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}}cos({{ B}}x+{{ C}})+{{ D}}\\
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
\end{array}[/tex]
[tex]\bf \begin{array}{llll} % right side info \bullet \textit{ stretches or shrinks}\\ \quad \textit{horizontally by amplitude } |{{ A}}|\\\\ \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}\\ \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]
now.. hmmm let's see, keeping in mind the template above
so, amplitude of 2, A=2, thus |A| = ±2, so, you can use either
and period of 4π, well, that simply means that
[tex]\bf \cfrac{2\pi }{B}=4\pi \implies \cfrac{2\pi }{4\pi }=B\implies \cfrac{1}{2}=B\\\\ -----------------------------\\\\ \begin{array}{llll} f(\theta)=&\pm 2cos&\left(\frac{1}{2}\theta \right)\\ &\ \uparrow &\ \uparrow \\ &A&B \end{array}[/tex]
[tex]\bf \begin{array}{llll} % right side info \bullet \textit{ stretches or shrinks}\\ \quad \textit{horizontally by amplitude } |{{ A}}|\\\\ \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}\\ \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]
now.. hmmm let's see, keeping in mind the template above
so, amplitude of 2, A=2, thus |A| = ±2, so, you can use either
and period of 4π, well, that simply means that
[tex]\bf \cfrac{2\pi }{B}=4\pi \implies \cfrac{2\pi }{4\pi }=B\implies \cfrac{1}{2}=B\\\\ -----------------------------\\\\ \begin{array}{llll} f(\theta)=&\pm 2cos&\left(\frac{1}{2}\theta \right)\\ &\ \uparrow &\ \uparrow \\ &A&B \end{array}[/tex]
