Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 230: 15

Answer

$$\sin\theta=\frac{\sqrt{102}}{12}$$ $$\cos\theta=-\frac{\sqrt{42}}{12}$$

Work Step by Step

$$\cos2\theta=-\frac{5}{12} \hspace{1.5cm}90^\circ\lt\theta\lt180^\circ$$ $$\sin\theta=?\hspace{2cm}\cos\theta=?$$ 1) First, we decode the signs of the $\sin\theta$ and $\cos\theta$ From the information given, we know that $90^\circ\lt\theta\lt180^\circ$. In the trigonometric circle, this is the realm of quadrant II. So $\theta$ must terminate in quadrant II. As $\theta$ terminates in quadrant II, it can, therefore, be stated that $\sin\theta\gt0$ and $\cos\theta\lt0$. 2) Using only the Double-Angle Identities for $\cos2\theta$, we can find out the value of $\sin\theta$ and $\cos\theta$ as follows. $\cos2\theta$ can be written as $$\cos2\theta=2\cos^2\theta-1$$ Thus, $$\cos^2\theta=\frac{\cos2\theta+1}{2}=\frac{-\frac{5}{12}+1}{2}=\frac{\frac{7}{12}}{2}=\frac{7}{24}$$ $$\cos\theta=-\frac{\sqrt7}{\sqrt{24}}=-\frac{\sqrt7}{2\sqrt6}=-\frac{\sqrt{42}}{12}\hspace{1.5cm}\cos\theta\lt0$$ Now, $\cos2\theta$ can also be written this way $$\cos2\theta=1-2\sin^2\theta$$ Thus, $$\sin^2\theta=\frac{1-\cos2\theta}{2}=\frac{1-\Big(-\frac{5}{12}\Big)}{2}=\frac{1+\frac{5}{12}}{2}=\frac{\frac{17}{12}}{2}=\frac{17}{24}$$ $$\sin\theta=\frac{\sqrt{17}}{\sqrt{24}}=\frac{\sqrt{17}}{2\sqrt{6}}=\frac{\sqrt{102}}{12}\hspace{1.5cm}\sin\theta\gt0$$
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