Standard +0.3 Part (i)(a) is a standard bookwork proof using given formulae. Part (i)(b) requires substituting the proven identity and solving a cubic in cos θ, which is routine but multi-step. Part (ii) is straightforward application of Pythagorean identity to express cot in terms of x. This is a typical C3 question testing standard techniques with no novel insight required, making it slightly easier than average.
7. (i) (a) Prove that
$$\cos 3 \theta \equiv 4 \cos ^ { 3 } \theta - 3 \cos \theta$$
(You may use the double angle formulae and the identity
$$\cos ( A + B ) \equiv \cos A \cos B - \sin A \sin B )$$
(b) Hence solve the equation
$$2 \cos 3 \theta + \cos 2 \theta + 1 = 0$$
giving answers in the interval \(0 \leqslant \theta \leqslant \pi\).
Solutions based entirely on graphical or numerical methods are not acceptable.
(ii) Given that \(\theta = \arcsin x\) and that \(0 < \theta < \frac { \pi } { 2 }\), show that
$$\cot \theta = \frac { \sqrt { \left( 1 - x ^ { 2 } \right) } } { x } , \quad 0 < x < 1$$
\(2\cos\theta(4\cos\theta - 3)(\cos\theta + 1) = 0\) so \(\cos\theta =\)
dM1
Solves by any valid method
\(\cos\theta = \frac{3}{4}\) (or \(0\) or \(-1\))
A1
\(\theta = 0.723\) and no extra answers in range, or \(\theta = \frac{\pi}{2}\) and \(\pi\) (or \(90°\) and \(180°\))
A1, B1
0.723 or answers rounding to this; no extra answers in range; do not accept degrees for A1
Part (ii):
Answer
Marks
Guidance
Answer/Working
Marks
Guidance
\(\sin\theta = x\) and so \(\cos\theta = \sqrt{1-x^2}\)
M1
States \(\cos\theta = \sqrt{1-x^2}\), or uses right-angled triangle with sides \(1\), \(x\) and \(\sqrt{1-x^2}\)
\(\cot\theta = \frac{\cos\theta}{\sin\theta}\)
M1
Implies \(\cot\theta = \frac{\cos\theta}{\sin\theta}\); not \(\frac{\cos}{\sin}\theta\) nor \(\frac{\cos}{\sin}\); or indicates \(\theta\) on diagram and implies \(\cot\theta = \frac{adjacent}{opposite}\)
\(= \frac{\sqrt{1-x^2}}{x}\)
A1*
Clear explanation, no errors, printed answer achieved. Needs both M marks
# Question 7:
## Part (i)(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos 3\theta = \cos 2\theta \cos\theta - \sin 2\theta \sin\theta$ | M1 | Correct statement using compound angle formula |
| $= (2\cos^2\theta - 1)\cos\theta - 2\sin\theta\cos\theta\sin\theta$ | M1 | Uses correct double angle formulae for $\sin 2\theta$ and $\cos 2\theta$ |
| $= 2\cos^3\theta - \cos\theta - 2(1-\cos^2\theta)\cos\theta$ | dM1 | Uses $\sin^2\theta = (1-\cos^2\theta)$ to replace all $\sin$ terms; dependent on both previous Ms |
| $= 4\cos^3\theta - 3\cos\theta$ | A1* | Deduces result with no errors | (4) |
## Part (i)(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $8\cos^3\theta - 6\cos\theta + (2\cos^2\theta - 1) + 1 = 0$ | M1 | Replaces $\cos 3\theta$ and $\cos 2\theta$ by expressions from (a) |
| $8\cos^3\theta + 2\cos^2\theta - 6\cos\theta = 0$ | A1 | Correct cubic shown with 3 terms |
| $2\cos\theta(4\cos\theta - 3)(\cos\theta + 1) = 0$ so $\cos\theta =$ | dM1 | Solves by any valid method |
| $\cos\theta = \frac{3}{4}$ (or $0$ or $-1$) | A1 | |
| $\theta = 0.723$ and no extra answers in range, or $\theta = \frac{\pi}{2}$ and $\pi$ (or $90°$ and $180°$) | A1, B1 | 0.723 or answers rounding to this; no extra answers in range; do not accept degrees for A1 | (6) |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sin\theta = x$ and so $\cos\theta = \sqrt{1-x^2}$ | M1 | States $\cos\theta = \sqrt{1-x^2}$, **or** uses right-angled triangle with sides $1$, $x$ and $\sqrt{1-x^2}$ |
| $\cot\theta = \frac{\cos\theta}{\sin\theta}$ | M1 | Implies $\cot\theta = \frac{\cos\theta}{\sin\theta}$; not $\frac{\cos}{\sin}\theta$ nor $\frac{\cos}{\sin}$; **or** indicates $\theta$ on diagram and implies $\cot\theta = \frac{adjacent}{opposite}$ |
| $= \frac{\sqrt{1-x^2}}{x}$ | A1* | Clear explanation, no errors, printed answer achieved. Needs both M marks | (3) [13] |
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7. (i) (a) Prove that
$$\cos 3 \theta \equiv 4 \cos ^ { 3 } \theta - 3 \cos \theta$$
(You may use the double angle formulae and the identity
$$\cos ( A + B ) \equiv \cos A \cos B - \sin A \sin B )$$
(b) Hence solve the equation
$$2 \cos 3 \theta + \cos 2 \theta + 1 = 0$$
giving answers in the interval $0 \leqslant \theta \leqslant \pi$.\\
Solutions based entirely on graphical or numerical methods are not acceptable.\\
(ii) Given that $\theta = \arcsin x$ and that $0 < \theta < \frac { \pi } { 2 }$, show that
$$\cot \theta = \frac { \sqrt { \left( 1 - x ^ { 2 } \right) } } { x } , \quad 0 < x < 1$$
\hfill \mbox{\textit{Edexcel C3 2014 Q7 [13]}}