C3 June 2014 Q7
7.
You must show your working.
(Solutions based entirely on graphical or numerical methods are not acceptable.)
(5)| Scheme | Marks |
|---|---|
| \(\operatorname{cosec}2x+\cot 2x=\dfrac{1}{\sin 2x}+\dfrac{\cos 2x}{\sin 2x}\) | M1 |
| \(=\dfrac{1+\cos 2x}{\sin 2x}\) | M1 |
| \(=\dfrac{1+2\cos^2x-1}{2\sin x\cos x}\) | |
| \(=\dfrac{2\cos^2x}{2\sin x\cos x}\) | M1 A1 |
| \(=\dfrac{\cos x}{\sin x}=\cot x\) | A1* |
| (5) |
Notes
M1 Writing \(\operatorname{cosec}2x=\dfrac{1}{\sin 2x}\) and \(\cot 2x=\dfrac{\cos 2x}{\sin 2x}\) or \(\dfrac{1}{\tan 2x}\)
M1 Writing the lhs as a single fraction \(\dfrac{a+b}{c}\). The denominator must be correct for their terms.
M1 Uses the appropriate double angle formulae/trig identities to produce a fraction in a form containing no addition or subtraction signs. A form \(\dfrac{p\times q}{s\times t}\) or similar
A1 A correct intermediate line. Accept \(\dfrac{2\cos^2x}{2\sin x\cos x}\) or \(\dfrac{2\sin x\cos x}{2\sin x\cos x\tan x}\) or similar
This cannot be scored if errors have been made
A1* Completes the proof by cancelling and using either \(\dfrac{\cos x}{\sin x}=\cot x\) or \(\dfrac{1}{\tan x}=\cot x\)
The cancelling could be implied by seeing \(\dfrac{2}{2}\dfrac{\cos x}{\sin x}\dfrac{\cos x}{\cos x}=\cot x\)
The proof cannot rely on expressions like \(\cot=\dfrac{\cos}{\sin}\) (with missing \(x\)'s) for the final A1
7.(a) Alt 1
| Scheme | Marks |
|---|---|
| \(\operatorname{cosec}2x+\cot 2x=\dfrac{1}{\sin 2x}+\dfrac{1}{\tan 2x}\) | 1st M1 |
| \(=\dfrac{1}{2\sin x\cos x}+\dfrac{1-\tan^2x}{2\tan x}\) | |
| \(=\dfrac{\tan x+(1-\tan^2x)\sin x\cos x}{2\sin x\cos x\tan x}\) or \(=\dfrac{2\tan x+2(1-\tan^2x)\sin x\cos x}{4\sin x\cos x\tan x}\) | 2nd M1 |
| \(=\dfrac{\tan x+\sin x\cos x-\tan^2x\sin x\cos x}{2\sin x\cos x\tan x}\) | |
| \(=\dfrac{\tan x+\sin x\cos x-\tan x\sin^2x}{2\sin x\cos x\tan x}\) | |
| \(=\dfrac{\tan x(1-\sin^2x)+\sin x\cos x}{2\sin x\cos x\tan x}\) | |
| \(=\dfrac{\tan x\cos^2x+\sin x\cos x}{2\sin x\cos x\tan x}\) | |
| \(=\dfrac{\sin x\cos x+\sin x\cos x}{2\sin x\cos x\tan x}\) | |
| \(=\dfrac{2\sin x\cos x}{2\sin x\cos x\tan x}\) oe | 3rd M1A1 |
| \(=\dfrac{1}{\tan x}=\cot x\) | A1* |
| (5) |
7.(a) Alt 2
| Scheme | Marks |
|---|---|
| Example of how main scheme could work in a roundabout route | |
| \(\operatorname{cosec}2x+\cot 2x=\cot x\Leftrightarrow\dfrac{1}{\sin 2x}+\dfrac{1}{\tan 2x}=\dfrac{1}{\tan x}\) | 1st M1 |
| \(\Leftrightarrow\tan 2x\tan x+\sin 2x\tan x=\sin 2x\tan 2x\) | 2nd M1 |
| \(\Leftrightarrow\dfrac{2\tan x}{1-\tan^2x}\times\tan x+2\sin x\cancel{\cos x}\times\dfrac{\sin x}{\cancel{\cos x}}=2\sin x\cos x\times\dfrac{2\tan x}{1-\tan^2x}\) | |
| \(\Leftrightarrow\dfrac{2\tan^2x}{1-\tan^2x}+2\sin^2x=\dfrac{4\sin^2x}{1-\tan^2x}\) | |
| \(\times(1-\tan^2x)\Leftrightarrow 2\tan^2x+2\sin^2x(1-\tan^2x)=4\sin^2x\) | |
| \(\Leftrightarrow 2\tan^2x-2\sin^2x\tan^2x=2\sin^2x\) | |
| \(\Leftrightarrow 2\tan^2x(1-\sin^2x)=2\sin^2x\) | 3rd M1 |
| \(\div 2\tan^2x\Leftrightarrow 1-\sin^2x=\cos^2x\) | A1 |
| As this is true, initial statement is true | A1* |
| (5) |
| Scheme | Marks |
|---|---|
| \(\operatorname{cosec}(4\theta+10^\circ)+\cot(4\theta+10^\circ)=\sqrt{3}\) | |
| \(\cot(2\theta\pm..^\circ)=\sqrt{3}\) | M1 |
| \(2\theta\pm\ldots=30^\circ\Rightarrow\theta=12.5^\circ\) | dM1, A1 |
| \(2\theta\pm\ldots=180+PV^\circ\Rightarrow\theta=..^\circ\) | dM1 |
| \(\theta=102.5^\circ\) | A1 |
| (5) | |
| (10 marks) |
Notes
M1 Attempt to use the solution to part (a) with \(2x=4\theta+10\Rightarrow\) to write or imply \(\cot(2\theta\pm\ldots^\circ)=\sqrt{3}\)
Watch for attempts which start \(\cot\alpha=\sqrt{3}\). The method mark here is not scored until the \(\alpha\) has been replaced by \(2\theta\pm\ldots^\circ\)
Accept a solution from \(\cot(2x\pm\ldots^\circ)=\sqrt{3}\) where \(\theta\) has been replaced by another variable.
dM1 Proceeds from the previous method and uses \(\tan..=\dfrac{1}{\cot..}\) and \(\arctan\left(\dfrac{1}{\sqrt{3}}\right)=30^\circ\) to solve \(2\theta\pm\ldots^\circ=30^\circ\Rightarrow\theta=..\)
A1 \(\theta=12.5^\circ\) or exact equivalent. Condone answers such as \(x=12.5^\circ\)
dM1 This mark is for the correct method to find a second solution to \(\theta\). It is dependent upon the first M only.
Accept \(2\theta\pm\ldots=180+PV^\circ\Rightarrow\theta=..^\circ\)
A1 \(\theta=102.5^\circ\) or exact equivalent. Condone answers such as \(x=102.5^\circ\)
Ignore any solutions outside the range. This mark is withheld for any extra solutions within the range.
If radians appear they could just lose the answer marks. So for example
\(2\theta\pm\ldots=\dfrac{\pi}{6}(0.524)\Rightarrow\theta=..\) is M1dM1A0 followed by
\(2\theta\pm\ldots=\pi+\text{'}\dfrac{\pi}{6}\text{'}\Rightarrow\theta=..\) dM1A0
Special case 1: For candidates in (b) who solve \(\cot(4\theta\pm\ldots^\circ)=\sqrt{3}\) the mark scheme is severe, so we are awarding a special case solution, scoring 00011.
\(\cot(4\theta+\beta^\circ)=\sqrt{3}\Rightarrow 4\theta+\beta=30^\circ\Rightarrow\theta=..\) is M0M0A0 where \(\beta=5^\circ\) or \(10^\circ\)
\(\Rightarrow 4\theta+\beta=210^\circ\Rightarrow\theta=..\) can score M1A1 Special case.
If \(\beta=5^\circ,\ \theta=51.25\) If \(\beta=10^\circ,\ \theta=50\)
Special case 2: Just answers in (b) with no working scores 1 1 0 0 0 for 12.5 and 102.5
BUT \(\cot(2\theta\pm 5^\circ)=\sqrt{3}\Rightarrow\theta=12.5^\circ,102.5^\circ\) scores all available marks.