Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Test - Page 250: 9a


$cos(2\theta) = -\frac{7}{25}$

Work Step by Step

If $90^{\circ} \lt \theta \lt 180^{\circ}$, then the angle $\theta$ is in quadrant II. If the hypotenuse is 5, and the adjacent side has a length of 3, then the length of the opposite side is $\sqrt{5^2-3^2} = 4$ $cos(2\theta) = cos^2~\theta -sin^2~\theta$ $cos(2\theta) = (-\frac{3}{5})^2 -(\frac{4}{5})^2$ $cos(2\theta) = \frac{9}{25} -\frac{16}{25}$ $cos(2\theta) = -\frac{7}{25}$
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