![\bf ~~~~~~~~~~~~\textit{function transformations} \\\\\\ f(x)=Asin(Bx+C)+D \qquad \qquad f(x)=Acos(Bx+C)+D \\\\ f(x)=Atan(Bx+C)+D \qquad \qquad f(x)=Asec(Bx+C)+D \\\\[-0.35em] ~\dotfill\\\\ \bullet \textit{ stretches or shrinks}\\ ~~~~~~\textit{horizontally by amplitude } A\cdot B\\\\ \bullet \textit{ flips it upside-down if }A\textit{ is negative}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hxxqtx78z5iykg98vwexgqiczmp183c02i.png)


with that template in mind, let's see.

now, since the period is 30, and cos(x) starts off at 1, recall cos(0) = 1, so then le'ts move the starting point over by simply doing a horizontal shift to the right by a quarter of the period, 30/4 = 15/2 units, that way the initial "hump" starts off at 0.
![\bf \stackrel{\textit{horizontal shift of }(15)/(2)}{\cfrac{15}{2}=\cfrac{C}{B}}\implies \cfrac{15}{2}=\cfrac{C}{~~(\pi )/(15)~~}\implies \cfrac{15}{2}=\cfrac{15C}{\pi }\implies 15\pi =30C \\\\\\ \cfrac{15\pi }{30}=C\implies \cfrac{\pi }{2}=C~\hfill \stackrel{\textit{horizontally to the right}}{-\cfrac{\pi }{2}=C} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill g(x)=2cos\left((\pi )/(15)x-(\pi )/(2) \right)~\hfill](https://img.qammunity.org/2020/formulas/mathematics/middle-school/mscr8ecf6vjvyxofda6bah264brk0f1daz.png)
Check the picture below.