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

Explanation:

Step1: Find (\sin s)

Using the identity (\sin^{2}\alpha+\cos^{2}\alpha = 1), for (\alpha=s), we have (\sin^{2}s+\cos^{2}s = 1). Given (\cos s=-\frac{3}{5}), then (\sin^{2}s=1-\cos^{2}s=1 - (-\frac{3}{5})^{2}=1-\frac{9}{25}=\frac{16}{25}). Since (s) is in quadrant II, (\sin s>0), so (\sin s=\frac{4}{5}).

Step2: Find (\cos t)

Using the identity (\sin^{2}\alpha+\cos^{2}\alpha = 1), for (\alpha = t), we have (\sin^{2}t+\cos^{2}t=1). Given (\sin t=\frac{2}{5}), then (\cos^{2}t=1-\sin^{2}t=1 - (\frac{2}{5})^{2}=1-\frac{4}{25}=\frac{21}{25}). Since (t) is in quadrant II, (\cos t<0), so (\cos t=-\frac{\sqrt{21}}{5}).

Step3: Use the cosine - of - a - sum formula

The cosine - of - a - sum formula is (\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{3}{5}), (\cos t=-\frac{\sqrt{21}}{5}), (\sin s=\frac{4}{5}), and (\sin t=\frac{2}{5}) into the formula: [ \begin{align*} \cos(s + t)&=\left(-\frac{3}{5}\right)\left(-\frac{\sqrt{21}}{5}\right)-\frac{4}{5}\times\frac{2}{5}\ &=\frac{3\sqrt{21}}{25}-\frac{8}{25}\ &=\frac{3\sqrt{21}-8}{25} \end{align*} ]

Answer:

(\frac{3\sqrt{21}-8}{25})