## Trigonometry (11th Edition) Clone

(a) $$\sin(s+t)=-\frac{36}{85}$$ (b) $$\tan(s+t)=\frac{36}{77}$$ (c) $(s+t)$ lies in quadrant III.
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 $\tan t$. $$\cos s=-\frac{15}{17}\hspace{1.5cm}\sin t=\frac{4}{5}\hspace{1.5cm}\text{s in quadrant II and t in quadrant I}$$ 1) Find $\sin s$, $\sin t$, $\cos s$, $\cos t$, $\tan s$ and $\tan t$. $s$ is in quadrant II, so $\sin s\gt0$ but $\cos s\lt0$. That means $\tan s=\frac{\sin s}{\cos s}\lt0$ $t$ is in quadrant I, so all trigonometric values are positive. $\sin^2 s=1-\cos^2 s=1-\Big(-\frac{15}{17}\Big)^2=1-\frac{225}{289}=\frac{64}{289}$ $\sin s=\frac{8}{17}$ $\cos^2 t=1-\sin^2t=1-\Big(\frac{4}{5}\Big)^2=1-\frac{16}{25}=\frac{9}{25}$ $\cos t=\frac{3}{5}$ $$\tan s=\frac{\sin s}{\cos s}=\frac{\frac{8}{17}}{-\frac{15}{17}}=-\frac{8}{15}$$ $$\tan t=\frac{\sin t}{\cos t}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}$$ (a) Find $\sin(s+t)$ $$\sin(s+t)=\sin s\cos t+\cos s\sin t\hspace{1.5cm}\text{(sine sum identity)}$$ $$\sin(s+t)=\frac{8}{17}\times\frac{3}{5}+\Big(-\frac{15}{17}\Big)\times\frac{4}{5}$$ $$\sin(s+t)=\frac{24}{85}-\frac{60}{85}$$ $$\sin(s+t)=-\frac{36}{85}$$ (b) Find $\tan(s+t)$ $$\tan(s+t)=\frac{\tan s+\tan t}{1-\tan s\tan t}\hspace{1.5cm}\text{(tangent sum identity)}$$ $$\tan(s+t)=\frac{-\frac{8}{15}+\frac{4}{3}}{1-\Big(-\frac{8}{15}\Big)\times\frac{4}{3}}$$ $$\tan(s+t)=\frac{\frac{12}{15}}{1+\frac{32}{45}}$$ $$\tan(s+t)=\frac{\frac{4}{5}}{\frac{77}{45}}=\frac{4\times45}{5\times77}$$ $$\tan(s+t)=\frac{36}{77}$$ (c) We see that while $\sin(s+t)\lt0$, $\tan (s+t)\gt0$. $\tan (s+t)=\frac{\sin(s+t)}{\cos(s+t)}$, this means $\cos(s+t)\lt0$ Both $\sin(s+t)\lt0$ and $\cos(s+t)\lt0$ show that $(s+t)$ lies in quadrant III.