CAIE Further Paper 2 2020 June — Question 8 15 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2020
SessionJune
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeIntegration using De Moivre identities
DifficultyChallenging +1.8 This is a substantial Further Maths question requiring de Moivre's theorem for multiple-angle identities, integration using those identities, and solving a trigonometric equation by substitution. Part (a) is a standard but lengthy de Moivre application (6 marks), part (b) requires careful integration and simplification, and part (c) demands recognizing that the given equation relates to the derived identities and solving for specific cosine values. The multi-part structure, algebraic manipulation demands, and need to connect all three parts elevate this above average difficulty, though it follows recognizable Further Maths patterns.
Spec1.08d Evaluate definite integrals: between limits4.02q De Moivre's theorem: multiple angle formulae
  1. Use de Moivre's theorem to show that \(\sin^6 \theta = -\frac{1}{32}(\cos 6\theta - 6\cos 4\theta + 15\cos 2\theta - 10)\). [6]
It is given that \(\cos^6 \theta = \frac{1}{32}(\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10)\).
  1. Find the exact value of \(\int_0^{\frac{1}{4}\pi}\left(\cos^6\left(\frac{1}{4}x\right) + \sin^6\left(\frac{1}{4}x\right)\right)dx\). [4]
  2. Express each root of the equation \(16c^6 + 16\left(1-c^2\right)^3 - 13 = 0\) in the form \(\cos k\pi\), where \(k\) is a rational number. [5]
Question 8:
AnswerMarks
8(a)− −1 = θ
z z 2isinB1
( ) ( ) ( ) ( )
AnswerMarks
z−z −1 6 = z6+z −6 −6 z4+z −4 +15 z2+z −2 −20M1 A1
( 2isinθ)6 =2cos6θ−6 ( 2cos4θ)+15 ( 2cos2θ)−20M1 A1
( )
sin6θ= 1 −cos6θ+6cos4θ−15cos2θ+10
AnswerMarks
32A1
6
AnswerMarks
8(b)1π 1π
3 cos6 1 x+sin6 1 xdx=13 3cosx+5dx
4 4 8
AnswerMarks
0 0M1 A1
( )
1 [ 3sinx+5x ]1 3 π=1 5π+3√3 .
AnswerMarks
8 0 8 3 2M1 A1
4
AnswerMarks Guidance
8(c)c = cosθ 1− c2 = sin2θ B1
( )
AnswerMarks
16c6+16 1−c2 3−13=06cos4θ−3=0M1 A1
4θ= 1π,5π,7π,11π
AnswerMarks
3 3 3 3M1
( ) ( ) ( ) ( )
c=cos 1 π ,cos 5 π ,cos 7 π ,cos 11π
AnswerMarks
12 12 12 12A1
5
Question 8:
--- 8(a) ---
8(a) | − −1 = θ
z z 2isin | B1
( ) ( ) ( ) ( )
z−z −1 6 = z6+z −6 −6 z4+z −4 +15 z2+z −2 −20 | M1 A1
( 2isinθ)6 =2cos6θ−6 ( 2cos4θ)+15 ( 2cos2θ)−20 | M1 A1
( )
sin6θ= 1 −cos6θ+6cos4θ−15cos2θ+10
32 | A1
6
--- 8(b) ---
8(b) | 1π 1π
3 cos6 1 x+sin6 1 xdx=13 3cosx+5dx
4 4 8
0 0 | M1 A1
( )
1 [ 3sinx+5x ]1 3 π=1 5π+3√3 .
8 0 8 3 2 | M1 A1
4
--- 8(c) ---
8(c) | c = cosθ 1− c2 = sin2θ | B1
( )
16c6+16 1−c2 3−13=06cos4θ−3=0 | M1 A1
4θ= 1π,5π,7π,11π
3 3 3 3 | M1
( ) ( ) ( ) ( )
c=cos 1 π ,cos 5 π ,cos 7 π ,cos 11π
12 12 12 12 | A1
5
\begin{enumerate}[label=(\alph*)]
\item Use de Moivre's theorem to show that $\sin^6 \theta = -\frac{1}{32}(\cos 6\theta - 6\cos 4\theta + 15\cos 2\theta - 10)$. [6]
\end{enumerate}

It is given that $\cos^6 \theta = \frac{1}{32}(\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10)$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the exact value of $\int_0^{\frac{1}{4}\pi}\left(\cos^6\left(\frac{1}{4}x\right) + \sin^6\left(\frac{1}{4}x\right)\right)dx$. [4]

\item Express each root of the equation $16c^6 + 16\left(1-c^2\right)^3 - 13 = 0$ in the form $\cos k\pi$, where $k$ is a rational number. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q8 [15]}}
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Q1 6 Q2 6 Q3 8 Q4 8 Q5 9 Q6 12 Q7 11 Q8 15