Pre-U Pre-U 9795/1 2017 June — Question 12

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2017
SessionJune
TopicAddition & Double Angle Formulae
TypeProve identity with double/compound angles
DifficultyChallenging +1.8 This is a sophisticated multi-part Further Maths question requiring proof of trigonometric identities using algebraic manipulation, de Moivre's theorem with complex numbers, and mathematical induction with inequalities. While each technique is standard for Further Maths, the combination and the non-routine nature of the recursive function F_n and the induction inequality make this significantly harder than average A-level questions.
Spec1.05l Double angle formulae: and compound angle formulae4.01a Mathematical induction: construct proofs4.02q De Moivre's theorem: multiple angle formulae
12 For each positive integer \(n\), the function \(\mathrm { F } _ { n }\) is defined for all real angles \(\theta\) by $$\mathrm { F } _ { n } ( \theta ) = c ^ { 2 n } + s ^ { 2 n }$$ where \(c = \cos \theta\) and \(s = \sin \theta\).
  1. Prove the identity $$\mathrm { F } _ { n + 2 } ( \theta ) - \frac { 1 } { 4 } \sin ^ { 2 } 2 \theta \times \mathrm { F } _ { n + 1 } ( \theta ) \equiv \mathrm { F } _ { n + 3 } ( \theta )$$ Let \(z\) denote the complex number \(c + \mathrm { i } s\).
  2. Using de Moivre's theorem,
    1. express \(z + z ^ { - 1 }\) and \(z - z ^ { - 1 }\) in terms of \(c\) and \(s\) respectively,
    2. prove the identity \(8 \left( c ^ { 6 } + s ^ { 6 } \right) \equiv 3 \cos 4 \theta + 5\) and deduce that $$c ^ { 6 } + s ^ { 6 } \equiv \cos ^ { 2 } 2 \theta + \frac { 1 } { 4 } \sin ^ { 2 } 2 \theta$$
    3. Prove by induction that, for all positive integers \(n\), $$c ^ { 2 n + 4 } + s ^ { 2 n + 4 } \leqslant \cos ^ { 2 } 2 \theta + \frac { 1 } { 2 ^ { n + 1 } } \sin ^ { 2 } 2 \theta$$ [You are given that the range of the function \(\mathrm { F } _ { n }\) is \(\frac { 1 } { 2 ^ { n - 1 } } \leqslant \mathrm {~F} _ { n } ( \theta ) \leqslant 1\).] {www.cie.org.uk} after the live examination series. }
Question 12:
AnswerMarks
12(i)Method I
1sin2( )
F (θ ) − 2θ F (θ )
n +2 4 n +1
( )( )
≡ c2 + s2 c2n+ 4 + s2n + 4
− 1(2sc)2 ( c2n+ 2 + s2n + 2 )
AnswerMarks Guidance
4M2 M1 all F terms
n
M1 sin2θ form
≡ c2n + 6 +c2s2n + 4 +s2c2n + 4 +s2n + 6
( )
AnswerMarks
−c2s2 c2n+ 2 + s2n + 2A1
≡ c2n + 6 + s2n + 6 ≡ F (θ )
AnswerMarks Guidance
n +3A1 AG
Method II
( )
AnswerMarks Guidance
≡ c2n + 4 +s2n + 4 −s2c2 c2n+ 2 +s2n + 2M1 Use of sin2θ form
≡ c2n + 4 +s2n + 4 −s2c2n + 4 −c2s2n + 4A1
( ) ( )
AnswerMarks
≡ 1−s2 c2n + 4 + 1−c2 s2n + 4M1
≡ c2n + 6 + s2n + 6 ≡ F (θ )
AnswerMarks Guidance
n +3A1 AG
12(ii)(a)Use of z = c + is and z−1 = c – is M1
z+z−1 =2c and z−z−1 = 2isA2 A1 for each
12(ii)(b)Method I
(2c)6 = ( z+z−1)6 =z6 +6z4 +15z2 +20
AnswerMarks
+15z−2 +6z−4 +z−6M1
= 2cos6θ + 12cos4θ + 30cos2θ + 20A1
−(2s)6 = ( z−z−1)6 =z6 −6z4 +15z2 −20
+15z−2 −6z−4 +z−6
AnswerMarks Guidance
= 2cos6θ − 12cos4θ + 30cos2θ − 20B1 FT (Must have – sign)
Subtracting:
( ) ( )
64 c6 +s6 =12 z4 +z−4 +40
AnswerMarks
= 12 . 2cos4θ + 40M1
( )
AnswerMarks Guidance
Dividing by 8: 8c6 +s6 = 3cos4θ + 5A1 AG
Use of cos4θ = 2cos22θ – 1 and 1 = cos22θ +
AnswerMarks
sin22θM1
⇒c6 +s6 = 3 ( 2cos22θ ) +(−3+5)( cos22θ+sin22θ )
8 8 8
= cos2 2θ+ 1sin2 2θ
AnswerMarks Guidance
4A1 AG
QuestionAnswer Marks
12(ii)(b)Method II
cos4θ = Re(c + is)4M1
( ) ( )2
= c4 −6c2s2 +s4 = c4 −6c2 1−c2 + 1−c2
AnswerMarks
= 8c4 −8c2 +1A1
( )3
AnswerMarks
c6 +s6 = c6 + 1−c2 = c6 +1−3c2 +3c4 −c6M1
= 3c4 −3c2 +1A1
( )
AnswerMarks Guidance
so that 8c6 +s6 = 3cos4θ + 5A1 AG
Use of cos4θ = cos22θ – sin22θ
AnswerMarks
and 1 = cos22θ + sin22θM1
( )
⇒ 8c6 +s6 = 3cos4θ + 5
= 3(cos22θ – sin22θ ) + 5(cos22θ + sin22θ )
⇒ c6 +s6= cos2 2θ+ 1sin2 2θ
AnswerMarks Guidance
4A1 AG
AnswerMarks Guidance
12(iii)Case for n = 1 established in (ii) (b): B1
1
Assume c2k+ 4 +s2k + 4≤cos22θ+ sin22θ
AnswerMarks Guidance
2k+1B1 i.e. the case for n = k
A clear statement of the result must be given,
possibly within what follows
Then c2k + 6 +s2k + 6 =
( )
c2k+ 4 +s2k + 4 −1sin22θc2k+ 2 +s2k + 2
AnswerMarks Guidance
4M1 attempt at n = k + 1 case using (i)’s
identity
≤cos22θ+ 1 sin22θ− 1 sin22θ ( c2k+ 2 +s2k + 2)
AnswerMarks Guidance
2k+1 4M1 use of the induction hypothesis
(i.e. the n = k case)
=cos22θ+ 1 sin22θ− 1 sin22θ  c2k+ 2 +s2k + 2 − 1  
AnswerMarks Guidance
2k+2 4  2k M1A1 splitting up the sin22θ term into
two equal parts
1
≤cos22θ+ sin22θ
2k+2
Proof follows by induction since sin22θ ≥ 0 and
1
given result that c2k+ 2 +s2k + 2≥
AnswerMarks
2kA1 
Question 12:
--- 12(i) ---
12(i) | Method I
1sin2( )
F (θ ) − 2θ F (θ )
n +2 4 n +1
( )( )
≡ c2 + s2 c2n+ 4 + s2n + 4
− 1(2sc)2 ( c2n+ 2 + s2n + 2 )
4 | M2 | M1 all F terms
n
M1 sin2θ form
≡ c2n + 6 +c2s2n + 4 +s2c2n + 4 +s2n + 6
( )
−c2s2 c2n+ 2 + s2n + 2 | A1
≡ c2n + 6 + s2n + 6 ≡ F (θ )
n +3 | A1 | AG
Method II
( )
≡ c2n + 4 +s2n + 4 −s2c2 c2n+ 2 +s2n + 2 | M1 | Use of sin2θ form
≡ c2n + 4 +s2n + 4 −s2c2n + 4 −c2s2n + 4 | A1
( ) ( )
≡ 1−s2 c2n + 4 + 1−c2 s2n + 4 | M1
≡ c2n + 6 + s2n + 6 ≡ F (θ )
n +3 | A1 | AG
12(ii)(a) | Use of z = c + is and z−1 = c – is | M1
z+z−1 =2c and z−z−1 = 2is | A2 | A1 for each
12(ii)(b) | Method I
(2c)6 = ( z+z−1)6 =z6 +6z4 +15z2 +20
+15z−2 +6z−4 +z−6 | M1
= 2cos6θ + 12cos4θ + 30cos2θ + 20 | A1
−(2s)6 = ( z−z−1)6 =z6 −6z4 +15z2 −20
+15z−2 −6z−4 +z−6
= 2cos6θ − 12cos4θ + 30cos2θ − 20 | B1 | FT (Must have – sign)
Subtracting:
( ) ( )
64 c6 +s6 =12 z4 +z−4 +40
= 12 . 2cos4θ + 40 | M1
( )
Dividing by 8: 8c6 +s6 = 3cos4θ + 5 | A1 | AG
Use of cos4θ = 2cos22θ – 1 and 1 = cos22θ +
sin22θ | M1
⇒c6 +s6 = 3 ( 2cos22θ ) +(−3+5)( cos22θ+sin22θ )
8 8 8
= cos2 2θ+ 1sin2 2θ
4 | A1 | AG
Question | Answer | Marks | Part Marks
12(ii)(b) | Method II
cos4θ = Re(c + is)4 | M1
( ) ( )2
= c4 −6c2s2 +s4 = c4 −6c2 1−c2 + 1−c2
= 8c4 −8c2 +1 | A1
( )3
c6 +s6 = c6 + 1−c2 = c6 +1−3c2 +3c4 −c6 | M1
= 3c4 −3c2 +1 | A1
( )
so that 8c6 +s6 = 3cos4θ + 5 | A1 | AG
Use of cos4θ = cos22θ – sin22θ
and 1 = cos22θ + sin22θ | M1
( )
⇒ 8c6 +s6 = 3cos4θ + 5
= 3(cos22θ – sin22θ ) + 5(cos22θ + sin22θ )
⇒ c6 +s6= cos2 2θ+ 1sin2 2θ
4 | A1 | AG
--- 12(iii) ---
12(iii) | Case for n = 1 established in (ii) (b): | B1 | noted explicitly (possibly at end)
1
Assume c2k+ 4 +s2k + 4≤cos22θ+ sin22θ
2k+1 | B1 | i.e. the case for n = k
A clear statement of the result must be given,
possibly within what follows
Then c2k + 6 +s2k + 6 =
( )
c2k+ 4 +s2k + 4 −1sin22θc2k+ 2 +s2k + 2
4 | M1 | attempt at n = k + 1 case using (i)’s
identity
≤cos22θ+ 1 sin22θ− 1 sin22θ ( c2k+ 2 +s2k + 2)
2k+1 4 | M1 | use of the induction hypothesis
(i.e. the n = k case)
=cos22θ+ 1 sin22θ− 1 sin22θ  c2k+ 2 +s2k + 2 − 1  
2k+2 4  2k  | M1A1 | splitting up the sin22θ term into
two equal parts
1
≤cos22θ+ sin22θ
2k+2
Proof follows by induction since sin22θ ≥ 0 and
1
given result that c2k+ 2 +s2k + 2≥
2k | A1
12 For each positive integer $n$, the function $\mathrm { F } _ { n }$ is defined for all real angles $\theta$ by

$$\mathrm { F } _ { n } ( \theta ) = c ^ { 2 n } + s ^ { 2 n }$$

where $c = \cos \theta$ and $s = \sin \theta$.\\
(i) Prove the identity

$$\mathrm { F } _ { n + 2 } ( \theta ) - \frac { 1 } { 4 } \sin ^ { 2 } 2 \theta \times \mathrm { F } _ { n + 1 } ( \theta ) \equiv \mathrm { F } _ { n + 3 } ( \theta )$$

Let $z$ denote the complex number $c + \mathrm { i } s$.\\
(ii) Using de Moivre's theorem,
\begin{enumerate}[label=(\alph*)]
\item express $z + z ^ { - 1 }$ and $z - z ^ { - 1 }$ in terms of $c$ and $s$ respectively,
\item prove the identity $8 \left( c ^ { 6 } + s ^ { 6 } \right) \equiv 3 \cos 4 \theta + 5$ and deduce that

$$c ^ { 6 } + s ^ { 6 } \equiv \cos ^ { 2 } 2 \theta + \frac { 1 } { 4 } \sin ^ { 2 } 2 \theta$$

(iii) Prove by induction that, for all positive integers $n$,

$$c ^ { 2 n + 4 } + s ^ { 2 n + 4 } \leqslant \cos ^ { 2 } 2 \theta + \frac { 1 } { 2 ^ { n + 1 } } \sin ^ { 2 } 2 \theta$$

[You are given that the range of the function $\mathrm { F } _ { n }$ is $\frac { 1 } { 2 ^ { n - 1 } } \leqslant \mathrm {~F} _ { n } ( \theta ) \leqslant 1$.]

{www.cie.org.uk} after the live examination series.

}
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2017 Q12}}
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