C3 June 2013 Q3
3. Given that\[2\cos(x+50)^\circ=\sin(x+40)^\circ\]
| Scheme | Marks |
|---|---|
| \(2\cos x\cos 50-2\sin x\sin 50=\sin x\cos 40+\cos x\sin 40\) | M1 |
| \(\sin x(\cos 40+2\sin 50)=\cos x(2\cos 50-\sin 40)\) | |
| \(\div\cos x\Rightarrow\tan x(\cos 40+2\sin 50)=2\cos 50-\sin 40\) | M1 |
| \(\tan x=\dfrac{2\cos 50-\sin 40}{\cos 40+2\sin 50},\) (or numerical answer awrt 0.28) | A1 |
| States or uses \(\cos 50=\sin 40\) and \(\cos 40=\sin 50\) and so \(\tan x^\circ=\tfrac{1}{3}\tan 40^\circ\ *\) cao | A1 * |
| (4) |
Notes
M1 Expand both expressions using \(\cos(x+50)=\cos x\cos 50-\sin x\sin 50\) and \(\sin(x+40)=\sin x\cos 40+\cos x\sin 40\). Condone a missing bracket on the lhs.
The terms of the expansions must be correct as these are given identities. You may condone a sign error on one of the expressions.
Allow if written separately and not in a connected equation.
M1 Divide by \(\cos x\) to reach an equation in \(\tan x\).
Below is an example of M1M1 with incorrect sign on left hand side
\(2\cos x\cos 50+2\sin x\sin 50=\sin x\cos 40+\cos x\sin 40\)
\(\Rightarrow 2\cos 50+2\tan x\sin 50=\tan x\cos 40+\sin 40\)
This is independent of the first mark.
A1 \(\tan x=\dfrac{2\cos 50-\sin 40}{\cos 40+2\sin 50}\)
Accept for this mark \(\tan x=\text{awrt }0.28\ldots\) as long as M1M1 has been achieved.
A1* States or uses \(\cos 50=\sin 40\) and \(\cos 40=\sin 50\) leading to showing
\(\tan x=\dfrac{2\cos 50-\sin 40}{\cos 40+2\sin 50}=\dfrac{\sin 40}{3\cos 40}=\dfrac{1}{3}\tan 40\)
This is a given answer and all steps above must be shown. The line above is acceptable.
Do not allow from \(\tan x=\text{awrt }0.28\ldots\)
Alt 1 3(a)
| Scheme | Marks |
|---|---|
| \(2\cos x\cos 50-2\sin x\sin 50=\sin x\cos 40+\cos x\sin 40\) | M1 |
| \(2\cos x\sin 40-2\sin x\cos 40=\sin x\cos 40+\cos x\sin 40\) | |
| \(\div\cos x\Rightarrow 2\sin 40-2\tan x\cos 40=\tan x\cos 40+\sin 40\) | M1 |
| \(\tan x=\dfrac{\sin 40}{3\cos 40}\) ( or numerical answer awrt 0.28), \(\ \Rightarrow\tan x=\dfrac{1}{3}\tan 40\) | A1,A1 |
Alt 2 3(a)
| Scheme | Marks |
|---|---|
| \(2\cos(x+50)=\sin(x+40)\Rightarrow 2\sin(40-x)=\sin(x+40)\) | |
| \(2\cos x\sin 40-2\sin x\cos 40=\sin x\cos 40+\cos x\sin 40\) | M1 |
| \(\div\cos x\Rightarrow 2\sin 40-2\tan x\cos 40=\tan x\cos 40+\sin 40\) | M1 |
| \(\tan x=\dfrac{\sin 40}{3\cos 40}\) ( or numerical answer awrt 0.28), \(\ \Rightarrow\tan x=\dfrac{1}{3}\tan 40\) | A1,A1 |
| Scheme | Marks |
|---|---|
| Deduces \(\tan 2\theta=\tfrac{1}{3}\tan 40\) | M1 |
| \(2\theta=15.6\) so \(\theta=\) awrt 7.8(1) One answer | A1 |
| Also \(2\theta=195.6,\ 375.6,\ 555.6\Rightarrow\theta=..\) | M1 |
| \(\theta=\) awrt 7.8, 97.8, 187.8, 277.8 All 4 answers | A1 |
| (4) | |
| (8 marks) |
Notes
M1 For linking part (a) with (b). Award for writing \(\tan 2\theta=\tfrac{1}{3}\tan 40\)
A1 Solves to find one solution of \(\theta\) which is usually (awrt) 7.8
M1 Uses the correct method to find at least another value of \(\theta\). It must be a full method but can be implied by any correct answer.
Accept \(\theta=\dfrac{180+\textit{their}\,\alpha}{2},\ (or)\ \dfrac{360+\textit{their}\,\alpha}{2},\ (or)\ \dfrac{540+\textit{their}\,\alpha}{2}\)
A1 Obtains all four answers awrt 1dp. \(\theta=7.8,\ 97.8,\ 187.8,\ 277.8\).
Ignore any extra solutions outside the range.
Withhold this mark for extras inside the range.
Condone a different variable. Accept \(x=7.8,\ 97.8,\ 187.8,\ 277.8\)
Answers fully given in radians, loses the first A mark.
Acceptable answers in rads are awrt 0.136, 1.71, 3.28, 4.85
Mixed units can only score the first M 1