Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 1 - Systems of Linear Equations - 1.1 Introduction to Systems of Linear Equations - 1.1 Exercises: 76

Answer

The solution of this system is unique and it is: $$x=\cos\theta-\sin\theta,\quad y= \cos\theta + \sin\theta.$$

Work Step by Step

Here we can treat sines and cosines as ordinary coefficients and we will solve this system as we would solve any system of two equations with two unknowns. Step 1: Multiply the first equation by $\sin\theta$ and the second equation by $\cos\theta$: \begin{align*} (\sin\theta\cos\theta)x+(\sin^2\theta)y=&\sin\theta\\ (-\sin\theta\cos\theta)x+(\cos^2\theta)y =& \cos\theta. \end{align*} Step 2: Add these two equations together to eliminate $x$ and find $y$: $$(\sin\theta\cos\theta)x +(-\sin\theta\cos\theta)x + (\sin^2\theta)y +(\cos^2\theta)y = \sin\theta+\cos\theta$$ which becomes $$y(\sin^2\theta+\cos^2\theta) = \sin\theta+\cos\theta.$$ Remember that for any $\theta$ we have that $\sin^2\theta+\cos^2\theta = 1$ so this simplifies to $$y=\sin\theta + \cos\theta$$ Step 3: Now let us do the similar thing as in step 1. Multiply the first equation by $\cos\theta$ and the second one by $\sin\theta$: \begin{align*} (\cos^2\theta)x+ (\sin\theta\cos\theta) y = &\cos\theta\\ (-\sin^2\theta)x+(\sin\theta\cos\theta)y = &\sin\theta \end{align*} Step 4: Now subtract these equations to eliminate $y$ and find $x$ : $$(\cos^2\theta)x-(-\sin^2\theta)x + (\sin\theta\cos\theta) y -(\sin\theta\cos\theta)y =\cos\theta- \sin\theta$$ This then gives $$x(\sin^2\theta+\cos^2\theta) = \cos\theta-\sin\theta.$$ Using the fact that $\sin^2\theta +\cos^2\theta = 1$ we again get simpler expression: $$x=\cos\theta-\sin\theta.$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.