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 5.4 Sum and Difference Identities for Sine and Tangent - 5.4 Exercises - Page 227: 56


(a) $$\sin(s+t)=-\frac{8\sqrt6+3}{25}$$ (b) $$\tan(s+t)=-\frac{8\sqrt6+3}{4-6\sqrt6}$$ (c) $(s+t)$ is in quadrant III.

Work Step by Step

To find $\sin(s+t)$ and $\tan(s+t)$, all values of $\sin s$, $\sin t$, $\cos s$, $\cos t$, $\tan s$ and $\tan t$ must be known. Therefore, the privileged job is to find all 6 essential values of $\sin s$, $\sin t$, $\cos s$, $\cos t$, $\tan s$ and $\tan t$. $$\cos s=-\frac{1}{5}\hspace{1.5cm}\sin t=\frac{3}{5}\hspace{1.5cm}\text{s and t are in quadrant II}$$ 1) Find $\sin s$, $\sin t$, $\cos s$, $\cos t$, $\tan s$ and $\tan t$. $s$ and $t$ are in quadrant II, so $\sin s\gt0$, $\sin t\gt0$ but $\cos s\lt0$ and $\cos t\lt0$. That means $\tan s=\frac{\sin s}{\cos s}\lt0$ and $\tan t=\frac{\sin t}{\cos t}\lt0$ $\sin^2 s=1-\cos^2 s=1-\Big(-\frac{1}{5}\Big)^2=1-\frac{1}{25}=\frac{24}{25}$ $\sin s=\frac{\sqrt{24}}{5}=\frac{2\sqrt6}{5}$ $\cos^2 t=1-\sin^2t=1-\Big(\frac{3}{5}\Big)^2=1-\frac{9}{25}=\frac{16}{25}$ $\cos t=-\frac{4}{5}$ $$\tan s=\frac{\sin s}{\cos s}=\frac{\frac{2\sqrt6}{5}}{-\frac{1}{5}}=-\frac{2\sqrt6}{1}=-2\sqrt6$$ $$\tan t=\frac{\sin t}{\cos t}=\frac{\frac{3}{5}}{-\frac{4}{5}}=-\frac{3}{4}$$ (a) Find $\sin(s+t)$ $$\sin(s+t)=\sin s\cos t+\cos s\sin t\hspace{1.5cm}\text{(sine sum identity)}$$ $$\sin(s+t)=\frac{2\sqrt6}{5}\times\Big(-\frac{4}{5}\Big)+\Big(-\frac{1}{5}\Big)\times\frac{3}{5}$$ $$\sin(s+t)=-\frac{8\sqrt6}{25}-\frac{3}{25}$$ $$\sin(s+t)=\frac{-8\sqrt6-3}{25}=-\frac{8\sqrt6+3}{25}$$ (b) Find $\tan(s+t)$ $$\tan(s+t)=\frac{\tan s+\tan t}{1-\tan s\tan t}\hspace{1.5cm}\text{(tangent sum identity)}$$ $$\tan(s+t)=\frac{-2\sqrt6-\frac{3}{4}}{1-(-2\sqrt6)\times\Big(-\frac{3}{4}\Big)}$$ $$\tan(s+t)=\frac{\frac{-8\sqrt6-3}{4}}{1-\frac{3\sqrt6}{2}}$$ $$\tan(s+t)=\frac{\frac{-8\sqrt6-3}{4}}{\frac{2-3\sqrt6}{2}}=\frac{(-8\sqrt6-3)\times2}{(2-3\sqrt6)\times4}$$ $$\tan(s+t)=\frac{-8\sqrt6-3}{2(2-3\sqrt6)}=-\frac{8\sqrt6+3}{4-6\sqrt6}$$ (c) We see that $\sin(s+t)\lt0$. $8\sqrt6+3\gt0$, and since $4\lt6\sqrt6$, $4-6\sqrt{6}\lt0$, so $\frac{8\sqrt6+3}{4-6\sqrt6}\lt0$. Therefore, $\tan(s+t)=-\frac{8\sqrt6+3}{4-6\sqrt6}\gt0$ $\tan (s+t)=\frac{\sin(s+t)}{\cos(s+t)}$, this means $\cos(s+t)\lt0$ As $\sin(s+t)\lt0$ and $\cos(s+t)\lt0$, $(s+t)$ would in quadrant III.
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