C3 January 2013 Q4

EdexcelOld spec8 marksTrigonometry

(a) Express \(6\cos\theta + 8\sin\theta\) in the form \(R\cos(\theta - \alpha)\), where \(R \gt 0\) and \(0 \lt \alpha \lt \dfrac{\pi}{2}\).
Give the value of \(\alpha\) to 3 decimal places. (4)

(b) \[\mathrm{p}(\theta) = \frac{4}{12 + 6\cos\theta + 8\sin\theta}, \qquad 0 \leqslant \theta \leqslant 2\pi\] Calculate

(i) the maximum value of \(\mathrm{p}(\theta)\),

(ii) the value of \(\theta\) at which the maximum occurs. (4)

Mark scheme (a)

Scheme	Marks
\(R^2 = 6^2 + 8^2 \Rightarrow R = 10\)	M1A1
\(\tan\alpha = \dfrac{8}{6} \Rightarrow \alpha = \text{awrt } 0.927\)	M1A1
	(4)

Notes

M1 Using Pythagoras’ Theorem with 6 and 8 to find R. Accept \(R^2 = 6^2 + 8^2\)
If \(\alpha\) has been found first accept \(R = \pm\dfrac{8}{\sin'\alpha'}\) or \(R = \pm\dfrac{6}{\cos'\alpha'}\)

A1 \(R = 10\). Many candidates will just write this down which is fine for the 2 marks.
Accept \(\pm 10\) but not -10

M1 For \(\tan\alpha = \pm\dfrac{8}{6}\) or \(\tan\alpha = \pm\dfrac{6}{8}\)
If R is used then only accept \(\sin\alpha = \pm\dfrac{8}{R}\) or \(\cos\alpha = \pm\dfrac{6}{R}\)

A1 \(\alpha = \text{awrt } 0.927\). Note that \(53.1^0\) is A0

Mark scheme (b)

Scheme	Marks
(i) \(\mathrm{p}(x) = \dfrac{4}{12 + 10\cos(\theta - 0.927)}\)
\(\mathrm{p}(x) = \dfrac{4}{12 - 10}\)	M1
Maximum = 2	A1
	(2)
(ii) \(\theta - '\text{their } \alpha' = \pi\)	M1
\(\theta = \text{awrt } 4.07\)	A1
	(2)
	(8 marks)

Notes

Note that (b)(i) and (b)(ii) can be marked together

(i) M1 Award for \(\mathrm{p}(x) = \dfrac{4}{12 - 'R'}\).

A1 Cao \(\mathrm{p}(x)_{\max} = 2\).
The answer is acceptable for both marks as long as no incorrect working is seen

(ii) M1 For setting \(\theta - '\text{their } \alpha' = \pi\) and proceeding to \(\theta\)=..
If working exclusively in degrees accept \(\theta - '\text{their } \alpha' = 180\)
Do not accept mixed units

A1 \(\theta = \text{awrt } 4.07\). If the final A mark in part (a) is lost for 53.1, then accept awrt 233.1