CAIE P2 2024 November — Question 7 11 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2024
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTrig Proofs
TypeSolve equation using proven identity
DifficultyStandard +0.8 Part (a) requires expanding compound angle formulas and manipulating to reach a non-obvious target form (4 marks suggests multiple steps). Part (b) applies the proven identity to solve an equation requiring inverse trig. Part (c) is the most challenging, requiring recognition that the expression can be evaluated using the identity from (a) with clever angle substitutions. While systematic, this demands more insight than typical A-level trig questions and tests proof technique across all parts.
Spec1.01a Proof: structure of mathematical proof and logical steps1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals
  1. Prove that \(\cos(\theta + 30°)\cos(\theta + 60°) = \frac{1}{4}\sqrt{3} - \frac{1}{2}\sin 2\theta\). [4]
  2. Solve the equation \(5\cos(2\alpha + 30°)\cos(2\alpha + 60°) = 1\) for \(0° < \alpha < 90°\). [4]
  3. Show that the exact value of \(\cos 20° \cos 50° + \cos 40° \cos 70°\) is \(\frac{1}{2}\sqrt{3}\). [3]
Question 7:
AnswerMarks Guidance
7(a)State (coscos30−sinsin30)(coscos60−sinsin60) B1
Expand and use correct exact valuesM1
Obtain 1 3(cos2+sin2)−sincos or similarly simplified equivalent
AnswerMarks
4A1
Conclude 1 3−1sin2
AnswerMarks Guidance
4 2A1 AG – necessary detail needed.
4
AnswerMarks Guidance
7(b)Use identity to obtain value forsin4 *M1
Obtainsin4=1 3−2 or 0.466…
AnswerMarks Guidance
2 5A1
Show correct process to obtain one value ofDM1
Obtain 6.9 and 38.1A1 Or greater accuracy; and no others between
0and90.
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(c)Substitute=−10to obtaincos20cos50=1 3−1sin(−20)
4 2B1
3 1
Substitute=10 to obtaincos40cos70= − sin20
AnswerMarks
4 2B1
Add and confirm 1 3 with clear indication thatsin(−20)=−sin20
AnswerMarks Guidance
2B1 AG – necessary detail needed.
Alternative solution for Question 7(c)
Rewrite as sin70cos50+sin50cos70 or cos20sin40+cos40sin20 or
AnswerMarks
sin70sin40+cos70cos40B1
Obtainsin120orsin60orcos30B1
Confirm 1 3
AnswerMarks Guidance
2B1 AG – necessary detail needed.
3
Question 7:
--- 7(a) ---
7(a) | State (coscos30−sinsin30)(coscos60−sinsin60) | B1
Expand and use correct exact values | M1
Obtain 1 3(cos2+sin2)−sincos or similarly simplified equivalent
4 | A1
Conclude 1 3−1sin2
4 2 | A1 | AG – necessary detail needed.
4
--- 7(b) ---
7(b) | Use identity to obtain value forsin4 | *M1
Obtainsin4=1 3−2 or 0.466…
2 5 | A1
Show correct process to obtain one value of | DM1
Obtain 6.9 and 38.1 | A1 | Or greater accuracy; and no others between
0and90.
4
Question | Answer | Marks | Guidance
--- 7(c) ---
7(c) | Substitute=−10to obtaincos20cos50=1 3−1sin(−20)
4 2 | B1
3 1
Substitute=10 to obtaincos40cos70= − sin20
4 2 | B1
Add and confirm 1 3 with clear indication thatsin(−20)=−sin20
2 | B1 | AG – necessary detail needed.
Alternative solution for Question 7(c)
Rewrite as sin70cos50+sin50cos70 or cos20sin40+cos40sin20 or
sin70sin40+cos70cos40 | B1
Obtainsin120orsin60orcos30 | B1
Confirm 1 3
2 | B1 | AG – necessary detail needed.
3
\begin{enumerate}[label=(\alph*)]
\item Prove that $\cos(\theta + 30°)\cos(\theta + 60°) = \frac{1}{4}\sqrt{3} - \frac{1}{2}\sin 2\theta$. [4]

\item Solve the equation $5\cos(2\alpha + 30°)\cos(2\alpha + 60°) = 1$ for $0° < \alpha < 90°$. [4]

\item Show that the exact value of $\cos 20° \cos 50° + \cos 40° \cos 70°$ is $\frac{1}{2}\sqrt{3}$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P2 2024 Q7 [11]}}
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Q1 5 Q2 4 Q3 3 Q3 4 Q4 5 Q4 3 Q5 17 Q6 7 Q7 11