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), since (\cos s=\frac{1}{5}) and (s) is in quadrant I, (\sin s=\sqrt{1 - \cos^{2}s}=\sqrt{1-\left(\frac{1}{5}\right)^{2}}=\sqrt{\frac{25 - 1}{25}}=\frac{2\sqrt{6}}{5}). For (t), since (\sin t=\frac{3}{5}) and (t) is in quadrant I, (\cos t=\sqrt{1-\sin^{2}t}=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{25 - 9}{25}}=\frac{4}{5}).

Step2: Use the cosine - of - a - difference identity

The cosine - of - a - difference identity 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{1}{5}), (\cos t=\frac{4}{5}), (\sin s=\frac{2\sqrt{6}}{5}), and (\sin t=\frac{3}{5}) into the formula: [ \begin{align*} \cos(s - t)&=\frac{1}{5}\times\frac{4}{5}+\frac{2\sqrt{6}}{5}\times\frac{3}{5}\ &=\frac{4+6\sqrt{6}}{25} \end{align*} ]

Answer:

(\frac{4 + 6\sqrt{6}}{25})