use the cosine of a sum and cosine of a difference identities to find \\( \\cos (s + t) \\) and \\( \\cos (s…

Explanation:

Step1: Find (\cos s)

Since (\sin s =-\frac{3}{5}) and (s) is in quadrant III. Using the identity (\sin^{2}\theta+\cos^{2}\theta = 1), we have (\cos^{2}s=1-\sin^{2}s). [ \begin{align*} \cos^{2}s&=1-\left(-\frac{3}{5}\right)^{2}\ &=1-\frac{9}{25}\ &=\frac{16}{25} \end{align*} ] In quadrant III, (\cos s<0), so (\cos s =-\frac{4}{5})

Step2: Find (\cos t)

Since (\sin t=\frac{5}{13}) and (t) is in quadrant I. Using the identity (\sin^{2}\theta+\cos^{2}\theta = 1), we have (\cos^{2}t=1-\sin^{2}t) [ \begin{align*} \cos^{2}t&=1-\left(\frac{5}{13}\right)^{2}\ &=1 - \frac{25}{169}\ &=\frac{144}{169} \end{align*} ] In quadrant I, (\cos t>0), so (\cos t=\frac{12}{13})

Step3: Use the cosine - of - a - sum formula (\cos(A + B)=\cos A\cos B-\sin A\sin B)

Here (A = s) and (B=t), so (\cos(s + t)=\cos s\cos t-\sin s\sin t) Substitute (\cos s=-\frac{4}{5}), (\cos t=\frac{12}{13}), (\sin s=-\frac{3}{5}), (\sin t=\frac{5}{13}) [ \begin{align*} \cos(s + t)&=\left(-\frac{4}{5}\right)\times\frac{12}{13}-\left(-\frac{3}{5}\right)\times\frac{5}{13}\ &=-\frac{48}{65}+\frac{15}{65}\ &=\frac{-48 + 15}{65}\ &=-\frac{33}{65} \end{align*} ]

Answer:

(-\frac{33}{65})