## Trigonometry (11th Edition) Clone

(a) $$\sin(s+t)=\frac{63}{65}$$ (b) $$\tan(s+t)=\frac{63}{16}$$ (c) $(s+t)$ lies in quadrant I.
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 $\tan t$. $$\cos s=\frac{3}{5}\hspace{1.5cm}\sin t=\frac{5}{13}\hspace{1.5cm}\text{s and t in quadrant I}$$ 1) Find $\sin s$, $\sin t$, $\cos s$, $\cos t$, $\tan s$ and $\tan t$. $s$ and $t$ are in quadrant I, so all trigonometric values would be positive. $\sin^2 s=1-\cos^2 s=1-\Big(\frac{3}{5}\Big)^2=1-\frac{9}{25}=\frac{16}{25}$ $\sin s=\frac{4}{5}$ $\cos^2 t=1-\sin^2t=1-\Big(\frac{5}{13}\Big)^2=1-\frac{25}{169}=\frac{144}{169}$ $\cos t=\frac{12}{13}$ $$\tan s=\frac{\sin s}{\cos s}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}$$ $$\tan t=\frac{\sin t}{\cos t}=\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12}$$ (a) Find $\sin(s+t)$ $$\sin(s+t)=\sin s\cos t+\cos s\sin t\hspace{1.5cm}\text{(sine sum identity)}$$ $$\sin(s+t)=\frac{4}{5}\times\frac{12}{13}+\frac{3}{5}\times\frac{5}{13}$$ $$\sin(s+t)=\frac{48}{65}+\frac{15}{65}$$ $$\sin(s+t)=\frac{63}{65}$$ (b) Find $\tan(s+t)$ $$\tan(s+t)=\frac{\tan s+\tan t}{1-\tan s\tan t}\hspace{1.5cm}\text{(tangent sum identity)}$$ $$\tan(s+t)=\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}\times\frac{5}{12}}$$ $$\tan(s+t)=\frac{\frac{21}{12}}{1-\frac{5}{9}}$$ $$\tan(s+t)=\frac{\frac{7}{4}}{\frac{4}{9}}=\frac{7\times9}{4\times4}$$ $$\tan(s+t)=\frac{63}{16}$$ (c) We see that $\sin(s+t)\gt0$ and $\tan (s+t)\gt0$. Since $\tan (s+t)=\frac{\sin(s+t)}{\cos(s+t)}$, this means $\cos(s+t)\gt0$ Both $\sin(s+t)\gt0$ and $\cos(s+t)\gt0$ show that $(s+t)$ lies in quadrant I.