CAIE P1 2003 November — Question 2 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2003
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicQuadratic trigonometric equations
TypeQuartic in sin or cos substitution
DifficultyModerate -0.3 This is a standard A-level trigonometric equation requiring routine substitution using the Pythagorean identity (cos²θ = 1 - sin²θ), solving a quadratic equation, then finding angles. Part (i) is guided algebraic manipulation, and part (ii) involves straightforward application of inverse trig functions. Slightly easier than average due to the scaffolding in part (i) and use of standard techniques throughout.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals

2
  1. Show that the equation \(4 \sin ^ { 4 } \theta + 5 = 7 \cos ^ { 2 } \theta\) may be written in the form \(4 x ^ { 2 } + 7 x - 2 = 0\), where \(x = \sin ^ { 2 } \theta\).
  2. Hence solve the equation \(4 \sin ^ { 4 } \theta + 5 = 7 \cos ^ { 2 } \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

AnswerMarks Guidance
(i) \(4s^4+s=7(1-s^2) \rightarrow 4x^2+7x-2=0\)B1 Use of \(s^2+c^2=1\). Answer given.
(ii) \(4s^4+7s^2-2=0\)
\(\rightarrow s^2 = \frac{1}{4}\) or \(s^2 = -2\)
\(\rightarrow \sin\theta = \pm\frac{1}{2}\)
AnswerMarks Guidance
\(\rightarrow \theta = 30°\) and \(150°\) and \(\theta = 210°\) and \(330°\)M1, A1, A1, A1 [4] Recognition of quadratic in \(s^2\). Co. For \(180° -\) "his value". For other 2 answers from "his value", providing no extra answers in the range or answers from \(s^2= -1\).
**(i)** $4s^4+s=7(1-s^2) \rightarrow 4x^2+7x-2=0$ | B1 | Use of $s^2+c^2=1$. Answer given.

**(ii)** $4s^4+7s^2-2=0$
$\rightarrow s^2 = \frac{1}{4}$ or $s^2 = -2$
$\rightarrow \sin\theta = \pm\frac{1}{2}$
$\rightarrow \theta = 30°$ and $150°$ and $\theta = 210°$ and $330°$ | M1, A1, A1, A1 [4] | Recognition of quadratic in $s^2$. Co. For $180° -$ "his value". For other 2 answers from "his value", providing no extra answers in the range or answers from $s^2= -1$.
2 (i) Show that the equation $4 \sin ^ { 4 } \theta + 5 = 7 \cos ^ { 2 } \theta$ may be written in the form $4 x ^ { 2 } + 7 x - 2 = 0$, where $x = \sin ^ { 2 } \theta$.\\
(ii) Hence solve the equation $4 \sin ^ { 4 } \theta + 5 = 7 \cos ^ { 2 } \theta$, for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.

\hfill \mbox{\textit{CAIE P1 2003 Q2 [5]}}