STEP III 1990 question 1 solution

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STEP III 1990 question 1 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 1990 question 1 solution

By expanding $(\cos\theta + i\sin\theta)^{9}$ and dividing the imaginary part by the real part you get:

$\displaystyle \tan9\theta = \frac{s^{9} - 36s^{7}c^{2} + 126s^{5}c^{4} - 84s^{3}c^{6} + 9sc^{8}}{c^{9} - 36c^{7}s^{2} + 126c^{5}s^{4} - 84c^{3}s^{6} + 9cs^{8}}$

Where s = $\sin\theta$ , c = $\cos\theta$ .

Dividing by $\cos^{9}\theta$ , you get:

$\displaystyle \tan9\theta = \frac{t^{9} - 36t^{7} + 126t^{5} - 84t^{3} + 9t}{1 - 36t^{2} + 126t^{4} - 84t^{6} + 9t^{8}}$

Where $t = \tan\theta$ .

By expanding the brackets, we see this is equivalent to $\displaystyle \frac{t(t^{2} - 2)(t^{6} - 33t^{4} + 27t^{2} - 3)}{(3t^{2} - 1)(3t^{6} - 27t^{4} + 33t^{2} - 1)}$ .

The roots of this equation are given by the roots of the numerator. The roots are $\tan\frac{k\pi}{9}, \; k = 0, \; 1, \; 2, \; \cdots, \; 8$ .

Since $\tan\frac{k\pi}{9} = - \tan\frac{(9 - k)\pi}{9}$ and using the difference of two squares along with the known values for $\tan0, \; \tan\frac{\pi}{3}, \; \tan\frac{2\pi}{3}$ , we get the roots as $t(t^{2} - 3)(t^{2} - \tan^{2}\frac{\pi}{9})(t^{2} - \tan^{2}\frac{2\pi}{9})(t^{2} - \tan^{2}\frac{4\pi}{9})$ .

This is because, for example $(t - \tan\frac{\pi}{9})(t - \tan\frac{8\pi}{9}) = (t - \tan\frac{\pi}{9})(t + \tan\frac{\pi}{9}) = (t^{2} - \tan^{2}\frac{\pi}{9})$ .

So $t^{6} - 33t^{4} + 27t^{2} - 3$ can be factorised into $(t^{2} - \tan^{2}\frac{\pi}{9})(t^{2} - \tan^{2}\frac{2\pi}{9})(t^{2} - \tan^{2}\frac{4\pi}{9})$ . Examining the constant on both we get $-(\tan\frac{\pi}{9}\tan\frac{2\pi}{9}\tan\frac{4\pi}{9})^{2} = - 3 \Rightarrow \tan\frac{\pi}{9}\tan\frac{2\pi}{9}\tan\frac{4\pi}{9} = \sqrt{3}$ .

Setting $\alpha = \tan^{2}\frac{\pi}{9}, \; \beta = \tan^{2}\frac{2\pi}{9}, \; \gamma = \tan^{2}\frac{4\pi}{9}$ , we also have:

$\alpha + \beta + \gamma = 33, \; \alpha\beta + \beta\gamma + \gamma\alpha = 27$

$\alpha^{3} + \beta^{3} + \gamma^{3} = (\alpha + \beta + \gamma)^{3} - 3(\alpha + \beta + \gamma)(\alpha\beta + \beta\gamma + \gamma\alpha) + 3\alpha\beta\gamma$