Question

use the given information to find (a) \\( \sin (s+t) \\), (b) \\( \tan (s+t) \\), and (c) the quadrant of \\( s+t \\).
\\( \sin s=\frac{1}{3} \\) and \\( \sin t=-\frac{2}{3} \\), s in quadrant ii and t in quadrant iv
(a) \\( \sin (s+t)= \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
(b) \\( \tan (s+t)= \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
(c) what is the quadrant of \\( s+t \\)?

Explanation:

Step1: Find $\cos s$ and $\cos t$

Using the identity $\sin^{2}\theta+\cos^{2}\theta = 1$.
For $s$ (in quadrant II, so $\cos s<0$):
$\cos s=-\sqrt{1-\sin^{2}s}=-\sqrt{1 - (\frac{1}{3})^{2}}=-\sqrt{\frac{9 - 1}{9}}=-\frac{2\sqrt{2}}{3}$
For $t$ (in quadrant IV, so $\cos t>0$):
$\cos t=\sqrt{1-\sin^{2}t}=\sqrt{1-(-\frac{2}{3})^{2}}=\sqrt{\frac{9 - 4}{9}}=\frac{\sqrt{5}}{3}$

Step2: Calculate $\sin(s + t)$

Using the formula $\sin(A + B)=\sin A\cos B+\cos A\sin B$
$\sin(s + t)=\sin s\cos t+\cos s\sin t$
Substitute $\sin s=\frac{1}{3},\cos s=-\frac{2\sqrt{2}}{3},\sin t=-\frac{2}{3},\cos t=\frac{\sqrt{5}}{3}$
$\sin(s + t)=\frac{1}{3}\times\frac{\sqrt{5}}{3}+(-\frac{2\sqrt{2}}{3})\times(-\frac{2}{3})=\frac{\sqrt{5}+4\sqrt{2}}{9}$

Step3: Calculate $\tan s$ and $\tan t$

$\tan\theta=\frac{\sin\theta}{\cos\theta}$
$\tan s=\frac{\sin s}{\cos s}=\frac{\frac{1}{3}}{-\frac{2\sqrt{2}}{3}}=-\frac{1}{2\sqrt{2}}=-\frac{\sqrt{2}}{4}$
$\tan t=\frac{\sin t}{\cos t}=\frac{-\frac{2}{3}}{\frac{\sqrt{5}}{3}}=-\frac{2}{\sqrt{5}}=-\frac{2\sqrt{5}}{5}$

Step4: Calculate $\tan(s + t)$

Using the formula $\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$
$\tan(s + t)=\frac{-\frac{\sqrt{2}}{4}-\frac{2\sqrt{5}}{5}}{1-(-\frac{\sqrt{2}}{4})(-\frac{2\sqrt{5}}{5})}=\frac{-\frac{5\sqrt{2}+8\sqrt{5}}{20}}{1-\frac{\sqrt{10}}{10}}=\frac{-(5\sqrt{2}+8\sqrt{5})}{20 - 2\sqrt{10}}=\frac{-(5\sqrt{2}+8\sqrt{5})(20 + 2\sqrt{10})}{(20 - 2\sqrt{10})(20 + 2\sqrt{10})}$
After simplification: $\tan(s + t)=\frac{-100\sqrt{2}-10\sqrt{20}-160\sqrt{5}-16\sqrt{50}}{400 - 40}=\frac{-100\sqrt{2}-20\sqrt{5}-160\sqrt{5}-80\sqrt{2}}{360}=\frac{-(180\sqrt{2}+180\sqrt{5})}{360}=-\frac{\sqrt{2}+\sqrt{5}}{2}$

Answer:

(a) $\frac{\sqrt{5}+4\sqrt{2}}{9}$
(b) $-\frac{\sqrt{2}+\sqrt{5}}{2}$
(c) Quadrant III