CAIE P1 2024 June — Question 3 6 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard trigonometric equations
TypeConvert to quadratic in sin/cos
DifficultyModerate -0.5 This is a straightforward two-part trigonometric equation question. Part (a) requires routine algebraic manipulation using tan θ = sin θ/cos θ and the Pythagorean identity, which is guided by the given target form. Part (b) involves solving a quadratic in sin θ and finding angles in a given range—standard A-level technique with no novel insight required. Slightly easier than average due to the scaffolding in part (a).
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
  1. Show that the equation \(\frac{7\tan\theta}{\cos\theta} + 12 = 0\) can be expressed as $$12\sin^2\theta - 7\sin\theta - 12 = 0.$$ [3]
  2. Hence solve the equation \(\frac{7\tan\theta}{\cos\theta} + 12 = 0\) for \(0° < \theta \leqslant 360°\). [3]
Question 3:
AnswerMarks
3(a)sin  sin 
7 cos12 0 leading to 7 12cos0
 
AnswerMarks Guidance
cos  cos M1* OE
sin
Use of tan .
cos
AnswerMarks Guidance
7sin12  1sin2  0DM1 Use of s2 c2 1.
⇒ 12sin27sin120A1 AG, WWW
Condone use of s, c and t and/or omission of θ
throughout working but the A1 is for cao.
3
AnswerMarks
3(b)12sin27sin120 leading to4sin33sin4
 M1
3  4
sin or
 
AnswerMarks Guidance
4  3B1 OE, WWW
 3
Can be implied by a correct value for sin1   
 4
e.g. 48.6°.
AnswerMarks Guidance
 228.6, 311.4B1 AWRT, WWW
No others in the range 0360.
Ignore any answers outside this range.
Condone 229°, 311°.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
--- 3(a) ---
3(a) | sin  sin 
7 cos12 0 leading to 7 12cos0
 
cos  cos  | M1* | OE
sin
Use of tan .
cos
7sin12  1sin2  0 | DM1 | Use of s2 c2 1.
⇒ 12sin27sin120 | A1 | AG, WWW
Condone use of s, c and t and/or omission of θ
throughout working but the A1 is for cao.
3
--- 3(b) ---
3(b) | 12sin27sin120 leading to4sin33sin4
  | M1
3  4
sin or
 
4  3 | B1 | OE, WWW
 3
Can be implied by a correct value for sin1   
 4
e.g. 48.6°.
 228.6, 311.4 | B1 | AWRT, WWW
No others in the range 0360.
Ignore any answers outside this range.
Condone 229°, 311°.
3
Question | Answer | Marks | Guidance
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\frac{7\tan\theta}{\cos\theta} + 12 = 0$ can be expressed as
$$12\sin^2\theta - 7\sin\theta - 12 = 0.$$ [3]

\item Hence solve the equation $\frac{7\tan\theta}{\cos\theta} + 12 = 0$ for $0° < \theta \leqslant 360°$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2024 Q3 [6]}}
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