Respuesta :
[tex]\bf sin(a)=\cfrac{\stackrel{opposite}{6}}{\stackrel{hypotenuse}{7}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{7^2-6^2}=a\implies \pm\sqrt{13}=a
\\\\\\
\textit{now, angle "a" is in the II quadrant, where the adjacent is negative}
\\\\\\
-\sqrt{13}=a\qquad \qquad \boxed{cos(a)=\cfrac{-\sqrt{13}}{7}}[/tex]
now, keep in mind that, the hypotenuse is just a radius unit, and thus is never negative, so if a fraction with it is negative, is the other unit. A good example of that is the second fraction here, -1/6, where the hypotenuse is 6, therefore the adjacent side is -1. Anyhow, let's find the opposite side to get the sin(b).
[tex]\bf cos(b)=\cfrac{\stackrel{adjacent}{-1}}{\stackrel{hypotenuse}{6}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a \\\\\\ \textit{now, angle "b" is in the III quadrant, where the opposite is negative} \\\\\\ -\sqrt{35}=b\qquad \qquad \boxed{sin(b)=\cfrac{-\sqrt{35}}{6}}[/tex]
now
[tex]\bf \textit{Sum and Difference Identities} \\ \quad \\ sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}}) \\ \quad \\ sin({{ \alpha}} - {{ \beta}})=sin({{ \alpha}})cos({{ \beta}})- cos({{ \alpha}})sin({{ \beta}}) \\ \quad \\ cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}}) \\ \quad \\ cos({{ \alpha}} - {{ \beta}})= cos({{ \alpha}})cos({{ \beta}}) + sin({{ \alpha}})sin({{ \beta}}) \\ \quad \\ [/tex]
[tex]\bf tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\qquad tan({{ \alpha}} - {{ \beta}}) = \cfrac{tan({{ \alpha}})- tan({{ \beta}})}{1+ tan({{ \alpha}})tan({{ \beta}})}[/tex]
[tex]\bf sin(a+b)=\cfrac{6}{7}\cdot \cfrac{-1}{6}+\cfrac{-\sqrt{13}}{7}\cdot \cfrac{-\sqrt{35}}{6}\implies \cfrac{-1}{7}+\cfrac{\sqrt{455}}{42} \\\\\\ \cfrac{-6+\sqrt{455}}{42}\\\\ -------------------------------\\\\ cos(a-b)=\cfrac{-\sqrt{13}}{7}\cdot \cfrac{-1}{6}+\cfrac{6}{7}\cdot \cfrac{-\sqrt{35}}{6}\implies \cfrac{\sqrt{13}}{42}-\cfrac{\sqrt{35}}{7} \\\\\\ \cfrac{\sqrt{13}-6\sqrt{35}}{42}[/tex]
[tex]\bf -------------------------------\\\\ tan(a)=\cfrac{\frac{6}{7}}{-\frac{\sqrt{13}}{7}}\implies -\cfrac{6}{\sqrt{13}}\implies -\cfrac{6\sqrt{13}}{13} \\\\\\ tan(b)=\cfrac{\frac{-\sqrt{35}}{6}}{\frac{-1}{6}}\implies -\sqrt{35}\\\\ -------------------------------\\\\[/tex]
[tex]\bf tan(a+b)=\cfrac{-\frac{6}{\sqrt{13}}-\sqrt{35}}{1-\left( -\frac{6}{\sqrt{13}} \right)\left( -\sqrt{35} \right)}\implies \cfrac{\frac{-6-\sqrt{455}}{\sqrt{13}}}{1-\frac{6\sqrt{35}}{\sqrt{13}}} \\\\\\ \cfrac{\frac{-6-\sqrt{455}}{\sqrt{13}}}{\frac{\sqrt{13}-6\sqrt{35}}{\sqrt{13}}}\implies \cfrac{-6-\sqrt{455}}{\sqrt{13}-6\sqrt{35}}[/tex]
and now, let's rationalize the denominator of that one, hmmm let's see
[tex]\bf \cfrac{-6-\sqrt{455}}{\sqrt{13}-6\sqrt{35}}\cdot \cfrac{\sqrt{13}+6\sqrt{35}}{\sqrt{13}+6\sqrt{35}} \\\\\\ \cfrac{-6\sqrt{13}-36\sqrt{35}-\sqrt{5915}-6\sqrt{15925}}{({\sqrt{13}-6\sqrt{35}})({\sqrt{13}+6\sqrt{35}})} \\\\\\ \cfrac{-6\sqrt{13}-36\sqrt{35}-13\sqrt{35}-210\sqrt{13}}{(\sqrt{13})^2-(6\sqrt{35})^2} \\\\\\ \cfrac{-216\sqrt{13}-49\sqrt{35}}{13-210}\implies \cfrac{-216\sqrt{13}-49\sqrt{35}}{-197} \\\\\\ \cfrac{216\sqrt{13}+49\sqrt{35}}{197}[/tex]
now, keep in mind that, the hypotenuse is just a radius unit, and thus is never negative, so if a fraction with it is negative, is the other unit. A good example of that is the second fraction here, -1/6, where the hypotenuse is 6, therefore the adjacent side is -1. Anyhow, let's find the opposite side to get the sin(b).
[tex]\bf cos(b)=\cfrac{\stackrel{adjacent}{-1}}{\stackrel{hypotenuse}{6}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a \\\\\\ \textit{now, angle "b" is in the III quadrant, where the opposite is negative} \\\\\\ -\sqrt{35}=b\qquad \qquad \boxed{sin(b)=\cfrac{-\sqrt{35}}{6}}[/tex]
now
[tex]\bf \textit{Sum and Difference Identities} \\ \quad \\ sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}}) \\ \quad \\ sin({{ \alpha}} - {{ \beta}})=sin({{ \alpha}})cos({{ \beta}})- cos({{ \alpha}})sin({{ \beta}}) \\ \quad \\ cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}}) \\ \quad \\ cos({{ \alpha}} - {{ \beta}})= cos({{ \alpha}})cos({{ \beta}}) + sin({{ \alpha}})sin({{ \beta}}) \\ \quad \\ [/tex]
[tex]\bf tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\qquad tan({{ \alpha}} - {{ \beta}}) = \cfrac{tan({{ \alpha}})- tan({{ \beta}})}{1+ tan({{ \alpha}})tan({{ \beta}})}[/tex]
[tex]\bf sin(a+b)=\cfrac{6}{7}\cdot \cfrac{-1}{6}+\cfrac{-\sqrt{13}}{7}\cdot \cfrac{-\sqrt{35}}{6}\implies \cfrac{-1}{7}+\cfrac{\sqrt{455}}{42} \\\\\\ \cfrac{-6+\sqrt{455}}{42}\\\\ -------------------------------\\\\ cos(a-b)=\cfrac{-\sqrt{13}}{7}\cdot \cfrac{-1}{6}+\cfrac{6}{7}\cdot \cfrac{-\sqrt{35}}{6}\implies \cfrac{\sqrt{13}}{42}-\cfrac{\sqrt{35}}{7} \\\\\\ \cfrac{\sqrt{13}-6\sqrt{35}}{42}[/tex]
[tex]\bf -------------------------------\\\\ tan(a)=\cfrac{\frac{6}{7}}{-\frac{\sqrt{13}}{7}}\implies -\cfrac{6}{\sqrt{13}}\implies -\cfrac{6\sqrt{13}}{13} \\\\\\ tan(b)=\cfrac{\frac{-\sqrt{35}}{6}}{\frac{-1}{6}}\implies -\sqrt{35}\\\\ -------------------------------\\\\[/tex]
[tex]\bf tan(a+b)=\cfrac{-\frac{6}{\sqrt{13}}-\sqrt{35}}{1-\left( -\frac{6}{\sqrt{13}} \right)\left( -\sqrt{35} \right)}\implies \cfrac{\frac{-6-\sqrt{455}}{\sqrt{13}}}{1-\frac{6\sqrt{35}}{\sqrt{13}}} \\\\\\ \cfrac{\frac{-6-\sqrt{455}}{\sqrt{13}}}{\frac{\sqrt{13}-6\sqrt{35}}{\sqrt{13}}}\implies \cfrac{-6-\sqrt{455}}{\sqrt{13}-6\sqrt{35}}[/tex]
and now, let's rationalize the denominator of that one, hmmm let's see
[tex]\bf \cfrac{-6-\sqrt{455}}{\sqrt{13}-6\sqrt{35}}\cdot \cfrac{\sqrt{13}+6\sqrt{35}}{\sqrt{13}+6\sqrt{35}} \\\\\\ \cfrac{-6\sqrt{13}-36\sqrt{35}-\sqrt{5915}-6\sqrt{15925}}{({\sqrt{13}-6\sqrt{35}})({\sqrt{13}+6\sqrt{35}})} \\\\\\ \cfrac{-6\sqrt{13}-36\sqrt{35}-13\sqrt{35}-210\sqrt{13}}{(\sqrt{13})^2-(6\sqrt{35})^2} \\\\\\ \cfrac{-216\sqrt{13}-49\sqrt{35}}{13-210}\implies \cfrac{-216\sqrt{13}-49\sqrt{35}}{-197} \\\\\\ \cfrac{216\sqrt{13}+49\sqrt{35}}{197}[/tex]
