Respuesta :
[tex]\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}
\end{array}[/tex]
[tex]\bf \begin{array}{llll} % right side info \bullet \textit{ stretches or shrinks}\\ \quad \textit{horizontally by amplitude } |{{ A}}|\\\\ \bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\\\ \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\ \end{array}[/tex]
[tex]\bf \begin{array}{llll} \bullet \textit{vertical shift by }{{ D}}\\ \qquad if\ {{ D}}\textit{ is negative, downwards}\\\\ \qquad if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{function period or frequency}\\ \qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\ \qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta) \end{array}[/tex]
so, with that template in mind, let's see
[tex]\bf y=5cos[9(\theta-30^o)]-2\implies y=5cos\left[9(\theta-\frac{\pi }{6}) \right]-2 \\\\\\ y=5cos(9\theta-\frac{3\pi }{2})-2\\\\ -------------------------------\\\\ \begin{array}{lllcclllll} y=&5cos(&9\theta&-\frac{3\pi }{2} )&-2\\ &\uparrow &\uparrow &\uparrow &\uparrow \\ &A&B&C&D \end{array}\\\\ -------------------------------\\\\[/tex]
[tex]\bf \textit{horizontal/phase shift}\implies \cfrac{C}{B}\implies \cfrac{-\frac{3\pi }{2}}{9}\implies \cfrac{-\frac{3\pi }{2}}{\frac{9}{1}} \\\\\\ -\cfrac{3\pi }{2}\cdot \cfrac{1}{9}\implies -\cfrac{\pi }{6}\\\\ -------------------------------\\\\ period\implies \cfrac{2\pi }{B}\implies \cfrac{2\pi }{9}[/tex]
[tex]\bf \begin{array}{llll} % right side info \bullet \textit{ stretches or shrinks}\\ \quad \textit{horizontally by amplitude } |{{ A}}|\\\\ \bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\\\ \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\ \end{array}[/tex]
[tex]\bf \begin{array}{llll} \bullet \textit{vertical shift by }{{ D}}\\ \qquad if\ {{ D}}\textit{ is negative, downwards}\\\\ \qquad if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{function period or frequency}\\ \qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\ \qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta) \end{array}[/tex]
so, with that template in mind, let's see
[tex]\bf y=5cos[9(\theta-30^o)]-2\implies y=5cos\left[9(\theta-\frac{\pi }{6}) \right]-2 \\\\\\ y=5cos(9\theta-\frac{3\pi }{2})-2\\\\ -------------------------------\\\\ \begin{array}{lllcclllll} y=&5cos(&9\theta&-\frac{3\pi }{2} )&-2\\ &\uparrow &\uparrow &\uparrow &\uparrow \\ &A&B&C&D \end{array}\\\\ -------------------------------\\\\[/tex]
[tex]\bf \textit{horizontal/phase shift}\implies \cfrac{C}{B}\implies \cfrac{-\frac{3\pi }{2}}{9}\implies \cfrac{-\frac{3\pi }{2}}{\frac{9}{1}} \\\\\\ -\cfrac{3\pi }{2}\cdot \cfrac{1}{9}\implies -\cfrac{\pi }{6}\\\\ -------------------------------\\\\ period\implies \cfrac{2\pi }{B}\implies \cfrac{2\pi }{9}[/tex]
