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

Explanation:

Step1: Find (\cos s)

Using the identity (\sin^{2}\alpha+\cos^{2}\alpha = 1), for (s) with (\sin s=-\frac{12}{13}) (in quadrant III where (\cos s<0)): [ \begin{align*} \cos^{2}s&=1-\sin^{2}s\ \cos^{2}s&=1 - (-\frac{12}{13})^{2}\ \cos^{2}s&=1-\frac{144}{169}\ \cos^{2}s&=\frac{169 - 144}{169}=\frac{25}{169}\ \cos s&=-\frac{5}{13} \end{align*} ]

Step2: Find (\cos t)

Using the identity (\sin^{2}\alpha+\cos^{2}\alpha = 1), for (t) with (\sin t=\frac{3}{5}) (in quadrant I where (\cos t>0)): [ \begin{align*} \cos^{2}t&=1-\sin^{2}t\ \cos^{2}t&=1-(\frac{3}{5})^{2}\ \cos^{2}t&=1-\frac{9}{25}\ \cos^{2}t&=\frac{25 - 9}{25}=\frac{16}{25}\ \cos t&=\frac{4}{5} \end{align*} ]

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{5}{13}), (\cos t=\frac{4}{5}), (\sin s=-\frac{12}{13}), (\sin t=\frac{3}{5}) [ \begin{align*} \cos(s + t)&=(-\frac{5}{13})\times\frac{4}{5}-(-\frac{12}{13})\times\frac{3}{5}\ &=-\frac{20}{65}+\frac{36}{65}\ &=\frac{-20 + 36}{65}\ &=\frac{16}{65} \end{align*} ]

Step4: Use the cosine - of - a - difference 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{5}{13}), (\cos t=\frac{4}{5}), (\sin s=-\frac{12}{13}), (\sin t=\frac{3}{5}) [ \begin{align*} \cos(s - t)&=(-\frac{5}{13})\times\frac{4}{5}+(-\frac{12}{13})\times\frac{3}{5}\ &=-\frac{20}{65}-\frac{36}{65}\ &=\frac{-20-36}{65}\ &=-\frac{56}{65} \end{align*} ]

Answer:

(\cos(s + t)=\frac{16}{65}) (\cos(s - t)=-\frac{56}{65})