June 2018 Paper 2 Q9

Notes

B1: Gives the correct fraction such as \(\dfrac{\cos(\theta + h) - \cos\theta}{h}\) or \(\dfrac{\cos(\theta + \delta\theta) - \cos\theta}{\delta\theta}\)
Allow \(\dfrac{\cos(\theta + h) - \cos\theta}{(\theta + h) - \theta}\) o.e. Note: \(\cos(\theta + h)\) or \(\cos(\theta + \delta\theta)\) may be expanded

M1: Uses the compound angle formula for \(\cos(\theta + h)\) to give \(\cos\theta\cos h \pm \sin\theta\sin h\)

A1: Achieves \(\dfrac{\cos\theta\cos h - \sin\theta\sin h - \cos\theta}{h}\) or equivalent

dM1: dependent on both the B and M marks being awarded
Complete attempt to apply the given limits to the gradient of their chord

Note: They must isolate \(\dfrac{\sin h}{h}\) and \(\left(\dfrac{\cos h - 1}{h}\right)\), and replace \(\dfrac{\sin h}{h}\) with 1 and replace \(\left(\dfrac{\cos h - 1}{h}\right)\) with 0

A1*: cso. Uses correct mathematical language of limiting arguments to prove \(\dfrac{\mathrm{d}}{\mathrm{d}\theta}(\cos\theta) = -\sin\theta\)

Note: Acceptable responses for the final A mark include:

• \(\dfrac{\mathrm{d}}{\mathrm{d}\theta}(\cos\theta) = \lim\limits_{h \to 0}\left(-\dfrac{\sin h}{h}\sin\theta + \left(\dfrac{\cos h - 1}{h}\right)\cos\theta\right) = -1\sin\theta + 0\cos\theta = -\sin\theta\)
• Gradient of chord \(= -\dfrac{\sin h}{h}\sin\theta + \left(\dfrac{\cos h - 1}{h}\right)\cos\theta\). As \(h \rightarrow 0\), gradient of chord tends to the gradient of the curve, so derivative is \(-\sin\theta\)
• Gradient of chord \(= -\dfrac{\sin h}{h}\sin\theta + \left(\dfrac{\cos h - 1}{h}\right)\cos\theta\). As \(h \rightarrow 0\), gradient of curve is \(-\sin\theta\)

Note: Give final A0 for the following example which shows no limiting arguments:
when \(h = 0,\ \dfrac{\mathrm{d}}{\mathrm{d}\theta}(\cos\theta) = -\dfrac{\sin h}{h}\sin\theta + \left(\dfrac{\cos h - 1}{h}\right)\cos\theta = -1\sin\theta + 0\cos\theta = -\sin\theta\)

Note: Do not allow the final A1 for stating \(\dfrac{\sin h}{h} = 1\) or \(\left(\dfrac{\cos h - 1}{h}\right) = 0\) and attempting to apply these

Note: In this question \(\delta\theta\) may be used in place of \(h\)

Note: Condone \(\mathrm{f}^{\prime}(\theta)\) where \(\mathrm{f}(\theta) = \cos\theta\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}\theta}\) where \(y = \cos\theta\) used in place of \(\dfrac{\mathrm{d}}{\mathrm{d}\theta}(\cos\theta)\)

Note: Condone \(x\) used in place of \(\theta\) if this is done consistently

Note: Give final A0 for

• \(\dfrac{\mathrm{d}}{\mathrm{d}\theta}(\cos x) = \lim\limits_{h \to 0}\left(-\dfrac{\sin h}{h}\sin\theta + \left(\dfrac{\cos h - 1}{h}\right)\cos\theta\right) = -1\sin\theta + 0\cos\theta = -\sin\theta\)
• \(\dfrac{\mathrm{d}}{\mathrm{d}\theta} = \ldots\)
• Defining \(\mathrm{f}(x) = \cos\theta\) and applying \(\mathrm{f}^{\prime}(x) = \ldots\)
• \(\dfrac{\mathrm{d}}{\mathrm{d}x}(\cos\theta)\)

Note: Give final A1 for a correct limiting argument in \(x\), followed by \(\dfrac{\mathrm{d}}{\mathrm{d}\theta}(\cos\theta) = -\sin\theta\)
e.g. \(\dfrac{\mathrm{d}}{\mathrm{d}\theta}(\cos x) = \lim\limits_{h \to 0}\left(-\dfrac{\sin h}{h}\sin x + \left(\dfrac{\cos h - 1}{h}\right)\cos x\right) = -1\sin x + 0\cos x = -\sin x\)
\(\Rightarrow \dfrac{\mathrm{d}}{\mathrm{d}\theta}(\cos\theta) = -\sin\theta\)

Note: Applying \(h \rightarrow 0,\ \sin h \rightarrow h,\ \cos h \rightarrow 1\) to give e.g.
\(\lim\limits_{h \to 0}\left(\dfrac{\cos\theta\cos h - \sin\theta\sin h - \cos\theta}{h}\right) = \left(\dfrac{\cos\theta(1) - \sin\theta(h) - \cos\theta}{h}\right) = \dfrac{-\sin\theta(h)}{h} = -\sin\theta\)
is final M0 A0 for incorrect application of limits

Note: \(\lim\limits_{h \to 0}\left(\dfrac{\cos\theta\cos h - \sin\theta\sin h - \cos\theta}{h}\right) = \lim\limits_{h \to 0}\left(-\dfrac{\sin h}{h}\sin\theta + \left(\dfrac{\cos h - 1}{h}\right)\cos\theta\right)\)
\(= \lim\limits_{h \to 0}\left(-(1)\sin\theta + 0\cos\theta\right) = -\sin\theta\). So for not removing \(\lim\limits_{h \to 0}\) when the limit was taken is final A0

Note: Alternative Method: Considers \(\dfrac{\cos(\theta + h) - \cos(\theta - h)}{(\theta + h) - (\theta - h)}\) which simplifies to \(\dfrac{-2\sin\theta\sin h}{2h}\)