5. (a) Using the identity \(\cos ( A + B ) \equiv \cos A \cos B - \sin A \sin B\), prove that
$$\cos 2 A \equiv 1 - 2 \sin ^ { 2 } A$$
(b) Show that
$$2 \sin 2 \theta - 3 \cos 2 \theta - 3 \sin \theta + 3 \equiv \sin \theta ( 4 \cos \theta + 6 \sin \theta - 3 )$$
(c) Express \(4 \cos \theta + 6 \sin \theta\) in the form \(R \sin ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\).
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Question 5:
Part (a):
Answer Marks
Guidance
Working/Answer Marks
Guidance
\(\cos 2A = \cos^2 A - \sin^2 A\) (+ use of \(\cos^2 A + \sin^2 A \equiv 1\)) M1
\(= (1-\sin^2 A) - \sin^2 A = 1 - 2\sin^2 A\) * A1
Part (b):
Answer Marks
Guidance
Working/Answer Marks
Guidance
\(2\sin 2\theta - 3\cos 2\theta - 3\sin\theta + 3 \equiv 4\sin\theta\cos\theta; -3(1-2\sin^2\theta) - 3\sin\theta + 3\) B1; M1
\(\equiv 4\sin\theta\cos\theta + 6\sin^2\theta - 3\sin\theta\) M1
\(\equiv \sin\theta(4\cos\theta + 6\sin\theta - 3)\) * A1
Part (c):
Answer Marks
Guidance
Working/Answer Marks
Guidance
\(4\cos\theta + 6\sin\theta \equiv R\sin\theta\cos\alpha + R\cos\theta\sin\alpha\); complete method for \(R\) M1
May be implied by correct answer
\([R^2 = 4^2 + 6^2,\ R\sin\alpha = 4,\ R\cos\alpha = 6]\); \(R = \sqrt{52}\) or \(7.21\) A1
Complete method for \(\alpha\); \(\alpha = 0.588\) (allow \(33.7°\)) M1 A1
Part (d):
Answer Marks
Guidance
Working/Answer Marks
Guidance
\(\sin\theta(4\cos\theta + 6\sin\theta - 3) = 0\) M1
\(\theta = 0\) B1
\(\sin(\theta + 0.588) = \frac{3}{\sqrt{52}} = 0.4160\ldots\ (24.6°)\) M1
\(\theta + 0.588 = (0.4291),\ 2.7125\) [or \(\theta + 33.7° = (24.6°),\ 155.4°\)] dM1
\(\theta = 2.12\) cao A1
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## Question 5:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\cos 2A = \cos^2 A - \sin^2 A$ (+ use of $\cos^2 A + \sin^2 A \equiv 1$) | M1 | |
| $= (1-\sin^2 A) - \sin^2 A = 1 - 2\sin^2 A$ * | A1 | |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $2\sin 2\theta - 3\cos 2\theta - 3\sin\theta + 3 \equiv 4\sin\theta\cos\theta; -3(1-2\sin^2\theta) - 3\sin\theta + 3$ | B1; M1 | |
| $\equiv 4\sin\theta\cos\theta + 6\sin^2\theta - 3\sin\theta$ | M1 | |
| $\equiv \sin\theta(4\cos\theta + 6\sin\theta - 3)$ * | A1 | |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $4\cos\theta + 6\sin\theta \equiv R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$; complete method for $R$ | M1 | May be implied by correct answer |
| $[R^2 = 4^2 + 6^2,\ R\sin\alpha = 4,\ R\cos\alpha = 6]$; $R = \sqrt{52}$ or $7.21$ | A1 | |
| Complete method for $\alpha$; $\alpha = 0.588$ (allow $33.7°$) | M1 A1 | |
### Part (d):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\sin\theta(4\cos\theta + 6\sin\theta - 3) = 0$ | M1 | |
| $\theta = 0$ | B1 | |
| $\sin(\theta + 0.588) = \frac{3}{\sqrt{52}} = 0.4160\ldots\ (24.6°)$ | M1 | |
| $\theta + 0.588 = (0.4291),\ 2.7125$ [or $\theta + 33.7° = (24.6°),\ 155.4°$] | dM1 | |
| $\theta = 2.12$ cao | A1 | |
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5. (a) Using the identity $\cos ( A + B ) \equiv \cos A \cos B - \sin A \sin B$, prove that
$$\cos 2 A \equiv 1 - 2 \sin ^ { 2 } A$$
(b) Show that
$$2 \sin 2 \theta - 3 \cos 2 \theta - 3 \sin \theta + 3 \equiv \sin \theta ( 4 \cos \theta + 6 \sin \theta - 3 )$$
(c) Express $4 \cos \theta + 6 \sin \theta$ in the form $R \sin ( \theta + \alpha )$, where $R > 0$ and $0 < \alpha < \frac { 1 } { 2 } \pi$.\\
(d) Hence, for $0 \leqslant \theta < \pi$, solve
$$2 \sin 2 \theta = 3 ( \cos 2 \theta + \sin \theta - 1 )$$
giving your answers in radians to 3 significant figures, where appropriate.\\
\hfill \mbox{\textit{Edexcel C3 2005 Q5 [15]}}