We have to graph the function:
[tex]y=-\frac{5}{2}+\cos \lbrack3(x-\frac{\pi}{6})\rbrack[/tex]We can start from known points of the cosine function and then find the values of y.
We know the exact values of cosine for the following angles:
[tex]\begin{gathered} \cos (0)=1 \\ \cos (\frac{\pi}{6})=\frac{\sqrt[]{3}}{2} \\ \cos (\frac{\pi}{4})=\frac{\sqrt[]{2}}{2} \\ \cos (\frac{\pi}{3})=\frac{1}{2} \\ \cos (\frac{\pi}{2})=0 \\ \cos (\frac{2\pi}{3})=-\frac{1}{2} \\ \cos (\frac{3\pi}{4})=\frac{-\sqrt[]{2}}{2} \\ \cos (\frac{5\pi}{6})=\frac{-\sqrt[]{3}}{2} \\ \cos (\pi)=-1 \end{gathered}[/tex]We have half the cycle here. We will complete the values later.
We then can find the value of x that matches the arguments of the known vlaues of the cosine as:
[tex]\begin{gathered} \alpha=3(x-\frac{\pi}{6}) \\ x=\frac{\alpha}{3}+\frac{\pi}{6} \end{gathered}[/tex]where α is the argument of the known values of cosine (0, π/6, π/4, ...).
We then can calculate the values of x for each one as:
[tex]\begin{gathered} x_1=\frac{0}{3}+\frac{\pi}{6}=\frac{\pi}{6} \\ x_2=\frac{1}{3}\cdot\frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{18}+\frac{\pi}{6}=\frac{4\pi}{18} \\ x_3=\frac{1}{3}\cdot\frac{\pi}{4}+\frac{\pi}{6}=\frac{\pi}{12}+\frac{\pi}{6}=\frac{3\pi}{12}=\frac{\pi}{4} \\ x_4=\frac{1}{3}\cdot\frac{\pi}{3}+\frac{\pi}{6}=\frac{\pi}{9}+\frac{\pi}{6}=\frac{5\pi}{18} \\ x_5=\frac{1}{3}\cdot\frac{\pi}{2}+\frac{\pi}{6}=\frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{3} \\ x_6=\frac{1}{3}\cdot\frac{2\pi}{3}+\frac{\pi}{6}=\frac{2\pi}{9}+\frac{\pi}{6}=\frac{7\pi}{18} \\ x_7=\frac{1}{3}\cdot\frac{3\pi}{4}+\frac{\pi}{6}=\frac{\pi}{4}+\frac{\pi}{6}=\frac{5\pi}{12} \\ x_8=\frac{1}{3}\cdot\frac{5\pi}{6}+\frac{\pi}{6}=\frac{5\pi}{18}+\frac{\pi}{6}=\frac{4\pi}{9} \\ x_9=\frac{1}{3}\pi+\frac{\pi}{6}=\frac{\pi}{2} \end{gathered}[/tex]We then can calculate the value of y for each of this points, using the known values of the cosine, as:
[tex]\begin{gathered} x=\frac{\pi}{6}\Rightarrow y=-\frac{5}{2}+1=-\frac{3}{2} \\ x=\frac{4\pi}{18}\Rightarrow y=-\frac{5}{2}+\frac{\sqrt[]{3}}{2}=\frac{\sqrt[]{3}-5}{2} \\ x=\frac{\pi}{4}\Rightarrow y=-\frac{5}{2}+\frac{\sqrt[]{2}}{2}=\frac{\sqrt[]{2}-5}{2} \\ x=\frac{5\pi}{18}\Rightarrow y=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=-2 \\ x=\frac{\pi}{3}\Rightarrow y=-\frac{5}{2}+0=-\frac{5}{2} \\ x=\frac{7\pi}{18}\Rightarrow y=-\frac{5}{2}-\frac{1}{2}=-\frac{6}{2}=-3 \\ x=\frac{5\pi}{12}\Rightarrow y=-\frac{5}{2}-\frac{\sqrt[]{2}}{2}=\frac{-5-\sqrt[]{2}}{2} \\ x=\frac{4\pi}{9}\Rightarrow y=-\frac{5}{2}-\frac{\sqrt[]{3}}{2}=\frac{-5-\sqrt[]{3}}{2} \\ x=\frac{\pi}{2}\Rightarrow y=-\frac{5}{2}-1=-\frac{7}{2} \end{gathered}[/tex]We can repeat this process for the rest of the cycle, but in this case, we will only graph the mean value (when cosine is 0) and the extreme values (when cosine is -1 or 1).
We can list this as:
[tex]\begin{gathered} \cos (\pi)=-1 \\ \cos (\frac{3\pi}{2})=0 \\ \cos (2\pi)=1 \end{gathered}[/tex]We can relate this values to x using the formula we used before:
[tex]\begin{gathered} x_{10}=\frac{1}{3}(\pi)+\frac{\pi}{6}=\frac{\pi}{3}+\frac{\pi}{6}=\frac{\pi}{2} \\ x_{11}=\frac{1}{3}(\frac{3\pi}{2})+\frac{\pi}{6}=\frac{\pi}{2}+\frac{\pi}{6}=\frac{2\pi}{3} \\ x_{12}=\frac{1}{3}(2\pi)+\frac{\pi}{6}=\frac{2\pi}{3}+\frac{\pi}{6}=\frac{5\pi}{6} \end{gathered}[/tex]Now, we calculate the values of y as:
[tex]\begin{gathered} x=\frac{\pi}{2}\Rightarrow y=-\frac{5}{2}-1=-\frac{7}{2} \\ x=\frac{2\pi}{3}\Rightarrow y=-\frac{5}{2}+0=-\frac{5}{2} \\ x=\frac{5\pi}{6}\Rightarrow y=-\frac{5}{2}+1=-\frac{3}{2} \end{gathered}[/tex]Using this particular values for the complete cycle we can complete the table as:
