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 s=\sqrt{1 - \cos^{2}s}). Since (\cos s=\frac{1}{5}), then (\sin s=\sqrt{1-\left(\frac{1}{5}\right)^{2}}=\sqrt{1-\frac{1}{25}}=\sqrt{\frac{24}{25}}=\frac{2\sqrt{6}}{5}).

Step2: Find (\cos t)

Using the identity (\sin^{2}\alpha+\cos^{2}\alpha = 1), for (\alpha = t), we have (\cos t=\sqrt{1-\sin^{2}t}). Since (\sin t=\frac{3}{5}), then (\cos t=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{1 - \frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{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{1}{5}), (\cos t=\frac{4}{5}), (\sin s=\frac{2\sqrt{6}}{5}), and (\sin t=\frac{3}{5}) into the formula: (\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}).

Answer:

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