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)

Use the identity (\sin^{2}\alpha+\cos^{2}\alpha = 1), so (\cos\alpha=-\sqrt{1 - \sin^{2}\alpha}) (since (s) and (t) are in quadrant III). For (s): (\cos s=-\sqrt{1-\left(-\frac{\sqrt{3}}{4}\right)^{2}}=-\sqrt{1-\frac{3}{16}}=-\sqrt{\frac{13}{16}}=-\frac{\sqrt{13}}{4}) For (t): (\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})

Step2: Use the cosine - difference identity (\cos(A - B)=\cos A\cos B+\sin A\sin B)

Here (A = s) and (B=t). (\cos(s - t)=\cos s\cos t+\sin s\sin t) Substitute (\sin s=-\frac{\sqrt{3}}{4}), (\sin t=-\frac{\sqrt{5}}{7}), (\cos s=-\frac{\sqrt{13}}{4}), and (\cos t=-\frac{2\sqrt{11}}{7}) [ \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})