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 $tan \ 3\theta = 1$ - CIE - A-Level Further Maths - Question 11 - 2011 - Paper 1

To find the values of $\theta$, we set:

$3\theta = \frac{\pi}{4} + n\pi, \ n \in \mathbb{Z}$

Thus,

$\theta = \frac{\pi}{12} + \frac{n\pi}{3}$

Considering values between 0 and $\pi$:

• For $n=0$: $\theta = \frac{\pi}{12}$
• For $n=1$: $\theta = \frac{5\pi}{12}$
• For $n=2$: $\theta = \frac{\pi}{12} + \frac{2\pi}{3} = \frac{9\pi}{12} = \frac{3\pi}{4}$

Final exact values: $\theta = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{3\pi}{4}$.

To solve the cubic equation:

$\theta^3 - 3\theta^2 - 3\theta + 1 = 0$

We can factor or use the Rational Root Theorem. Finding roots, we can express them:

1. The root $\theta = \frac{\pi}{12}$ gives $tan(k\pi) = \tan(\frac{\pi}{12})$;
2. The root $\theta = \frac{5\pi}{12}$ gives $tan(k\pi) = \tan(\frac{5\pi}{12})$;
3. The root $\theta = \frac{3\pi}{4}$ gives $tan(k\pi) = \tan(\frac{3\pi}{4})$;

Hence, we can deduce: $k = \frac{1}{12}, \frac{5}{12}, \frac{3}{4}$.

Calculating:

1. For $k = \frac{1}{12}$: $tan(\frac{1}{12} \pi) = \tan(\frac{\pi}{12}) = 2 - \sqrt{3}$

2. For $k = \frac{5}{12}$: $tan(\frac{5}{12} \pi) = '\tan(\frac{5\pi}{12}) = 2 + \sqrt{3}$

3. For $k = \frac{3}{4}$: $tan(\frac{3}{4} \pi) = \tan(\frac{3\pi}{4}) = -1$

Thus, the exact values are: $tan(\frac{\pi}{12}) = 2 - \sqrt{3}, \ tan(\frac{5\pi}{12}) = 2 + \sqrt{3}, \ tan(\frac{3\pi}{4}) = -1$.