Answer only one of the following two alternatives. EITHER Use de Moivre’s theorem to prove that $$\tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}$$ State the exact values of $\theta$, between 0 and $\pi$, that satisfy $\tan 3\theta = 1$. Express each root of the equation $\theta^3 - 3\theta^2 - 3\theta + 1 = 0$ in the form $\tan(k\pi)$, where $k$ is a positive rational number. For each of these values of $k$, find the exact value of $\tan(k\pi)$. OR The curve C has equation $$y = \frac{x^2 + \lambda x - 6x^2}{x + 3}$$ where $\lambda$ is a constant such that $\lambda -\frac{2}{3}$. (i) Find $\frac{dy}{dx}$ and deduce that C has two stationary points then. (ii) Find the equations of the asymptotes of C. (iii) Draw a sketch of C for the case $0 3$.

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Answer only one of the following two alternatives - CIE - A-Level Further Maths - Question 11 - 2011 - Paper 1

Question 11

Answer only one of the following two alternatives. EITHER Use de Moivre’s theorem to prove that $$\tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \t... show full transcript

Worked Solution & Example Answer:Answer only one of the following two alternatives - CIE - A-Level Further Maths - Question 11 - 2011 - Paper 1

Step 1

Use de Moivre’s theorem to prove that tan 3θ = (3tan θ - tan³ θ) / (1 - 3tan² θ)

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Answer

To prove the identity using de Moivre’s theorem, recall that:

$\tan n\theta = \frac{\sin(n\theta)}{\cos(n\theta)}$

Using de Moivre's theorem, we can express $\sin$ and $\cos$ terms: $\cos 3\theta + i \sin 3\theta = (\cos \theta + i \sin \theta)^3$

By expanding using the binomial theorem, $= \cos^3 \theta + 3\cos^2 \theta \cdot i\sin \theta - 3\cos \theta \cdot \sin^2 \theta - i\sin^3 \theta$

Equating real and imaginary parts, we have: $\tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}$

Step 2

State the exact values of θ, between 0 and π, that satisfy tan 3θ = 1.

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Answer

We set: $\tan 3\theta = 1$ This occurs when: $3\theta = \frac{\pi}{4} + n\pi, \text{ for } n \in \mathbb{Z}$ Therefore: $\theta = \frac{\pi}{12} + \frac{n\pi}{3}$ The acceptable values in the interval ( $0, \pi$ ) are: $\theta = \frac{\pi}{12}, \frac{5\pi}{12}, \text{ and } \frac{\pi}{4}.$

Step 3

Express each root of the equation θ³ - 3θ² - 3θ + 1 = 0 in the form tan(kπ), where k is a positive rational number.

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Answer

We factor the cubic equation: $\theta^3 - 3\theta^2 - 3\theta + 1 = 0$ Through synthetic division or numerical methods, the roots can be approximately found as:

One root is $\tan(k\pi) = \frac{\pi}{12}$
Another root suggests $\tan(k\pi) = \frac{5\pi}{12}$
The final root simplifies to $\tan(k\pi) = \frac{\pi}{4}$ , which can be expressed as rational multiples of $\pi$ .

Step 4

For each of these values of k, find the exact value of tan(kπ).

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Answer

The exact values of $\tan(k\pi)$ can be computed as follows:

For $k = \frac{1}{12}$ , we find: $\tan\left(\frac{\pi}{12}\right) = 2 - \sqrt{3}$
For $k = \frac{5}{12}$ : $\tan\left(\frac{5\pi}{12}\right) = 2 + \sqrt{3}$
For $k = \frac{1}{4}$ : $\tan\left(\frac{\pi}{4}\right) = 1$

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Questions answered

Answer only one of the following two alternatives - CIE - A-Level Further Maths - Question 11 - 2011 - Paper 1

Worked Solution & Example Answer:Answer only one of the following two alternatives - CIE - A-Level Further Maths - Question 11 - 2011 - Paper 1

Use de Moivre’s theorem to prove that tan 3θ = (3tan θ - tan³ θ) / (1 - 3tan² θ)

State the exact values of θ, between 0 and π, that satisfy tan 3θ = 1.

Express each root of the equation θ³ - 3θ² - 3θ + 1 = 0 in the form tan(kπ), where k is a positive rational number.

For each of these values of k, find the exact value of tan(kπ).

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