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

Explanation:

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

Since (\sin^{2}\alpha+\cos^{2}\alpha = 1), for (s) with (\sin s=-\frac{\sqrt{2}}{6}) and (s) in quadrant IV ((\cos s>0)): (\cos s=\sqrt{1 - \sin^{2}s}=\sqrt{1-\left(-\frac{\sqrt{2}}{6}\right)^{2}}=\sqrt{1-\frac{2}{36}}=\sqrt{\frac{34}{36}}=\frac{\sqrt{34}}{6}) For (t) with (\sin t =-\frac{\sqrt{5}}{7}) and (t) in quadrant IV ((\cos t>0)): (\cos t=\sqrt{1-\sin^{2}t}=\sqrt{1-\left(-\frac{\sqrt{5}}{7}\right)^{2}}=\sqrt{1 - \frac{5}{49}}=\sqrt{\frac{44}{49}}=\frac{2\sqrt{11}}{7})

Step2: Use the cosine - of - a - difference identity (\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{\sqrt{34}}{6}), (\cos t=\frac{2\sqrt{11}}{7}), (\sin s=-\frac{\sqrt{2}}{6}), (\sin t=-\frac{\sqrt{5}}{7}) [ \begin{align*} \cos(s - t)&=\frac{\sqrt{34}}{6}\times\frac{2\sqrt{11}}{7}+\left(-\frac{\sqrt{2}}{6}\right)\times\left(-\frac{\sqrt{5}}{7}\right)\ &=\frac{2\sqrt{374}}{42}+\frac{\sqrt{10}}{42}\ &=\frac{2\sqrt{374}+\sqrt{10}}{42} \end{align*} ]

Answer:

(\frac{2\sqrt{374}+\sqrt{10}}{42})