Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.4 - Rotation of Axes - Exercise Set - Page 1010: 12


$x= \frac{1}{2}(\sqrt 3x'-y')$ and $y= \frac{1}{2}(x'+\sqrt 3y')$

Work Step by Step

Step 1. From the given equation, we have $A=7, B=-6\sqrt 3, C=13$. Thus we have $cot2\theta=\frac{A-C}{B}=\frac{7-13}{-6\sqrt 3}=\frac{\sqrt 3}{3}$, which gives $2\theta=\frac{\pi}{3}$ and the rotation angle $\theta=\frac{\pi}{6}=30^\circ$ Step 2. Using the transformation formula $x=x'cos\theta-y'sin\theta$ and $y=x'sin\theta+y'cos\theta$, we have $x=x'cos30^\circ-y'sin30^\circ=\frac{1}{2}(\sqrt 3x'-y')$ $y=x'sin30^\circ+y'cos30^\circ=\frac{1}{2}(x'+\sqrt 3y')$
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