## Trigonometry (11th Edition) Clone

(a) $$\sin(s+t)=\frac{4\sqrt2+\sqrt5}{9}$$ (b) $$\tan(s+t)=-\frac{8\sqrt5+5\sqrt2}{20-2\sqrt{10}}$$ (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 $\tan t$. $$\sin s=\frac{2}{3}\hspace{1.5cm}\sin t=-\frac{1}{3}\hspace{1.5cm}\text{s in quadrant II and t in quadrant IV}$$ 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 IV, so $\sin t\lt0$ but $\cos t\gt0$. That means $\tan t=\frac{\sin t}{\cos t}\lt0$ $\cos^2 s=1-\sin^2 s=1-\Big(\frac{2}{3}\Big)^2=1-\frac{4}{9}=\frac{5}{9}$ $\cos s=-\frac{\sqrt5}{3}$ $\cos^2 t=1-\sin^2t=1-\Big(-\frac{1}{3}\Big)^2=1-\frac{1}{9}=\frac{8}{9}$ $\cos t=\frac{\sqrt8}{3}=\frac{2\sqrt2}{3}$ $$\tan s=\frac{\sin s}{\cos s}=\frac{\frac{2}{3}}{-\frac{\sqrt5}{3}}=-\frac{2}{\sqrt5}=-\frac{2\sqrt5}{5}$$ $$\tan t=\frac{\sin t}{\cos t}=\frac{-\frac{1}{3}}{\frac{2\sqrt2}{3}}=-\frac{1}{2\sqrt2}=-\frac{\sqrt2}{4}$$ (a) Find $\sin(s+t)$ $$\sin(s+t)=\sin s\cos t+\cos s\sin t\hspace{1.5cm}\text{(sine sum identity)}$$ $$\sin(s+t)=\frac{2}{3}\times\frac{2\sqrt2}{3}+\Big(-\frac{\sqrt5}{3}\Big)\times\Big(-\frac{1}{3}\Big)$$ $$\sin(s+t)=\frac{4\sqrt2}{9}+\frac{\sqrt5}{9}$$ $$\sin(s+t)=\frac{4\sqrt2+\sqrt5}{9}$$ (b) Find $\tan(s+t)$ $$\tan(s+t)=\frac{\tan s+\tan t}{1-\tan s\tan t}\hspace{1.5cm}\text{(tangent sum identity)}$$ $$\tan(s+t)=\frac{-\frac{2\sqrt5}{5}-\frac{\sqrt2}{4}}{1-\Big(-\frac{2\sqrt5}{5}\Big)\times\Big(-\frac{\sqrt2}{4}\Big)}$$ $$\tan(s+t)=\frac{\frac{-8\sqrt5-5\sqrt2}{20}}{1-\frac{\sqrt{10}}{10}}$$ $$\tan(s+t)=\frac{\frac{-8\sqrt5-5\sqrt2}{20}}{\frac{10-\sqrt{10}}{10}}=\frac{(-8\sqrt5-5\sqrt2)\times10}{(10-\sqrt{10})\times20}$$ $$\tan(s+t)=\frac{-8\sqrt5-5\sqrt2}{2(10-\sqrt{10})}=-\frac{8\sqrt5+5\sqrt2}{20-2\sqrt{10}}$$ (c) We see that $\sin(s+t)\gt0$. $8\sqrt5+5\sqrt2\gt0$, and since $20\gt2\sqrt{10}$, $20-2\sqrt{10}\gt0$, so $\tan(s+t)=-\frac{8\sqrt5+5\sqrt2}{20-2\sqrt{10}}\lt0$ $\tan (s+t)=\frac{\sin(s+t)}{\cos(s+t)}$, this means $\cos(s+t)\lt0$ As $\sin(s+t)\gt0$ yet $\cos(s+t)\lt0$ shows that $(s+t)$ lies in quadrant II.