )
3 1−cos22θ +8cos2θ=0 → 3cos22θ−8cos2θ−3 (=0 ) | M1 | Use s2 =1−c2 and simplify to 3-term quadratic in 2θ
1
cos2θ=− soi
3 | A1 | Ignore other solution
2θ = 109.(47)º or 250.(53)º | A1 | One solution is sufficient, may be implied by either of
the next solns
θ = 54.7º or 125.3º | A1A1ft | Ft for 180º ‒ other solution
Use of double angles leads to
3c4 −7c2 +2=0⇒c=±1/√3 for M1A1A1 then
A1A1 for each angle
Similar marking if 3sin22θ=−8cos2θ is squared
leading to 9sin42θ+64sin22θ−64=0
5
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | √3=a+tan0 → a=√3 | B1 | b = 8 or –4 (or –10, 14 etc) scores M1A0
0=tan(−bπ/6)+√3 taken as far as tan−1, angle units consistent | M1 | ( )
A0 if tan−1 − 3 is not exact; (b=2 no working scores
B2)
b=2 | A1
3
Question | Answer | Marks | Guidance\begin{enumerate}[label=(\alph*)]
\item Solve the equation $3\sin^2 2\theta + 8\cos 2\theta = 0$ for $0° < \theta < 180°$. [5]
\item
\includegraphics{figure_7b}
The diagram shows part of the graph of $y = a + \tan bx$, where $x$ is measured in radians and $a$ and $b$ are constants. The curve intersects the $x$-axis at $\left(-\frac{1}{6}\pi, 0\right)$ and the $y$-axis at $(0, \sqrt{3})$. Find the values of $a$ and $b$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2019 Q7 [8]}}