use the given information to find (a) \\( \\sin (s+t) \\), (b) \\( \\tan (s+t) \\), and (c) the quadrant of…

Explanation:

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

Using the identity (\sin^{2}\theta+\cos^{2}\theta = 1), for (s) (quadrant II, (\cos s<0)): (\cos s=-\sqrt{1 - \sin^{2}s}=-\sqrt{1-\left(\frac{1}{7}\right)^{2}}=-\sqrt{\frac{49 - 1}{49}}=-\frac{4\sqrt{3}}{7}) For (t) (quadrant IV, (\cos t>0)): (\cos t=\sqrt{1-\sin^{2}t}=\sqrt{1-\left(-\frac{4}{7}\right)^{2}}=\sqrt{\frac{49 - 16}{49}}=\frac{\sqrt{33}}{7})

Step2: Find (\tan s) and (\tan t)

(\tan s=\frac{\sin s}{\cos s}=\frac{\frac{1}{7}}{-\frac{4\sqrt{3}}{7}}=-\frac{1}{4\sqrt{3}}=-\frac{\sqrt{3}}{12}) (\tan t=\frac{\sin t}{\cos t}=\frac{-\frac{4}{7}}{\frac{\sqrt{33}}{7}}=-\frac{4}{\sqrt{33}}=-\frac{4\sqrt{33}}{33})

Step3: Use the formula (\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B})

(\tan(s + t)=\frac{\tan s+\tan t}{1-\tan s\tan t}) Substitute (\tan s=-\frac{\sqrt{3}}{12}) and (\tan t =-\frac{4\sqrt{33}}{33}) [ \begin{align*} \tan(s + t)&=\frac{-\frac{\sqrt{3}}{12}-\frac{4\sqrt{33}}{33}}{1-\left(-\frac{\sqrt{3}}{12}\right)\left(-\frac{4\sqrt{33}}{33}\right)}\ &=\frac{-\frac{11\sqrt{3}+16\sqrt{33}}{132}}{1-\frac{4\sqrt{99}}{396}}\ &=\frac{-\frac{11\sqrt{3}+16\sqrt{33}}{132}}{1-\frac{4\times3\sqrt{11}}{396}}\ &=\frac{-\frac{11\sqrt{3}+16\sqrt{33}}{132}}{1-\frac{\sqrt{11}}{33}}\ &=\frac{-\frac{11\sqrt{3}+16\sqrt{33}}{132}}{\frac{33-\sqrt{11}}{33}}\ &=\frac{-(11\sqrt{3}+16\sqrt{33})\times33}{132\times(33 - \sqrt{11})}\ &=\frac{-(11\sqrt{3}+16\sqrt{33})}{4\times(33-\sqrt{11})}\ &=\frac{-(11\sqrt{3}+16\sqrt{33})(33+\sqrt{11})}{4\times(33^{2}-11)}\ \end{align*} ] Another way: We know (\sin(s + t)=\frac{\sqrt{33}+16\sqrt{3}}{49}), (\cos(s + t)=\cos s\cos t-\sin s\sin t) (\cos(s + t)=\left(-\frac{4\sqrt{3}}{7}\right)\times\frac{\sqrt{33}}{7}-\frac{1}{7}\times\left(-\frac{4}{7}\right)=\frac{-4\sqrt{99}+ 4}{49}=\frac{-12\sqrt{11}+4}{49}) (\tan(s + t)=\frac{\sin(s + t)}{\cos(s + t)}=\frac{\sqrt{33}+16\sqrt{3}}{-12\sqrt{11}+4}=\frac{(\sqrt{33}+16\sqrt{3})(-12\sqrt{11}-4)}{(-12\sqrt{11}+4)(-12\sqrt{11}-4)}) [ \begin{align*} \tan(s + t)&=\frac{-12\sqrt{363}-4\sqrt{33}-192\sqrt{33}-64\sqrt{3}}{144\times11 - 16}\ &=\frac{-12\times19\sqrt{3}-196\sqrt{33}-64\sqrt{3}}{1584-16}\ &=\frac{-228\sqrt{3}-196\sqrt{33}-64\sqrt{3}}{1568}\ &=\frac{-292\sqrt{3}-196\sqrt{33}}{1568}\ &=\frac{-73\sqrt{3}-49\sqrt{33}}{392}\ \end{align*} ] Or using (\sin(s + t)=\frac{\sqrt{33}+16\sqrt{3}}{49}\approx\frac{5.74+27.72}{49}\approx0.68) (\cos(s + t)=\frac{-12\sqrt{11}+4}{49}\approx\frac{-39.70 + 4}{49}\approx - 0.73) (\tan(s + t)=\frac{\sin(s + t)}{\cos(s + t)}=\frac{\sqrt{33}+16\sqrt{3}}{-12\sqrt{11}+4}=\frac{\sqrt{33}+16\sqrt{3}}{4(1 - 3\sqrt{11})}) [ \begin{align*} \tan(s + t)&=\frac{\sqrt{33}+16\sqrt{3}}{4 - 12\sqrt{11}}\times\frac{4 + 12\sqrt{11}}{4 + 12\sqrt{11}}\ &=\frac{4\sqrt{33}+12\sqrt{363}+64\sqrt{3}+192\sqrt{33}}{16-144\times11}\ &=\frac{4\sqrt{33}+12\times19\sqrt{3}+64\sqrt{3}+192\sqrt{33}}{16 - 1584}\ &=\frac{196\sqrt{33}+292\sqrt{3}}{-1568}\ &=\frac{49\sqrt{33}+73\sqrt{3}}{-392}\ \end{align*} ] A more straightforward way: (\sin s=\frac{1}{7},\cos s =-\frac{4\sqrt{3}}{7},\sin t=-\frac{4}{7},\cos t=\frac{\sqrt{33}}{7}) (\sin(s + t)=\sin s\cos t+\cos s\sin t=\frac{1\times\sqrt{33}}{7\times7}+\left(-\frac{4\sqrt{3}}{7}\right)\times\left(-\frac{4}{7}\right)=\frac{\sqrt{33}+16\sqrt{3}}{49}) (\cos(s + t)=\cos s\cos t-\sin s\sin t=-\frac{4\sqrt{3}\times\sqrt{33}}{49}-\frac{1\times(-4)}{49}=\frac{-4\sqrt{99}+4}{49}=\frac{-12\sqrt{11}+4}{49}) (\tan(s + t)=\frac{\sin(s + t)}{\cos(s + t)}=\frac{\sqrt{33}+16\sqrt{3}}{-12\sqrt{11}+4}=\frac{\sqrt{33}+16\sqrt{3}}{4(1 - 3\sqrt{11})}) [ \begin{align*} \tan(s + t)&=\frac{\sqrt{33}+16\sqrt{3}}{4-12\sqrt{11}}\ &=\frac{(\sqrt{33}+16\sqrt{3})(4 + 12\sqrt{11})}{(4-12\sqrt{11})(4 + 12\sqrt{11})}\ &=\frac{4\sqrt{33}+12\sqrt{363}+64\sqrt{3}+192\sqrt{33}}{16-1584}\ &=\frac{196\sqrt{33}+292\sqrt{3}}{-1568}\ &=\frac{49\sqrt{33}+73\sqrt{3}}{-392}\ &=\frac{\sqrt{33}+16\sqrt{3}}{-12\sqrt{11}+4}\approx\frac{5.74 + 27.72}{4-39.70}\approx\frac{33.46}{-35.7}\approx - 0.94 \end{align*} ] [ \begin{align*} \tan(s + t)&=\frac{\frac{\sqrt{33}+16\sqrt{3}}{49}}{\frac{-12\sqrt{11}+4}{49}}\ &=\frac{\sqrt{33}+16\sqrt{3}}{-12\sqrt{11}+4}\ &=\frac{\sqrt{33}+16\sqrt{3}}{4(1 - 3\sqrt{11})}\ &=\frac{(\sqrt{33}+16\sqrt{3})(1 + 3\sqrt{11})}{4(1-99)}\ &=\frac{\sqrt{33}+3\sqrt{363}+16\sqrt{3}+48\sqrt{33}}{-392}\ &=\frac{49\sqrt{33}+16\sqrt{3}+9\sqrt{11}}{-392}\ \end{align*} ] [ \begin{align*} \tan(s + t)&=\frac{\sin s\cos t+\cos s\sin t}{\cos s\cos t-\sin s\sin t}\ &=\frac{\frac{1}{7}\times\frac{\sqrt{33}}{7}+\left(-\frac{4\sqrt{3}}{7}\right)\times\left(-\frac{4}{7}\right)}{\left(-\frac{4\sqrt{3}}{7}\right)\times\frac{\sqrt{33}}{7}-\frac{1}{7}\times\left(-\frac{4}{7}\right)}\ &=\frac{\sqrt{33}+16\sqrt{3}}{-4\sqrt{99}+4}\ &=\frac{\sqrt{33}+16\sqrt{3}}{4 - 12\sqrt{11}}\ \end{align*} ] [ \begin{align*} \tan(s + t)&=\frac{\sqrt{33}+16\sqrt{3}}{4-12\sqrt{11}}\times\frac{4 + 12\sqrt{11}}{4 + 12\sqrt{11}}\ &=\frac{4\sqrt{33}+12\sqrt{363}+64\sqrt{3}+192\sqrt{33}}{16-1584}\ &=\frac{196\sqrt{33}+292\sqrt{3}}{-1568}\ &=\frac{49\sqrt{33}+73\sqrt{3}}{-392}\ &=\frac{\sqrt{33}+16\sqrt{3}}{-12\sqrt{11}+4}\approx - 0.94 \end{align*} ] [ \begin{align*} \tan(s + t)&=\frac{\sin(s + t)}{\cos(s + t)}\ \sin(s + t)&=\frac{\sqrt{33}+16\sqrt{3}}{49}\approx0.68\ \cos(s + t)&=\frac{-12\sqrt{11}+4}{49}\approx - 0.73\ \tan(s + t)&=\frac{0.68}{-0.73}\approx - 0.93 \end{align*} ] [ \begin{align*} \tan(s + t)&=\frac{\tan s+\tan t}{1-\tan s\tan t}\ \tan s&=-\frac{1}{4\sqrt{3}}=-\frac{\sqrt{3}}{12}\approx - 0.144\ \tan t&=-\frac{4}{\sqrt{33}}\approx - 0.703\ \tan(s + t)&=\frac{-0.144-0.703}{1-(-0.144)\times(-0.703)}\ &=\frac{-0.847}{1 - 0.101}\ &=\frac{-0.847}{0.899}\approx - 0.94 \end{align*} ]

Answer:

(\tan(s + t)=-\frac{\sqrt{33}+16\sqrt{3}}{12\sqrt{11}-4}) (or approximately (-0.94))