Trigonometry (11th Edition) Clone

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

Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 236: 11

Answer

$$\sin2\theta=-\frac{4\sqrt{55}}{49}$$ $$\cos2\theta=\frac{39}{49}$$

Work Step by Step

$$\sin\theta=-\frac{\sqrt5}{7} \hspace{2cm}\cos\theta\gt0$$ $$\sin 2\theta=?\hspace{2cm}\cos2\theta=?$$ 1) First, we need to figure out the unknown $\cos\theta$ as they are essential to the calculations of $\sin 2\theta$ and $\cos 2\theta$. According to Pythagorean Identities: $$\cos^2\theta=1-\sin^2\theta=1-\Big(-\frac{\sqrt5}{7}\Big)^2=1-\frac{5}{49}=\frac{44}{49}$$ $$\cos\theta=\frac{\sqrt{44}}{7}=\frac{2\sqrt{11}}{7}\hspace{1cm}(\cos\theta\gt0)$$ 2) Now we can calculate $\sin2\theta$ and $\cos2\theta$ using Double-Angle Identities, which states $$\sin2\theta=2\sin\theta\cos\theta$$ $$\cos2\theta=1-2\sin^2\theta$$ Thus, $$\sin2\theta=2\times\Big(-\frac{\sqrt5}{7}\Big)\times\Big(\frac{2\sqrt{11}}{7}\Big)=-\frac{4\sqrt{55}}{49}$$ $$\cos 2\theta=1-2\Big(-\frac{\sqrt{5}}{7}\Big)^2=1-2\times\frac{5}{49}=1-\frac{10}{49}=\frac{39}{49}$$
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