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Mathematics

1 answer:

-Dominant- [34]3 years ago

4 0

Answer:

45inches

Step-by-step explanation:

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$\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}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}
\end{array}$

$\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}$

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

$\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}$

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