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

Explanation:

Step1: Find (\cos s)

Using the identity (\sin^{2}\alpha+\cos^{2}\alpha = 1), for (\alpha=s), we have (\cos^{2}s=1-\sin^{2}s). Given (\sin s=-\frac{\sqrt{3}}{4}), then (\sin^{2}s=\frac{3}{16}), so (\cos^{2}s = 1-\frac{3}{16}=\frac{16 - 3}{16}=\frac{13}{16}). Since (s) is in quadrant III, (\cos s<0), so (\cos s=-\frac{\sqrt{13}}{4}).

Step2: Find (\cos t)

Using the identity (\sin^{2}\alpha+\cos^{2}\alpha = 1), for (\alpha=t), we have (\cos^{2}t=1-\sin^{2}t). Given (\sin t=-\frac{\sqrt{5}}{7}), then (\sin^{2}t=\frac{5}{49}), so (\cos^{2}t=1 - \frac{5}{49}=\frac{49-5}{49}=\frac{44}{49}). Since (t) is in quadrant III, (\cos t<0), so (\cos t=-\frac{2\sqrt{11}}{7}).

Step3: Use the cosine - of - a - sum formula

The cosine - of - a - sum formula is (\cos(s + t)=\cos s\cos t-\sin s\sin t). Substitute (\cos s=-\frac{\sqrt{13}}{4}), (\cos t=-\frac{2\sqrt{11}}{7}), (\sin s=-\frac{\sqrt{3}}{4}), and (\sin t=-\frac{\sqrt{5}}{7}) into the formula: [ \begin{align*} \cos(s + t)&=\left(-\frac{\sqrt{13}}{4}\right)\left(-\frac{2\sqrt{11}}{7}\right)-\left(-\frac{\sqrt{3}}{4}\right)\left(-\frac{\sqrt{5}}{7}\right)\ &=\frac{2\sqrt{143}}{28}-\frac{\sqrt{15}}{28}\ &=\frac{2\sqrt{143}-\sqrt{15}}{28} \end{align*} ]

Answer:

(\frac{2\sqrt{143}-\sqrt{15}}{28})