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{8}{17} \\) and \\( \cos t=\frac{4}{5} \\), s and t in quadrant iv
(a) \\( \sin (s+t)=-\frac{84}{85} \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
(b) \\( \tan (s+t)=\frac{84}{13} \\)
(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 i
quadrant ii
quadrant iii
quadrant iv

Explanation:

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

Since \(s\) and \(t\) are in quadrant IV, \(\sin s<0\) and \(\sin t < 0\).
Using the identity \(\sin^{2}\alpha+\cos^{2}\alpha = 1\), for \(s\):
\(\sin s=-\sqrt{1-\cos^{2}s}=-\sqrt{1 - (\frac{8}{17})^{2}}=-\sqrt{1-\frac{64}{289}}=-\sqrt{\frac{289 - 64}{289}}=-\sqrt{\frac{225}{289}}=-\frac{15}{17}\)
For \(t\):
\(\sin t=-\sqrt{1-\cos^{2}t}=-\sqrt{1-(\frac{4}{5})^{2}}=-\sqrt{1-\frac{16}{25}}=-\sqrt{\frac{25 - 16}{25}}=-\sqrt{\frac{9}{25}}=-\frac{3}{5}\)

Step2: Find \(\tan s\) and \(\tan t\)

Using the identity \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\), for \(s\):
\(\tan s=\frac{\sin s}{\cos s}=\frac{-\frac{15}{17}}{\frac{8}{17}}=-\frac{15}{8}\)
For \(t\):
\(\tan t=\frac{\sin t}{\cos t}=\frac{-\frac{3}{5}}{\frac{4}{5}}=-\frac{3}{4}\)

Step3: Find \(\sin(s + t)\)

Using the formula \(\sin(A + B)=\sin A\cos B+\cos A\sin B\)
\(\sin(s + t)=\sin s\cos t+\cos s\sin t\)
\(=(-\frac{15}{17})\times\frac{4}{5}+\frac{8}{17}\times(-\frac{3}{5})\)
\(=-\frac{60}{85}-\frac{24}{85}=-\frac{84}{85}\)

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

Using the formula \(\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\)
\(\tan(s + t)=\frac{\tan s+\tan t}{1-\tan s\tan t}=\frac{-\frac{15}{8}-\frac{3}{4}}{1-(-\frac{15}{8})\times(-\frac{3}{4})}\)
First, simplify the numerator: \(-\frac{15}{8}-\frac{3}{4}=-\frac{15 + 6}{8}=-\frac{21}{8}\)
Second, simplify the denominator: \(1-\frac{45}{32}=\frac{32-45}{32}=-\frac{13}{32}\)
Then \(\tan(s + t)=\frac{-\frac{21}{8}}{-\frac{13}{32}}=\frac{21}{8}\times\frac{32}{13}=\frac{84}{13}\)

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

Since \(\sin(s + t)=-\frac{84}{85}<0\) and \(\tan(s + t)=\frac{84}{13}>0\)
In quadrant III, \(\sin\theta<0\) and \(\tan\theta>0\)

Answer:

(a) \(\sin(s + t)=-\frac{84}{85}\)
(b) \(\tan(s + t)=\frac{84}{13}\)
(c) Quadrant III