## Trigonometry (11th Edition) Clone

(a) $$\sin(s+t)=-\frac{63}{65}$$ (b) $$\tan(s+t)=-\frac{63}{16}$$ (c) $(s+t)$ lies in quadrant IV.
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 $\tan t$. $$\sin s=\frac{3}{5}\hspace{1.5cm}\sin t=-\frac{12}{13}\hspace{1.5cm}\text{s in quadrant I and t in quadrant III}$$ 1) Find $\sin s$, $\sin t$, $\cos s$, $\cos t$, $\tan s$ and $\tan t$. $s$ is in quadrant I, so all $\sin s$, $\cos s$ and $\tan s$ are positive. $t$ is in quadrant III, so $\sin t\lt0$, $\cos t\lt0$ and $\tan t=\frac{\sin t}{\cos t}\gt0$ $\cos^2 s=1-\sin^2 s=1-\Big(\frac{3}{5}\Big)^2=1-\frac{9}{25}=\frac{16}{25}$ $\cos s=\frac{4}{5}$ $\cos^2 t=1-\sin^2t=1-\Big(-\frac{12}{13}\Big)^2=1-\frac{144}{169}=\frac{25}{169}$ $\cos t=-\frac{5}{13}$ $$\tan s=\frac{\sin s}{\cos s}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$$ $$\tan t=\frac{\sin t}{\cos t}=\frac{-\frac{12}{13}}{-\frac{5}{13}}=\frac{12}{5}$$ (a) Find $\sin(s+t)$ $$\sin(s+t)=\sin s\cos t+\cos s\sin t\hspace{1.5cm}\text{(sine sum identity)}$$ $$\sin(s+t)=\frac{3}{5}\times\Big(-\frac{5}{13}\Big)+\frac{4}{5}\times\Big(-\frac{12}{13}\Big)$$ $$\sin(s+t)=-\frac{15}{65}-\frac{48}{65}$$ $$\sin(s+t)=-\frac{63}{65}$$ (b) Find $\tan(s+t)$ $$\tan(s+t)=\frac{\tan s+\tan t}{1-\tan s\tan t}\hspace{1.5cm}\text{(tangent sum identity)}$$ $$\tan(s+t)=\frac{\frac{3}{4}+\frac{12}{5}}{1-\frac{3}{4}\times\frac{12}{5}}$$ $$\tan(s+t)=\frac{\frac{63}{20}}{1-\frac{9}{5}}$$ $$\tan(s+t)=\frac{\frac{63}{20}}{-\frac{4}{5}}=-\frac{63\times5}{20\times4}$$ $$\tan(s+t)=-\frac{63}{16}$$ (c) We see that $\sin(s+t)\lt0$ and $\tan (s+t)\lt0$. $\tan (s+t)=\frac{\sin(s+t)}{\cos(s+t)}$, this means $\cos(s+t)\gt0$ $\sin(s+t)\lt0$ but $\cos(s+t)\gt0$ show that $(s+t)$ lies in quadrant IV.