STEP II 2007 question 4 solution

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STEP II 2007 question 4 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 2007 question 4 solution

$\alpha \sin(A-B) + \beta\cos(A+B) = \gamma\sin(A+B)$

$\Rightarrow \alpha(\sin A\cos B - \sin B\cos A) + \beta(\cos A\cos B - \sin A\sin B) = \gamma(\sin A\cos B + \sin B\cos A)$

$\Rightarrow \alpha(\tan A - \tan B) + \beta(1 - \tan A\tan B) = \gamma(\tan A + \tan B)$

$\Rightarrow \tan A\tan B + \frac{\gamma -\alpha}{\beta} \tan A + \frac{\gamma +\alpha}{\beta} \tan B - 1 = 0$

$\Rightarrow (\tan A-m)(\tan B-n) = 0$

where

$\displaystyle m = -\left( \frac{\gamma +\alpha}{\beta} \right) , \quad n = -\left( \frac{\gamma -\alpha}{\beta} \right)$

and

$mn = -1 \Leftrightarrow \left( \frac{\gamma +\alpha}{\beta} \right)\left( \frac{\gamma -\alpha}{\beta} \right) = -1 \Leftrightarrow \alpha^2 = \beta^2 + \gamma^2 .$

(i)

$\alpha=2,\quad \beta=\sqrt{3} ,\quad \gamma=1 \Rightarrow m = -\sqrt{3},\quad n=1/\sqrt{3}$

$A = x, B = \frac{1}{4}\pi$

$\therefore \text{Given equation} \Leftrightarrow (\tan x+\sqrt{3})(\tan \frac{1}{4}\pi-\frac{1}{\sqrt{3}} ) = 0$

$\Rightarrow \tan x = -\sqrt{3} \Rightarrow x = \frac{2\pi}{3} ,\;\; \frac{5\pi}{3} .$

(ii)

$m = -\sqrt{3} ,\quad n = 1/\sqrt{3} ,\quad A = x ,\quad B = \frac{1}{6}\pi$

$\therefore \text{Given equation} \Leftrightarrow (\tan x+\sqrt{3})(\tan \frac{1}{6}\pi-\frac{1}{\sqrt{3}} ) = 0$

But $\tan \frac{1}{6}\pi=\frac{1}{\sqrt{3}}$ and so the equation is an identity.

(iii)

$m = -\sqrt{3} ,\quad n = 1/\sqrt{3} ,\quad A = 2x + \frac{1}{6}\pi ,\quad B = x - \frac{1}{6}\pi$

$\therefore \text{Given equation} \Leftrightarrow (\tan A+\sqrt{3})(\tan B-\frac{1}{\sqrt{3}} ) = 0$

$\Rightarrow A = \frac{2\pi}{3} ,\;\; \frac{5\pi}{3} ,\;\; \frac{8\pi}{3} ,\;\; \frac{11\pi}{3} ,\quad B = \frac{\pi}{6} ,\;\; \frac{7\pi}{6}$

$\Rightarrow x = \frac{\pi}{4} ,\;\; \frac{\pi}{3} ,\;\; \frac{3\pi}{4} ,\;\; \frac{5\pi}{4} ,\;\; \frac{4\pi}{3} ,\;\; \frac{7\pi}{4} .$

Solution by generalebriety.

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