STEP III 2007 question 1 solution

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STEP III 2007 question 1 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 2007 question 1 solution

Solution

$\displaystyle \tan (\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{\tan (\theta_1 + \theta_2) + \tan (\theta_3 + \theta_4)}{1 - \tan (\theta_1 + \theta_2)\tan (\theta_3 + \theta_4)}$

$\displaystyle = \frac{\frac{t_1+t_2}{1-t_1t_2} + \frac{t_3+t_4}{1-t_3t_4}}{1 - \left(\frac{t_1+t_2}{1-t_1t_2}\right) \left(\frac{t_3+t_4}{1-t_3t_4}\right)}$

$\displaystyle = \frac{(t_1+t_2)(1-t_3t_4) + (t_3+t_4)(1-t_1t_2)}{ (1-t_1t_2)(1-t_3t_4) - (t_1+t_2)(t_3+t_4)} .$

$\displaystyle at^4 + bt^3 + ct^2 + dt + e = a(t-t_1)(t-t_2)(t-t_3)(t-t_4)$

$\displaystyle \Rightarrow t_1+t_2+t_3+t_4 = -b/a,\quad t_1t_2 + t_1t_3 + t_1t_4 + t_2t_3 + t_2t_4 + t_3t_4 = c/a,$ $\displaystyle t_1t_2t_3 + t_1t_2t_4 + t_1t_3t_4 + t_2t_3t_4 = -d/a, \quad t_1t_2t_3t_4 = e/a.$

$\displaystyle \tan (\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{(t_1+t_2+t_3+t_4) - (t_1t_2t_3 + t_1t_2t_4 + t_1t_3t_4 + t_2t_3t_4)}{t_1t_2t_3t_4 + 1 - (t_1t_2 + t_1t_3 + t_1t_4 + t_2t_3 + t_2t_4 + t_3t_4)}$

$\displaystyle = \frac{(-b/a) + (d/a)}{(e/a) + 1 - (c/a)}$

$\displaystyle = \frac{d-b}{a+e-c} . \quad (*)$

$\displaystyle \tan \theta_i = t_i \Rightarrow \sin \theta_i = \frac{t_i}{\sqrt{1 + t_i^2}} ,\quad \cos \theta_i = \frac{1}{\sqrt{1+t_i^2}} .$

Substituting into the given equation:

$\displaystyle p\cos 2\theta + \cos (\theta -\alpha ) + p = 0$

$\displaystyle 2p\cos^2\theta - p + \cos\theta\cos\alpha - \sin\theta\sin\alpha + p = 0$

$\displaystyle \frac{2p}{1+t^2} + \frac{\cos\alpha}{\sqrt{1+t^2}} - \frac{t\sin\alpha}{\sqrt{1 + t^2}} = 0$

$\displaystyle 2p = -(\cos\alpha - t\sin\alpha ) \sqrt{1+t^2}$

$\displaystyle 4p^2 = (\cos^2\alpha - t\sin 2\alpha + t^2\sin^2\alpha )(1+t^2)$

$\displaystyle (\sin^2\alpha )t^4 - (\sin 2\alpha )t^3 + t^2 - (\sin 2\alpha )t + (\cos^2\alpha - 4p^2) = 0$

Substituting into (*):

$\displaystyle \tan (\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{-\sin 2\alpha + \sin 2\alpha}{a+e-c} = 0$

$\displaystyle \tan (\theta_1 + \theta_2 + \theta_3 + \theta_4) = 0 \Rightarrow \theta_1 + \theta_2 + \theta_3 + \theta_4 = n\pi$

as required.

Solution by generalebriety.

Alternative

We are asked to find $\tan(\theta_1+\theta_2+\theta_3+\theta_4)$ in terms of $t_i = \tan(\theta_k)$ . Use De Moivre:

$\cos(\theta_1+\theta_2+\theta_3+\theta_4)+i \sin(\theta_1+\theta_2+\theta_3+\theta_4)$

$\displaystyle =\prod_k (\cos \theta_k + i \sin \theta_k) =\left(\prod \cos \theta_k\right) \left(\prod (1+i t_k)\right)$

Then $\tan(\theta_1+\theta_2+\theta_3+\theta_4)$ is just the ratio of the real and imaginary parts of $\prod (1+i t_k)$ ; the most difficult part at this point is simply writing out the answer - excuse me if I don't bother LaTeXing it.

Comparing $\prod (1+i t_k)$ with $\prod (x-t_k)$ also lets you more or less write down the relation between $\tan(\theta_1+\theta_2+\theta_3+\theta_4)$ and the coefficients of the quartic with roots t_1, t_2, t_3, t_4 as well.

Of course, this approach generalises nicely to arbitrary numbers of $\theta_k$ as well.

Alternative by DFranklin.

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