8. (i) Given that \(\cos ( x + 30 ) ^ { \circ } = 3 \cos ( x - 30 ) ^ { \circ }\), prove that tan \(x ^ { \circ } = - \frac { \sqrt { 3 } } { 2 }\).
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Question 8:
Part (i)
Answer Marks
Guidance
\(\cos(x+30)° = 3\cos(x-30)°\) \(\cos x\cos 30 - \sin x\sin 30 = 3(\cos x\cos 30 + \sin x\sin 30)\) M1
Expanding both sides
\(\frac{\sqrt{3}}{2}\cos x - \frac{1}{2}\sin x = \frac{3\sqrt{3}}{2}\cos x + \frac{3}{2}\sin x\) A1
Correct expansion
\(-2\sin x = \sqrt{3}\cos x\) (rearranging) M1
Collecting terms
\(\tan x = -\frac{\sqrt{3}}{2}\) M1 A1
Correct result shown
Part (ii)(a)
Answer Marks
Guidance
\(\frac{1-\cos 2\theta}{\sin 2\theta} = \frac{1-(1-2\sin^2\theta)}{2\sin\theta\cos\theta}\) M1
Using double angle formulae
\(= \frac{2\sin^2\theta}{2\sin\theta\cos\theta}\) A1
Simplifying numerator
\(= \frac{\sin\theta}{\cos\theta} = \tan\theta\) A1
Completion
Part (ii)(b)
Answer Marks
Guidance
\(\theta=180°\): LHS \(= \sin 360° = 0\); RHS \(= 2 - 2\cos 360° = 2-2=0\) ✓ B1
Both sides evaluated correctly
Part (ii)(c)
Answer Marks
Guidance
\(\sin 2\theta = 2 - 2\cos 2\theta \Rightarrow \frac{1-\cos 2\theta}{\sin 2\theta} = \frac{1}{\tan\theta} \cdot \frac{1}{...}\) or using part (a): \(\tan\theta = \frac{1}{2}\) M1
Using result from (a)
Wait — from \(\sin 2\theta = 2-2\cos 2\theta\), divide by \(\sin 2\theta\): \(1 = \frac{2(1-\cos 2\theta)}{\sin 2\theta} = 2\tan\theta\) M1 A1
Correct manipulation
\(\tan\theta = \frac{1}{2}\), \(\theta = \arctan(\frac{1}{2}) \approx 26.6°\) and \(\theta = 206.6°\) A1
Both correct solutions
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# Question 8:
## Part (i)
| $\cos(x+30)° = 3\cos(x-30)°$ | | |
| $\cos x\cos 30 - \sin x\sin 30 = 3(\cos x\cos 30 + \sin x\sin 30)$ | M1 | Expanding both sides |
| $\frac{\sqrt{3}}{2}\cos x - \frac{1}{2}\sin x = \frac{3\sqrt{3}}{2}\cos x + \frac{3}{2}\sin x$ | A1 | Correct expansion |
| $-2\sin x = \sqrt{3}\cos x$ (rearranging) | M1 | Collecting terms |
| $\tan x = -\frac{\sqrt{3}}{2}$ | M1 A1 | Correct result shown |
## Part (ii)(a)
| $\frac{1-\cos 2\theta}{\sin 2\theta} = \frac{1-(1-2\sin^2\theta)}{2\sin\theta\cos\theta}$ | M1 | Using double angle formulae |
| $= \frac{2\sin^2\theta}{2\sin\theta\cos\theta}$ | A1 | Simplifying numerator |
| $= \frac{\sin\theta}{\cos\theta} = \tan\theta$ | A1 | Completion |
## Part (ii)(b)
| $\theta=180°$: LHS $= \sin 360° = 0$; RHS $= 2 - 2\cos 360° = 2-2=0$ ✓ | B1 | Both sides evaluated correctly |
## Part (ii)(c)
| $\sin 2\theta = 2 - 2\cos 2\theta \Rightarrow \frac{1-\cos 2\theta}{\sin 2\theta} = \frac{1}{\tan\theta} \cdot \frac{1}{...}$ or using part (a): $\tan\theta = \frac{1}{2}$ | M1 | Using result from (a) |
| Wait — from $\sin 2\theta = 2-2\cos 2\theta$, divide by $\sin 2\theta$: $1 = \frac{2(1-\cos 2\theta)}{\sin 2\theta} = 2\tan\theta$ | M1 A1 | Correct manipulation |
| $\tan\theta = \frac{1}{2}$, $\theta = \arctan(\frac{1}{2}) \approx 26.6°$ and $\theta = 206.6°$ | A1 | Both correct solutions |
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8. (i) Given that $\cos ( x + 30 ) ^ { \circ } = 3 \cos ( x - 30 ) ^ { \circ }$, prove that tan $x ^ { \circ } = - \frac { \sqrt { 3 } } { 2 }$.\\
(ii) (a) Prove that $\frac { 1 - \cos 2 \theta } { \sin 2 \theta } \equiv \tan \theta$.\\
(b) Verify that $\theta = 180 ^ { \circ }$ is a solution of the equation $\sin 2 \theta = 2 - 2 \cos 2 \theta$.\\
(c) Using the result in part (a), or otherwise, find the other two solutions, $0 < \theta < 360 ^ { \circ }$, of the equation using $\sin 2 \theta = 2 - 2 \cos 2 \theta$.\\
\hfill \mbox{\textit{Edexcel C4 Q8 [13]}}