CAIE FP1 2015 November — Question 10 12 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2015
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeDe Moivre to derive tan/cot identities
DifficultyChallenging +1.3 This is a structured multi-part Further Maths question requiring de Moivre's theorem application, algebraic manipulation to derive a tan multiple angle formula, and then using special angle properties. While it involves several steps and Further Maths content (making it harder than average A-level), the question provides clear scaffolding through its parts and follows a standard template for this topic. The techniques are well-practiced in FP1 courses, though the algebraic manipulation requires care.
Spec4.02q De Moivre's theorem: multiple angle formulae4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots
Using de Moivre's theorem, show that $$\tan 5\theta = \frac{5\tan\theta - 10\tan^3\theta + \tan^5\theta}{1 - 10\tan^2\theta + 5\tan^4\theta}.$$ [5] Hence show that the equation \(x^2 - 10x + 5 = 0\) has roots \(\tan^2\left(\frac{1}{5}\pi\right)\) and \(\tan^2\left(\frac{2}{5}\pi\right)\). [4] Deduce a quadratic equation, with integer coefficients, having roots \(\sec^2\left(\frac{1}{5}\pi\right)\) and \(\sec^2\left(\frac{2}{5}\pi\right)\). [3]
Question 10:
AnswerMarks
10(cosθ +isinθ)5 =cos5θ +isin5θ
(c+is)5 =c5+5c4si−10c3s2−10ic2s3i+5cs4+s5i
5c4s−10c2s3+s5
tan5θ =
c5−10c3s2+5cs4
c5
Divide numerator and denominator by (stated or shown):
5tanθ −10tan3θ +tan5θ
⇒tan5θ = (AG)
1−10tan2θ +5tan4θ
1 2 3 4
tan5θ =0⇒θ = π, π, π, π, π
5 5 5 5
1  2  3  4 
t5−10t3+5t =0 has roots tan π, tan π, tan π, tan π, tanπ
5  5  5  5 
1  2  3  4 
⇒t4−10t2+5=0 has roots tan π, tan π, tan π, tan π.
5  5  5  5 
 1  2 
⇒ t2−tan2 π  t2−tan2 π =0
   
 5  5 
1  4  2  3 
since tan π=−tan π and tan π=−tan π.
5  5  5  5 
1  2 
⇒ x2 −10x+5=0 has roots tan2 π and tan2 π. (AG)
5  5 
sec2α =1+tan2α
y=1+x⇒x= y−1⇒(y−1)2−10(y−1)+5=0
AnswerMarks
⇒ y2−12y+16=0B1
M1A1
M1
A1
[5]
B1
B1
M1
A1
[4]
M1
M1
A1
[3]

Total

12
AnswerMarks Guidance
Page 9Mark Scheme Syllabus
Cambridge International A Level – October/November 20159231 11
Qn &
AnswerMarks Guidance
PartSolution Marks
11 Ei j k 8 4
   
E.g. −1 2 0 =4~2
   
−1 0 4 2 1
   
14
  
0.2
  
 0  1  4
= (AG)
42 +22 +12 21
3 4i+2j+k 1
p=  = (4i+2j+k)
21 21  7
1 −3
   
Line AP: r=0+t 2 
   
0 1
   
0
1 1 
For Q 1−3t =0⇒t = ⇒q= 2
3 3
 
1
 
−1  0 
  1 
E.g. AB= 2 , BQ= −4
3
   
0 1
   
i j k 2
 
−1 2 0 =1
 
0 −4 1 4
 
42
  
2.1
  
 1  4  8+2+4 14 2
cos−1 =cos−1 =cos−1 =cos−1
(AG)
AnswerMarks
21. 21 21 21 3M1A1
M1A1
[4]
B1
M1A1
M1A1
[5]
B1
M1A1
M1A1
[5]

Total

14
AnswerMarks Guidance
Page 10Mark Scheme Syllabus
Cambridge International A Level – October/November 20159231 11
Qn &
AnswerMarks Guidance
PartSolution Marks
11OClosed curve through pole with correct orientation.
Completely correct.
2× 1 a2∫ π (1−2cosθ +cos2θ)dθ =a2∫ π  3 −2cosθ + 1 cos2θ  dθ
2 1 π 1 π2 2 
2 2
3 1  π
=a2 θ −2sinθ + sin2θ
 
2 4  1
π
2
3 
=a2 π +2
4 
 ds  2
  =a2(1−2cosθ +cos2θ +sin2θ)
dθ 
1 1
=2a2(1−cosθ)=2a2.2sin2 θ =4a2sin2 θ (AG)
2 2
s =2×∫ π 2asin 1 θ dθ
1 π 2
2
 1  π
=4a −2cos θ
 
 2  1
π
2
AnswerMarks
=4 2aB1
B1
[2]
M1M1
M1A1
A1
[5]
B1
M1A1
M1
A1
M1A1
[7]

Total

14
Question 10:
10 | (cosθ +isinθ)5 =cos5θ +isin5θ
(c+is)5 =c5+5c4si−10c3s2−10ic2s3i+5cs4+s5i
5c4s−10c2s3+s5
tan5θ =
c5−10c3s2+5cs4
c5
Divide numerator and denominator by (stated or shown):
5tanθ −10tan3θ +tan5θ
⇒tan5θ = (AG)
1−10tan2θ +5tan4θ
1 2 3 4
tan5θ =0⇒θ = π, π, π, π, π
5 5 5 5
1  2  3  4 
t5−10t3+5t =0 has roots tan π, tan π, tan π, tan π, tanπ
5  5  5  5 
1  2  3  4 
⇒t4−10t2+5=0 has roots tan π, tan π, tan π, tan π.
5  5  5  5 
 1  2 
⇒ t2−tan2 π  t2−tan2 π =0
   
 5  5 
1  4  2  3 
since tan π=−tan π and tan π=−tan π.
5  5  5  5 
1  2 
⇒ x2 −10x+5=0 has roots tan2 π and tan2 π. (AG)
5  5 
sec2α =1+tan2α
y=1+x⇒x= y−1⇒(y−1)2−10(y−1)+5=0
⇒ y2−12y+16=0 | B1
M1A1
M1
A1
[5]
B1
B1
M1
A1
[4]
M1
M1
A1
[3]
Total
12
Page 9 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2015 | 9231 | 11
Qn &
Part | Solution | Marks
11 E | i j k 8 4
   
E.g. −1 2 0 =4~2
   
−1 0 4 2 1
   
14
  
0.2
  
 0  1  4
= (AG)
42 +22 +12 21
3 4i+2j+k 1
p=  = (4i+2j+k)
21 21  7
1 −3
   
Line AP: r=0+t 2 
   
0 1
   
0
1 1 
For Q 1−3t =0⇒t = ⇒q= 2
3 3
 
1
 
−1  0 
  1 
E.g. AB= 2 , BQ= −4
3
   
0 1
   
i j k 2
 
−1 2 0 =1
 
0 −4 1 4
 
42
  
2.1
  
 1  4  8+2+4 14 2
cos−1 =cos−1 =cos−1 =cos−1
(AG)
21. 21 21 21 3 | M1A1
M1A1
[4]
B1
M1A1
M1A1
[5]
B1
M1A1
M1A1
[5]
Total
14
Page 10 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2015 | 9231 | 11
Qn &
Part | Solution | Marks
11O | Closed curve through pole with correct orientation.
Completely correct.
2× 1 a2∫ π (1−2cosθ +cos2θ)dθ =a2∫ π  3 −2cosθ + 1 cos2θ  dθ
2 1 π 1 π2 2 
2 2
3 1  π
=a2 θ −2sinθ + sin2θ
 
2 4  1
π
2
3 
=a2 π +2
4 
 ds  2
  =a2(1−2cosθ +cos2θ +sin2θ)
dθ 
1 1
=2a2(1−cosθ)=2a2.2sin2 θ =4a2sin2 θ (AG)
2 2
s =2×∫ π 2asin 1 θ dθ
1 π 2
2
 1  π
=4a −2cos θ
 
 2  1
π
2
=4 2a | B1
B1
[2]
M1M1
M1A1
A1
[5]
B1
M1A1
M1
A1
M1A1
[7]
Total
14
Using de Moivre's theorem, show that
$$\tan 5\theta = \frac{5\tan\theta - 10\tan^3\theta + \tan^5\theta}{1 - 10\tan^2\theta + 5\tan^4\theta}.$$ [5]

Hence show that the equation $x^2 - 10x + 5 = 0$ has roots $\tan^2\left(\frac{1}{5}\pi\right)$ and $\tan^2\left(\frac{2}{5}\pi\right)$. [4]

Deduce a quadratic equation, with integer coefficients, having roots $\sec^2\left(\frac{1}{5}\pi\right)$ and $\sec^2\left(\frac{2}{5}\pi\right)$. [3]

\hfill \mbox{\textit{CAIE FP1 2015 Q10 [12]}}
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