Join TSR
 
 About Us | FAQs | Sign in
 
HomeForumsArticlesRevision NotesPersonal StatementsUniversity GuidesHealth & RelationshipsPostgraduate StudyCareers
Join the UK's largest student community  
Advanced
Search

Join The Student Room Today

Be part of the UK's largest and fastest growing student community.

It's free to join and a lot of fun - Get inspired, express your ideas, interact and share

Join The Student Room Today

STEP II 2007 question 4 solution

From The Student Room

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.

Retrieved from "http://thestudentroom.co.uk/wiki/STEP_II_2007_question_4_solution"
Article Tools:  Create New Article  |  Report This Article  |  This Article's History  |