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

Explanation:

Step1: Find (\cos s) and (\cos t) using (\sin^{2}\alpha+\cos^{2}\alpha = 1)

For (s): Since (\sin s=-\frac{\sqrt{2}}{6}) and (s) is in quadrant IV ((\cos s>0)), we have (\cos s=\sqrt{1 - \sin^{2}s}) [ \begin{align*} \cos s&=\sqrt{1-\left(-\frac{\sqrt{2}}{6}\right)^{2}}\ &=\sqrt{1-\frac{2}{36}}\ &=\sqrt{\frac{34}{36}}\ &=\frac{\sqrt{34}}{6} \end{align*} ] For (t): Since (\sin t =-\frac{\sqrt{5}}{7}) and (t) is in quadrant IV ((\cos t>0)), we have (\cos t=\sqrt{1-\sin^{2}t}) [ \begin{align*} \cos t&=\sqrt{1-\left(-\frac{\sqrt{5}}{7}\right)^{2}}\ &=\sqrt{1 - \frac{5}{49}}\ &=\sqrt{\frac{44}{49}}\ &=\frac{2\sqrt{11}}{7} \end{align*} ]

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

Here (A = s) and (B=t) [ \begin{align*} \cos(s + t)&=\cos s\cos t-\sin s\sin 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})