CAIE P1 2015 June — Question 5 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2015
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTrig Proofs
TypeSolve equation using proven identity
DifficultyModerate -0.3 Part (i) is a straightforward algebraic manipulation requiring division of numerator and denominator by cos θ, a standard technique worth only 1 mark. Part (ii) uses the proven identity to create a simple quadratic in tan θ, then requires solving within a restricted domain. While multi-step, this follows a clear path with routine techniques and no novel insight, making it slightly easier than a typical A-level question.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities
  1. Prove the identity \(\frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} \equiv \frac{\tan \theta - 1}{\tan \theta + 1}\). [1]
  2. Hence solve the equation \(\frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} = \frac{\tan \theta}{6}\), for \(0° \leqslant \theta \leqslant 180°\). [4]
Question 5:
(ii) ---
5 (i)
AnswerMarks
(ii)sinθ −cosθ
.
sinθ +cosθ
Divides top and bottom by cos θ
t−1
→ t+1
sinθ −cosθ 1
= tanθ
sinθ +cosθ 6
t−1 t
=
→ t+1 6
→ t2 −5t+6=0
→ t = 2 or t = 3
AnswerMarks
→ θ = 63.4º or 71.6ºB1
[1]
B1
M1
A1 A1
AnswerMarks
[4]Answer given.
Using the identity.
Forms a 3 term quadratic with
terms all on same side.
co co
AnswerMarks Guidance
Page 5Mark Scheme Syllabus
Cambridge International AS/A Level – May/June 20159709 12
6
(i)
(ii)
AnswerMarks
(iii)h=60(1−coskt)
Max h when cos = −1 → 120
h = 0 and t = 30, or h =120 and t = 15
→ cos30k = 1 or cos15k = –1
→ 30k = 2π or 15k = π
2π π
→ k = =
30 15
90 = 60(1 – cos kt)
→ coskt = −30 = −0.5
60
2π 4π
→ kt = 3 or → kt = 3
→ Either t = 10 or 20 or both
AnswerMarks
→ t = 10 minutesB1
[1]
M1
A1
[2]
B1
B1
B1
AnswerMarks
[3]Co
Substituting a correct pair of values
into the equation.
co ag
co – but there must be evidence of
correct subtraction.
Question 5:
--- 5 (i)
(ii) ---
5 (i)
(ii) | sinθ −cosθ
.
sinθ +cosθ
Divides top and bottom by cos θ
t−1
→ t+1
sinθ −cosθ 1
= tanθ
sinθ +cosθ 6
t−1 t
=
→ t+1 6
→ t2 −5t+6=0
→ t = 2 or t = 3
→ θ = 63.4º or 71.6º | B1
[1]
B1
M1
A1 A1
[4] | Answer given.
Using the identity.
Forms a 3 term quadratic with
terms all on same side.
co co
Page 5 | Mark Scheme | Syllabus | Paper
Cambridge International AS/A Level – May/June 2015 | 9709 | 12
6
(i)
(ii)
(iii) | h=60(1−coskt)
Max h when cos = −1 → 120
h = 0 and t = 30, or h =120 and t = 15
→ cos30k = 1 or cos15k = –1
→ 30k = 2π or 15k = π
2π π
→ k = =
30 15
90 = 60(1 – cos kt)
→ coskt = −30 = −0.5
60
2π 4π
→ kt = 3 or → kt = 3
→ Either t = 10 or 20 or both
→ t = 10 minutes | B1
[1]
M1
A1
[2]
B1
B1
B1
[3] | Co
Substituting a correct pair of values
into the equation.
co ag
co – but there must be evidence of
correct subtraction.
\begin{enumerate}[label=(\roman*)]
\item Prove the identity $\frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} \equiv \frac{\tan \theta - 1}{\tan \theta + 1}$. [1]
\item Hence solve the equation $\frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} = \frac{\tan \theta}{6}$, for $0° \leqslant \theta \leqslant 180°$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2015 Q5 [5]}}
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