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: 15


$x= \frac{\sqrt 5}{5}(2x'-y')$ and $y= \frac{\sqrt 5}{5}(x'+2y')$

Work Step by Step

Step 1. From the given equation, we have $A=1, B=4, C=-2$. Thus we have $cot2\theta=\frac{A-C}{B}=\frac{1+2}{4}=\frac{3}{4}$. Thus $2\theta$ is in Quadrant I and $cos2\theta=\frac{3}{\sqrt {3^2+4^2}}=\frac{3}{5}$ (form a right triangle with sides $3,4,5$ here). Step 2. We can find values of $sin\theta$ and $\cos\theta$ as $sin\theta=\sqrt {\frac{1-cos2\theta}{2}}=\sqrt {\frac{1-(\frac{3}{5})}{2}}=\frac{\sqrt 5}{5}$ and $cos\theta=\sqrt {1-sin^2\theta}=\frac{2\sqrt 5}{5}$ Step 3. Using the transformation formula $x=x'cos\theta-y'sin\theta$ and $y=x'sin\theta+y'cos\theta$, we have $x=x'(\frac{2\sqrt 5}{5})-y'(\frac{\sqrt 5}{5})=\frac{\sqrt 5}{5}(2x'-y')$ $y=x'(\frac{\sqrt 5}{5})+y'(\frac{2\sqrt 5}{5})=\frac{\sqrt 5}{5}(x'+2y')$
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