CAIE P1 2015 November — Question 7 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2015
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicQuadratic trigonometric equations
TypeShow then solve: tan/sin/cos identity manipulation
DifficultyModerate -0.3 Part (a) is a standard trigonometric equation requiring routine manipulation using tan θ = sin θ/cos θ and the Pythagorean identity to convert to a quadratic in cos θ, then solving. Part (b) involves straightforward substitution into y = a cos x - b to find intercepts. Both parts are typical P1 exercises requiring only standard techniques with no novel insight, making this slightly easier than average.
Spec1.02f Solve quadratic equations: including in a function of unknown1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals

7
  1. Show that the equation \(\frac { 1 } { \cos \theta } + 3 \sin \theta \tan \theta + 4 = 0\) can be expressed as $$3 \cos ^ { 2 } \theta - 4 \cos \theta - 4 = 0$$ and hence solve the equation \(\frac { 1 } { \cos \theta } + 3 \sin \theta \tan \theta + 4 = 0\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
  2. \includegraphics[max width=\textwidth, alt={}, center]{5c1ab2aa-3609-4245-b87a-98ecedc83a11-3_581_773_1400_721} The diagram shows part of the graph of \(y = a \cos x - b\), where \(a\) and \(b\) are constants. The graph crosses the \(x\)-axis at the point \(C \left( \cos ^ { - 1 } c , 0 \right)\) and the \(y\)-axis at the point \(D ( 0 , d )\). Find \(c\) and \(d\) in terms of \(a\) and \(b\).

Question 7:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
(a) \(1 + 3\sin^2\theta + 4\cos\theta = 0\)M1 Attempt to multiply by \(\cos\theta\)
\(1 + 3(1 - \cos^2\theta) + 4\cos\theta + 0\)M1 Use \(c^2 + s^2 = 1\)
\(3\cos^2\theta - 4\cos\theta - 4 = 0\) AGA1
\(\cos\theta = -2/3\)B1 Ignore other solution
\(\theta = 131.8\) *or* \(228.2\)B1B1\(\checkmark\) Ft for \(360 - 1^{\text{st}}\) soln. \(-1\) extra solns in range; Radians \(2.30\) & \(3.98\) scores SCB1
[6]
(b) \(c = b/a\) caoB1
\(d = a - b\)B1 Allow \(D = (0,\ a-b)\)
[2]
## Question 7:

| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** $1 + 3\sin^2\theta + 4\cos\theta = 0$ | M1 | Attempt to multiply by $\cos\theta$ |
| $1 + 3(1 - \cos^2\theta) + 4\cos\theta + 0$ | M1 | Use $c^2 + s^2 = 1$ |
| $3\cos^2\theta - 4\cos\theta - 4 = 0$ **AG** | A1 | |
| $\cos\theta = -2/3$ | B1 | Ignore other solution |
| $\theta = 131.8$ *or* $228.2$ | B1B1$\checkmark$ | Ft for $360 - 1^{\text{st}}$ soln. $-1$ extra solns in range; Radians $2.30$ & $3.98$ scores SCB1 |
| **[6]** | | |
| **(b)** $c = b/a$ cao | B1 | |
| $d = a - b$ | B1 | Allow $D = (0,\ a-b)$ |
| **[2]** | | |

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7
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\frac { 1 } { \cos \theta } + 3 \sin \theta \tan \theta + 4 = 0$ can be expressed as

$$3 \cos ^ { 2 } \theta - 4 \cos \theta - 4 = 0$$

and hence solve the equation $\frac { 1 } { \cos \theta } + 3 \sin \theta \tan \theta + 4 = 0$ for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{5c1ab2aa-3609-4245-b87a-98ecedc83a11-3_581_773_1400_721}

The diagram shows part of the graph of $y = a \cos x - b$, where $a$ and $b$ are constants. The graph crosses the $x$-axis at the point $C \left( \cos ^ { - 1 } c , 0 \right)$ and the $y$-axis at the point $D ( 0 , d )$. Find $c$ and $d$ in terms of $a$ and $b$.
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2015 Q7 [8]}}