CAIE P1 2024 November — Question 8 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard trigonometric equations
TypeEquation with non-equation preliminary part (sketch/proof/identity)
DifficultyModerate -0.3 Part (a) is a routine trigonometric identity manipulation using Pythagorean identity in the second quadrant (3 marks). Part (b) is a standard quadratic-in-sin equation requiring substitution of cos²θ = 1-sin²θ and solving within a given range (5 marks). Both parts follow textbook patterns with no novel insight required, making this slightly easier than average for A-level.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals
  1. It is given that \(\beta\) is an angle between \(90°\) and \(180°\) such that \(\sin \beta = a\). Express \(\tan^2 \beta - 3 \sin \beta \cos \beta\) in terms of \(a\). [3]
  2. Solve the equation \(\sin^2 \theta + 2 \cos^2 \theta = 4 \sin \theta + 3\) for \(0° < \theta < 360°\). [5]
Question 8:
AnswerMarks
8(a)sin2
Use tan2=
AnswerMarks Guidance
cos2B1 sin2
E.g. tan2= and then replaces sin2
cos2
with a2 or cos2 with 1 – a2.
AnswerMarks
cos=− 1−a2B1
a2
Obtain +3a 1−a2
AnswerMarks
1−a2B1
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
8(b)Use correct identity to obtain 3-term quadratic equation in sin *M1
Obtain sin2+4sin+1 =0 A1
Attempt to solve quadraticDM1 −4  12
At least as far as .
2
–15.5o implies attempt at solving quadratic.
AnswerMarks Guidance
Obtain 195.5A1
Obtain 344.5A1FT Following first answer; and no others for
0360 but must be in 4th quadrant.
SC B1 for 3.41c and 6.01c.
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 8:
--- 8(a) ---
8(a) | sin2
Use tan2=
cos2 | B1 | sin2
E.g. tan2= and then replaces sin2
cos2
with a2 or cos2 with 1 – a2.
cos=− 1−a2 | B1
a2
Obtain +3a 1−a2
1−a2 | B1
3
Question | Answer | Marks | Guidance
--- 8(b) ---
8(b) | Use correct identity to obtain 3-term quadratic equation in sin | *M1
Obtain sin2+4sin+1 =0  | A1
Attempt to solve quadratic | DM1 | −4  12
At least as far as .
2
–15.5o implies attempt at solving quadratic.
Obtain 195.5 | A1
Obtain 344.5 | A1FT | Following first answer; and no others for
0360 but must be in 4th quadrant.
SC B1 for 3.41c and 6.01c.
5
Question | Answer | Marks | Guidance
\begin{enumerate}[label=(\alph*)]
\item It is given that $\beta$ is an angle between $90°$ and $180°$ such that $\sin \beta = a$.

Express $\tan^2 \beta - 3 \sin \beta \cos \beta$ in terms of $a$. [3]

\item Solve the equation $\sin^2 \theta + 2 \cos^2 \theta = 4 \sin \theta + 3$ for $0° < \theta < 360°$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2024 Q8 [8]}}
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