C3 June 2015 Q8
8.
| Scheme | Marks |
|---|---|
| \(\sec 2A + \tan 2A = \dfrac{1}{\cos 2A} + \dfrac{\sin 2A}{\cos 2A}\) | B1 |
| \(= \dfrac{1 + \sin 2A}{\cos 2A}\) | M1 |
| \(= \dfrac{1 + 2\sin A\cos A}{\cos^2 A - \sin^2 A}\) | M1 |
| \(= \dfrac{\cos^2 A + \sin^2 A + 2\sin A\cos A}{\cos^2 A - \sin^2 A}\) | |
| \(= \dfrac{(\cos A + \sin A)(\cos A + \sin A)}{(\cos A + \sin A)(\cos A - \sin A)}\) | M1 |
| \(= \dfrac{\cos A + \sin A}{\cos A - \sin A}\) | A1* |
| (5) |
Notes
B1: A correct identity for \(\sec 2A = \dfrac{1}{\cos 2A}\) OR \(\tan 2A = \dfrac{\sin 2A}{\cos 2A}\).
It need not be in the proof and it could be implied by the sight of \(\sec 2A = \dfrac{1}{\cos^2 A - \sin^2 A}\)
M1: For setting their expression as a single fraction. The denominator must be correct for their fractions and at least two terms on the numerator.
This is usually scored for \(\dfrac{1 + \cos 2A\tan 2A}{\cos 2A}\) or \(\dfrac{1 + \sin 2A}{\cos 2A}\)
M1: For getting an expression in just \(\sin A\) and \(\cos A\) by using the double angle identities \(\sin 2A = 2\sin A\cos A\) and \(\cos 2A = \cos^2 A - \sin^2 A\), \(2\cos^2 A - 1\) or \(1 - 2\sin^2 A\).
Alternatively for getting an expression in just \(\sin A\) and \(\cos A\) by using the double angle identities \(\sin 2A = 2\sin A\cos A\) and \(\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}\) with \(\tan A = \dfrac{\sin A}{\cos A}\).
For example \(= \dfrac{1}{\cos^2 A - \sin^2 A} + \dfrac{2\sin A/\cos A}{1 - \sin^2 A/\cos^2 A}\) is B1M0M1 so far
M1: In the main scheme it is for replacing 1 by \(\cos^2 A + \sin^2 A\) and factorising both numerator and denominator
A1*: Cancelling to produce given answer with no errors.
Allow a consistent use of another variable such as \(\theta\), but mixing up variables will lose the A1*.
Alt I From RHS
| Scheme | Marks |
|---|---|
| \(\dfrac{\cos A + \sin A}{\cos A - \sin A} = \dfrac{\cos A + \sin A}{\cos A - \sin A}, \dfrac{\cos A + \sin A}{\cos A + \sin A}\) | |
| \(= \dfrac{\cos^2 A + \sin^2 A + 2\sin A\cos A}{\cos^2 A - \sin^2 A}\) | |
| \(= \dfrac{1 + \sin 2A}{\cos 2A}\) | (Pythagoras) M1 (Double Angle) M1 |
| \(= \dfrac{1}{\cos 2A} + \dfrac{\sin 2A}{\cos 2A}\) | (Single Fraction) M1 |
| \(= \sec 2A + \tan 2A\) | B1(Identity), A1* |
Alt II Both sides
| Scheme | Marks |
|---|---|
| Assume true \(\sec 2A + \tan 2A = \dfrac{\cos A + \sin A}{\cos A - \sin A}\) | |
| \(\dfrac{1}{\cos 2A} + \dfrac{\sin 2A}{\cos 2A} = \dfrac{\cos A + \sin A}{\cos A - \sin A}\) | B1 (identity) |
| \(\dfrac{1 + \sin 2A}{\cos 2A} = \dfrac{\cos A + \sin A}{\cos A - \sin A}\) | M1 (single fraction) |
| \(\dfrac{1 + 2\sin A\cos A}{\cos^2 A - \sin^2 A} = \dfrac{\cos A + \sin A}{\cos A - \sin A}\) | M1(double angles) |
| \(\times(\cos A - \sin A) \Rightarrow \dfrac{1 + 2\sin A\cos A}{\cos A + \sin A} = \cos A + \sin A\) | |
| \(1 + 2\sin A\cos A = \cos^2 A + 2\sin A\cos A + \sin^2 A = 1 + 2\sin A\cos A\) True | M1(Pythagoras)A1* |
Alt III Very difficult
| Scheme | Marks |
|---|---|
| \(\sec 2A + \tan 2A = \dfrac{1}{\cos 2A} + \tan 2A\) | (Identity) B1 |
| \(= \dfrac{1}{\cos 2A} + \dfrac{2\tan A}{1 - \tan^2 A}\) | |
| \(= \dfrac{1 - \tan^2 A + 2\tan A\cos 2A}{\cos 2A(1 - \tan^2 A)}\) | (Single fraction) M1 |
| \(= \dfrac{1 - \tan^2 A + 2\tan A(\cos^2 A - \sin^2 A)}{(\cos^2 A - \sin^2 A)(1 - \tan^2 A)}\) | |
| \(= \dfrac{1 - \dfrac{\sin^2 A}{\cos^2 A} + 2\dfrac{\sin A}{\cos A}(\cos^2 A - \sin^2 A)}{(\cos^2 A - \sin^2 A)\left(1 - \dfrac{\sin^2 A}{\cos^2 A}\right)}\) | (Double Angle and in just sin and cos) M1 |
| \(\times\cos^2 A = \dfrac{\cos^2 A - \sin^2 A + 2\sin A\cos A(\cos^2 A - \sin^2 A)}{(\cos^2 A - \sin^2 A)(\cos^2 A - \sin^2 A)}\) | |
| \(= \dfrac{\cancel{(\cos^2 A - \sin^2 A)}(1 + 2\sin A\cos A)}{\cancel{(\cos^2 A - \sin^2 A)}(\cos^2 A - \sin^2 A)}\) | |
| Final two marks as in main scheme | M1A1* |
| Scheme | Marks |
|---|---|
| \(\sec 2\theta + \tan 2\theta = \dfrac{1}{2} \Rightarrow \dfrac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} = \dfrac{1}{2}\) | |
| \(\Rightarrow 2\cos\theta + 2\sin\theta = \cos\theta - \sin\theta\) | |
| \(\Rightarrow \tan\theta = -\dfrac{1}{3}\) | M1 A1 |
| \(\Rightarrow \theta = \text{awrt } 2.820, 5.961\) | dM1A1 |
| (4) | |
| (9 marks) |
Notes
M1: For using part (a), cross multiplying, dividing by \(\cos\theta\) to reach \(\tan\theta = k\)
Condone \(\tan 2\theta = k\) for this mark only
A1: \(\tan\theta = -\dfrac{1}{3}\)
dM1: Scored for \(\tan\theta = k\) leading to at least one value (with 1 dp accuracy) for \(\theta\) between 0 and \(2\pi\). You may have to use a calculator to check. Allow answers in degrees for this mark.
A1: \(\theta = \text{awrt } 2.820, 5.961\) with no extra solutions within the range. Condone 2.82 for 2.820.
You may condone different/ mixed variables in part (b)
There are some long winded methods. Eg. M1, dM1 applied as in main scheme
\(\Rightarrow (2\cos\theta + 2\sin\theta)^2 = (\cos\theta - \sin\theta)^2 \Rightarrow 4 + 4\sin 2\theta = 1 - \sin 2\theta\)
\(\Rightarrow \sin 2\theta = -\dfrac{3}{5}\) is M1 (for \(\sin 2\theta = k\)) A1
\(\Rightarrow \theta = 2.820, 5.961\) for dM1 (for \(\theta = \dfrac{\arcsin k}{2}\)) A1
\(\cos\theta + 3\sin\theta = 0 \Rightarrow \left(\sqrt{10}\right)\cos(\theta - 1.25) = 0\) M1 for..\(\cos(\theta - \alpha) = 0, \alpha = \arctan\left(\pm\dfrac{3}{1} \text{ or } \pm\dfrac{1}{3}\right)\) A1
\(\Rightarrow \theta = 2.820, 5.961\) dM1 A1
\(\cos\theta + 3\sin\theta = 0 \Rightarrow \left(\sqrt{10}\right)\sin(\theta + 0.32) = 0\) M1 A1
\(\Rightarrow \theta = 2.820, 5.961\) dM1 A1
\(\cos\theta = -3\sin\theta \Rightarrow \cos^2\theta = 9\sin^2\theta \Rightarrow \sin^2\theta = \dfrac{1}{10} \Rightarrow \sin\theta = (\pm)\sqrt{\dfrac{1}{10}}\) M1 A1
\(\Rightarrow \theta = 2.820, 5.961\) dM1 A1
\(\cos\theta = -3\sin\theta \Rightarrow \cos^2\theta = 9\sin^2\theta \Rightarrow \cos^2\theta = \dfrac{9}{10} \Rightarrow \cos\theta = (\pm)\sqrt{\dfrac{9}{10}}\) M1 A1
\(\Rightarrow \theta = 2.820, 5.961\) dM1 A1