Use the given conditions to find the exact value of the expression. cos α = 24/25, sin α < 0, cos (α+11π/6)Please help me find the answer and explain how you got your answer. Any help is much appreciated. Thanks!

sin(α) < 0, is just another way to say that sin(α) is a negative value, and that only happens on the III and IV quadrants.

now, we know what the cosine is, so let's use that, notice that, the cosine is a positive value, and if the sine is negative, that only happens on the IV quadrant

[tex]\bf cos(\alpha)=\cfrac{\stackrel{adjacent}{24}}{\stackrel{opposite}{25}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \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{25^2-24^2}=b\implies \pm\sqrt{49}=b\implies \pm 7=b\implies \stackrel{IV~quadrant}{-7=b}[/tex]

 now let's use the sum identities, recall that [tex]\bf sin(\alpha)=\cfrac{opposite}{hypotenuse}[/tex]

[tex]\bf \textit{Sum and Difference Identities} \\\\ cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta)\\\\ -------------------------------[/tex]

[tex]\bf cos\left(\alpha+\frac{11\pi }{6} \right)=cos(\alpha)cos\left(\frac{11\pi }{6} \right)-sin(\alpha)sin\left(\frac{11\pi }{6} \right) \\\\\\ \left( \cfrac{24}{25} \right)\left( \cfrac{\sqrt{3}}{2} \right)-\left( \cfrac{-7}{25} \right)\left( -\cfrac{1}{2} \right)\implies \cfrac{24\sqrt{3}}{50}-\left( \cfrac{7}{50} \right) \\\\\\ \cfrac{24\sqrt{3}}{50}-\cfrac{7}{50}\implies \cfrac{24\sqrt{3}-7}{50}[/tex]