Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 8 - Section 8.1 - Complex Numbers - 8.1 Problem Set - Page 427: 96


$sin\theta = \frac{12}{13}$, $cos\theta =-\frac{5}{13}$

Work Step by Step

For point (-5, 12), it is in the 2nd quadrant, the absolute value (modulus) is $\sqrt{(-5)^2 + 12^2} = 13$, hence, $sin\theta = \frac{12}{13}$ and $cos\theta =-\frac{5}{13}$
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