Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 5 - Trigonometric Functions - Test - Page 562: 8

Answer

See graph and explanations.

Work Step by Step

1. Plot the point $(2,-7)$ (red) as shown in the figure. 2. Connect the point with the origin, the positive angle formed with the x-axis is $\theta$ 3. We have $sin\theta=\frac{-7}{\sqrt {2^2+7^2}}=-\frac{7}{\sqrt {53}}=-\frac{7\sqrt {53}}{53}$ 4. We have $cos\theta=\frac{2}{\sqrt {2^2+7^2}}=\frac{2}{\sqrt {53}}=\frac{2\sqrt {53}}{53}$ 5. We have $tan\theta=-\frac{7}{2}$ 6. We have $cot\theta=-\frac{2}{7}$ 7. We have $sec\theta=\frac{1}{cos\theta}=\frac{\sqrt {53}}{2}$ 8. We have $csc\theta=\frac{1}{sin\theta}=-\frac{\sqrt {53}}{7}$
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