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

Explanation:

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

Using the Pythagorean identity (\sin^{2}\alpha+\cos^{2}\alpha = 1). For (s) in quadrant II, (\sin s=\sqrt{1-\cos^{2}s}). Given (\cos s=-\frac{4}{5}), then (\sin s=\sqrt{1 - (-\frac{4}{5})^{2}}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}). For (t) in quadrant II, (\cos t=-\sqrt{1-\sin^{2}t}). Given (\sin t=\frac{1}{5}), then (\cos t=-\sqrt{1 - (\frac{1}{5})^{2}}=-\sqrt{1-\frac{1}{25}}=-\sqrt{\frac{24}{25}}=-\frac{2\sqrt{6}}{5}).

Step2: Use the cosine of a sum identity (\cos(A + B)=\cos A\cos B-\sin A\sin B)

For (\cos(s + t)), substitute (A = s) and (B = t). (\cos(s + t)=\cos s\cos t-\sin s\sin t) (=\left(-\frac{4}{5}\right)\left(-\frac{2\sqrt{6}}{5}\right)-\frac{3}{5}\times\frac{1}{5}) (=\frac{8\sqrt{6}}{25}-\frac{3}{25}=\frac{8\sqrt{6}-3}{25})

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

For (\cos(s - t)), substitute (A = s) and (B = t). (\cos(s - t)=\cos s\cos t+\sin s\sin t) (=\left(-\frac{4}{5}\right)\left(-\frac{2\sqrt{6}}{5}\right)+\frac{3}{5}\times\frac{1}{5}) (=\frac{8\sqrt{6}}{25}+\frac{3}{25}=\frac{8\sqrt{6}+3}{25})

Answer:

(\cos(s + t)=\frac{8\sqrt{6}-3}{25}) (\cos(s - t)=\frac{8\sqrt{6}+3}{25})