Question

use the given information to find (a) \\( \sin (s+t) \\), (b) \\( \tan (s+t) \\), and (c) the quadrant of \\( s+t \\).
\\( \cos s=-\frac{12}{13} \\) and \\( \sin t=-\frac{4}{5} \\), s and t in quadrant iii
(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 \\)?
quadrant iv
quadrant ii
quadran
quadrant

Explanation:

Step1: Find \(\sin s\) and \(\cos t\)

Since \(\sin^{2}\alpha+\cos^{2}\alpha = 1\), for \(s\) with \(\cos s=-\frac{12}{13}\) (in quadrant III), \(\sin s=-\sqrt{1-\cos^{2}s}=-\sqrt{1 - (\frac{- 12}{13})^{2}}=-\sqrt{1-\frac{144}{169}}=-\sqrt{\frac{169 - 144}{169}}=-\frac{5}{13}\)
For \(t\) with \(\sin t=-\frac{4}{5}\) (in quadrant III), \(\cos t=-\sqrt{1-\sin^{2}t}=-\sqrt{1-(\frac{-4}{5})^{2}}=-\sqrt{1-\frac{16}{25}}=-\sqrt{\frac{25 - 16}{25}}=-\frac{3}{5}\)

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 \(\sin s=-\frac{5}{13},\cos t=-\frac{3}{5},\cos s=-\frac{12}{13},\sin t =-\frac{4}{5}\)
\(\sin(s + t)=(-\frac{5}{13})(-\frac{3}{5})+(-\frac{12}{13})(-\frac{4}{5})=\frac{15}{65}+\frac{48}{65}=\frac{15 + 48}{65}=\frac{63}{65}\)

Step3: Calculate \(\tan s\) and \(\tan t\)

\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\), so \(\tan s=\frac{\sin s}{\cos s}=\frac{-\frac{5}{13}}{-\frac{12}{13}}=\frac{5}{12}\), \(\tan t=\frac{\sin t}{\cos t}=\frac{-\frac{4}{5}}{-\frac{3}{5}}=\frac{4}{3}\)

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

Using the formula \(\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\)
\(\tan(s + t)=\frac{\tan s+\tan t}{1-\tan s\tan t}=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\times\frac{4}{3}}=\frac{\frac{5 + 16}{12}}{1-\frac{20}{36}}=\frac{\frac{21}{12}}{\frac{36- 20}{36}}=\frac{\frac{21}{12}}{\frac{16}{36}}=\frac{21}{12}\times\frac{36}{16}=\frac{63}{16}\)

Step5: Determine the quadrant of \(s + t\)

Since \(\sin(s + t)=\frac{63}{65}>0\) and \(\tan(s + t)=\frac{63}{16}>0\)
In quadrant I, \(\sin\theta>0\) and \(\tan\theta>0\)

Answer:

(a) \(\frac{63}{65}\)
(b) \(\frac{63}{16}\)
(c) Quadrant I