Respuesta :
[tex]\bf \textit{Sum and Difference Identities} \\\\ cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta)[/tex]
we know that angle A is in the III Quadrant, where sine is negative and cosine is negative.
we also know that angle B is in the II Quadrant, where sine is positive and cosine is negative.
the hypotenuse is never negative, since it's simply a radius length, so in a fraction, the hypotenuse will be the positive value.
[tex]\bf sin(A)=\cfrac{\stackrel{opposite}{-8}}{\underset{hypotenuse}{17}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \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{17^2-(-8)^2}=a\implies \pm\sqrt{225}=a\implies \pm 15 = a\implies \stackrel{III~Quadrant}{-15=a} \\\\\\ therefore\qquad \qquad cos(A)=\cfrac{-15}{17} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf sec(B)=\cfrac{\stackrel{hypotenuse}{5}}{\underset{adjacent}{-4}}\qquad \impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{5^2-(-4)^2}=b\implies \pm\sqrt{9}=b \\\\\\ \pm 3 = b\implies \stackrel{II~Quadrant}{3=b}~\hfill cos(B)=\cfrac{\stackrel{adjacent}{-4}}{\underset{hypotenuse}{5}}~\hfill sin(B)=\cfrac{\stackrel{opposite}{3}}{\underset{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf cos(A+B)=cos(A)cos(B)-sin(A)sin(B) \\\\\\ cos(A+B)=\cfrac{-15}{17}\cdot \cfrac{-4}{5}-\left(\cfrac{-8}{17}\cdot \cfrac{3}{5} \right)\implies cos(A+B)=\cfrac{60}{85}-\left( \cfrac{-24}{85} \right) \\\\\\ cos(A+B)=\cfrac{60}{85}+\cfrac{24}{85}\implies cos(A+B)=\cfrac{84}{85}[/tex]
