CAIE P2 2024 June — Question 7 10 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2024
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTrig Proofs
TypeSolve equation using proven identity
DifficultyStandard +0.3 Part (a) is a straightforward identity proof using standard definitions (cosec 2θ = 1/sin 2θ and double angle formula). Part (b) requires substituting the proven identity and solving a quadratic in sin θ, which is routine A-level technique. Part (c) involves using a half-angle identity and standard integration. All parts follow predictable patterns with no novel insights required, making this slightly easier than average but still requiring competent algebraic manipulation across multiple steps.
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. Prove that \(2\sin\theta\cosec 2\theta \equiv \sec\theta\). [2]
  2. Solve the equation \(\tan^2\theta + 7\sin\theta\cosec 2\theta = 8\) for \(-\pi < \theta < \pi\). [5]
  3. Find \(\int 8\sin^2\frac{1}{2}x\cosec^2 x \, dx\). [3]
Question 7:
AnswerMarks
7(a)Express left-hand side in terms of sin and cos using
1
cosec2
AnswerMarks
sin2M1
1
Obtain and confirm sec
AnswerMarks Guidance
cosA1 Answer given – necessary detail needed.
2
AnswerMarks Guidance
7(b)Attempt to obtain quadratic equation in sec or cos only *M1
Obtain sec21 7sec8 involving one trigonometric ratio
AnswerMarks Guidance
2A1 Or equivalent, may be unsimplified, but reduce to
2sec27sec180
18cos27cos20.
Attempt to solve 3-term quadratic equation for sec, using a
AnswerMarks Guidance
correct method, to find at least one value of DM1 Or equivalent using cos.
Obtain any two of the four correct solutions 0.952, 1.76A1 Or greater accuracy.
Obtain remaining two correct solutionsA1 Or greater accuracy; and no others between π and π.
5
AnswerMarks
7(c)Identify integrand as 2sec2 1x
2B1
Integrate ksec2 1x to obtain 2ktan1x
AnswerMarks
2 2M1
Obtain correct 4tan1x
AnswerMarks Guidance
2A1 Condone omission of ...c.
3
Question 7:
--- 7(a) ---
7(a) | Express left-hand side in terms of sin and cos using
1
cosec2
sin2 | M1
1
Obtain and confirm sec
cos | A1 | Answer given – necessary detail needed.
2
--- 7(b) ---
7(b) | Attempt to obtain quadratic equation in sec or cos only | *M1
Obtain sec21 7sec8 involving one trigonometric ratio
2 | A1 | Or equivalent, may be unsimplified, but reduce to
2sec27sec180
18cos27cos20.
Attempt to solve 3-term quadratic equation for sec, using a
correct method, to find at least one value of  | DM1 | Or equivalent using cos.
Obtain any two of the four correct solutions 0.952, 1.76 | A1 | Or greater accuracy.
Obtain remaining two correct solutions | A1 | Or greater accuracy; and no others between π and π.
5
--- 7(c) ---
7(c) | Identify integrand as 2sec2 1x
2 | B1
Integrate ksec2 1x to obtain 2ktan1x
2 2 | M1
Obtain correct 4tan1x
2 | A1 | Condone omission of ...c.
3
\begin{enumerate}[label=(\alph*)]
\item Prove that $2\sin\theta\cosec 2\theta \equiv \sec\theta$. [2]
\item Solve the equation $\tan^2\theta + 7\sin\theta\cosec 2\theta = 8$ for $-\pi < \theta < \pi$. [5]
\item Find $\int 8\sin^2\frac{1}{2}x\cosec^2 x \, dx$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P2 2024 Q7 [10]}}
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