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

Explanation:

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

Using the Pythagorean identity (\sin^{2}\alpha+\cos^{2}\alpha = 1), for (s) with (\sin s=-\frac{3}{5}) (in quadrant III, so (\cos s<0)): [ \begin{align*} \cos^{2}s&=1-\sin^{2}s\ \cos^{2}s&=1 - (-\frac{3}{5})^{2}\ \cos^{2}s&=1-\frac{9}{25}\ \cos^{2}s&=\frac{16}{25}\ \cos s&=-\frac{4}{5} \end{align*} ] For (t) with (\sin t=\frac{5}{13}) (in quadrant I, so (\cos t>0)): [ \begin{align*} \cos^{2}t&=1-\sin^{2}t\ \cos^{2}t&=1-(\frac{5}{13})^{2}\ \cos^{2}t&=1 - \frac{25}{169}\ \cos^{2}t&=\frac{144}{169}\ \cos t&=\frac{12}{13} \end{align*} ]

Step2: Use the cosine of a difference identity (\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)&=(-\frac{4}{5})\times\frac{12}{13}+(-\frac{3}{5})\times\frac{5}{13}\ &=-\frac{48}{65}-\frac{15}{65}\ &=-\frac{63}{65} \end{align*} ]

Answer:

(-\frac{63}{65})