2. (a) Show that
$$\frac { 1 - \cos 2 x } { 2 \sin 2 x } \equiv k \tan x \quad x \neq ( 90 n ) ^ { \circ } \quad n \in \mathbb { Z }$$
where \(k\) is a constant to be found.
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Question 2:
Part (a)
Answer Marks
Guidance
Answer/Working Marks
Guidance
States or uses \(\sin 2x = 2\sin x\cos x\) B1
\(\frac{1-\cos 2x}{2\sin 2x} = \frac{1-(1-2\sin^2 x)}{4\sin x\cos x} = \frac{2\sin^2 x}{4\sin x\cos x} = \frac{1}{2}\tan x\) M1 A1
M1: Uses \(1-\cos 2x = \pm 2\sin^2 x\) and \(2\sin 2x = A\sin x\cos x\) and proceeds to \(k\tan x\). A1: Fully correct with correct bracketing if seen.
(3 marks) Part (b) — Way 1
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(\frac{9(1-\cos 2\theta)}{2\sin 2\theta} = 2\sec^2\theta \Rightarrow 9\times\frac{1}{2}\tan\theta = 2\sec^2\theta\) M1
Uses part (a) to form equation of form \(A\tan\theta = B\sec^2\theta\)
Attempts \(1+\tan^2\theta = \sec^2\theta \Rightarrow 4\tan^2\theta - 9\tan\theta + 4 = 0\) dM1 A1
dM1: Replaces \(\sec^2\theta\) by \(\pm1\pm\tan^2\theta\) to form 3TQ in \(\tan\theta\). A1: Correct 3TQ e.g. \(9\tan\theta = 4 + 4\tan^2\theta\)
\(\tan\theta = \frac{9\pm\sqrt{17}}{8}\) \((1.64038...,\ 0.60961...)\) M1
Correct attempt to find at least one value of \(\tan\theta\). Allow calculator solutions; condone decimals to at least 2sf.
\(\theta =\) awrt \(31.4°\), \(58.6°\) A1 A1
A1: One of either awrt \(31.4°\) or awrt \(58.6°\). A1: Both awrt \(31.4°\) and \(58.6°\). Withhold if extra solutions in range.
(6 marks) Part (b) — Way 2
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(\frac{9(1-\cos 2\theta)}{2\sin 2\theta} = 2\sec^2\theta \Rightarrow 9\times\frac{1}{2}\tan\theta = 2\sec^2\theta\) M1
\(\frac{9}{2}\tan\theta = 2\sec^2\theta \Rightarrow \frac{9\sin\theta}{\cos\theta} = \frac{4}{\cos^2\theta} \Rightarrow 9\sin\theta\cos\theta = 4 \Rightarrow \sin 2\theta = \frac{8}{9}\) dM1 A1
Uses \(\sec^2\theta = \frac{1}{\cos^2\theta}\), \(\tan\theta = \frac{\sin\theta}{\cos\theta}\), \(\sin 2\theta = ...\sin\theta\cos\theta\) to produce \(\sin 2\theta = C\). A1: \(\sin 2\theta = \frac{8}{9}\)
\(\sin 2\theta = \frac{8}{9} \Rightarrow 2\theta = 62.7333...,\ 117.266... \Rightarrow \theta = ...\) M1
Full method to find at least one \(\theta\) from \(\sin 2\theta = C\), \(-1
\(\theta =\) awrt \(31.4°\), \(58.6°\) A1 A1
As above
(9 marks total)
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# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| States or uses $\sin 2x = 2\sin x\cos x$ | B1 | |
| $\frac{1-\cos 2x}{2\sin 2x} = \frac{1-(1-2\sin^2 x)}{4\sin x\cos x} = \frac{2\sin^2 x}{4\sin x\cos x} = \frac{1}{2}\tan x$ | M1 A1 | M1: Uses $1-\cos 2x = \pm 2\sin^2 x$ and $2\sin 2x = A\sin x\cos x$ and proceeds to $k\tan x$. A1: Fully correct with correct bracketing if seen. |
**(3 marks)**
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## Part (b) — Way 1
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{9(1-\cos 2\theta)}{2\sin 2\theta} = 2\sec^2\theta \Rightarrow 9\times\frac{1}{2}\tan\theta = 2\sec^2\theta$ | M1 | Uses part (a) to form equation of form $A\tan\theta = B\sec^2\theta$ |
| Attempts $1+\tan^2\theta = \sec^2\theta \Rightarrow 4\tan^2\theta - 9\tan\theta + 4 = 0$ | dM1 A1 | dM1: Replaces $\sec^2\theta$ by $\pm1\pm\tan^2\theta$ to form 3TQ in $\tan\theta$. A1: Correct 3TQ e.g. $9\tan\theta = 4 + 4\tan^2\theta$ |
| $\tan\theta = \frac{9\pm\sqrt{17}}{8}$ $(1.64038...,\ 0.60961...)$ | M1 | Correct attempt to find at least one value of $\tan\theta$. Allow calculator solutions; condone decimals to at least 2sf. |
| $\theta =$ awrt $31.4°$, $58.6°$ | A1 A1 | A1: One of either awrt $31.4°$ **or** awrt $58.6°$. A1: Both awrt $31.4°$ **and** $58.6°$. Withhold if extra solutions in range. |
**(6 marks)**
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## Part (b) — Way 2
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{9(1-\cos 2\theta)}{2\sin 2\theta} = 2\sec^2\theta \Rightarrow 9\times\frac{1}{2}\tan\theta = 2\sec^2\theta$ | M1 | |
| $\frac{9}{2}\tan\theta = 2\sec^2\theta \Rightarrow \frac{9\sin\theta}{\cos\theta} = \frac{4}{\cos^2\theta} \Rightarrow 9\sin\theta\cos\theta = 4 \Rightarrow \sin 2\theta = \frac{8}{9}$ | dM1 A1 | Uses $\sec^2\theta = \frac{1}{\cos^2\theta}$, $\tan\theta = \frac{\sin\theta}{\cos\theta}$, $\sin 2\theta = ...\sin\theta\cos\theta$ to produce $\sin 2\theta = C$. A1: $\sin 2\theta = \frac{8}{9}$ |
| $\sin 2\theta = \frac{8}{9} \Rightarrow 2\theta = 62.7333...,\ 117.266... \Rightarrow \theta = ...$ | M1 | Full method to find at least one $\theta$ from $\sin 2\theta = C$, $-1<C<1$. e.g. $2\theta = \sin^{-1}C$, $\theta = \frac{1}{2}\sin^{-1}C$ |
| $\theta =$ awrt $31.4°$, $58.6°$ | A1 A1 | As above |
**(9 marks total)**
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2. (a) Show that
$$\frac { 1 - \cos 2 x } { 2 \sin 2 x } \equiv k \tan x \quad x \neq ( 90 n ) ^ { \circ } \quad n \in \mathbb { Z }$$
where $k$ is a constant to be found.\\
(b) Hence solve, for $0 < \theta < 90 ^ { \circ }$
$$\frac { 9 ( 1 - \cos 2 \theta ) } { 2 \sin 2 \theta } = 2 \sec ^ { 2 } \theta$$
giving your answers to one decimal place.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)\\
\hfill \mbox{\textit{Edexcel P3 2021 Q2 [9]}}