Trigonometry (11th Edition) Clone

(a) $$\sin(s+t)=\frac{77}{85}$$ (b) $$\tan(s+t)=-\frac{77}{36}$$ (c) $(s+t)$ lies in quadrant II.
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 $\tan t$. $$\cos s=-\frac{8}{17}\hspace{1.5cm}\cos t=-\frac{3}{5}\hspace{1.5cm}\text{s and t in quadrant III}$$ 1) Find $\sin s$, $\sin t$, $\cos s$, $\cos t$, $\tan s$ and $\tan t$. $s$ and $t$ are in quadrant III, so $\sin t\lt0$, $\cos t\lt0$, $\sin s\lt0$ and $\cos s\lt0$; but $\tan t=\frac{\sin t}{\cos t}\gt0$ and $\tan s=\frac{\sin s}{\cos s}\gt0$ $\sin^2 s=1-\cos^2 s=1-\Big(-\frac{8}{17}\Big)^2=1-\frac{64}{289}=\frac{225}{289}$ $\sin s=-\frac{15}{17}$ $\sin^2 t=1-\cos^2t=1-\Big(-\frac{3}{5}\Big)^2=1-\frac{9}{25}=\frac{16}{25}$ $\sin t=-\frac{4}{5}$ $$\tan s=\frac{\sin s}{\cos s}=\frac{-\frac{15}{17}}{-\frac{8}{17}}=\frac{15}{8}$$ $$\tan t=\frac{\sin t}{\cos t}=\frac{-\frac{4}{5}}{-\frac{3}{5}}=\frac{4}{3}$$ (a) Find $\sin(s+t)$ $$\sin(s+t)=\sin s\cos t+\cos s\sin t\hspace{1.5cm}\text{(sine sum identity)}$$ $$\sin(s+t)=\Big(-\frac{15}{17}\Big)\times\Big(-\frac{3}{5}\Big)+\Big(-\frac{8}{17}\Big)\times\Big(-\frac{4}{5}\Big)$$ $$\sin(s+t)=\frac{45}{85}+\frac{32}{85}$$ $$\sin(s+t)=\frac{77}{85}$$ (b) Find $\tan(s+t)$ $$\tan(s+t)=\frac{\tan s+\tan t}{1-\tan s\tan t}\hspace{1.5cm}\text{(tangent sum identity)}$$ $$\tan(s+t)=\frac{\frac{15}{8}+\frac{4}{3}}{1-\frac{15}{8}\times\frac{4}{3}}$$ $$\tan(s+t)=\frac{\frac{77}{24}}{1-\frac{5}{2}}$$ $$\tan(s+t)=\frac{\frac{77}{24}}{-\frac{3}{2}}=-\frac{77\times2}{24\times3}$$ $$\tan(s+t)=-\frac{77}{36}$$ (c) We see that $\sin(s+t)\gt0$ and $\tan (s+t)\lt0$. $\tan (s+t)=\frac{\sin(s+t)}{\cos(s+t)}$, this means $\cos(s+t)\lt0$ $\sin(s+t)\gt0$ but $\cos(s+t)\lt0$ show that $(s+t)$ lies in quadrant II.