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lianna [129]
3 years ago
13

For the function given state the period f(t) =6sin(3t-pi/6)-1

Mathematics
1 answer:
lapo4ka [179]3 years ago
8 0
\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}

\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 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}


now, with that template in mind, let's take a peek at yours

\bf \begin{array}{lllcclll}
f(t)=&6sin(&3t&-\frac{\pi }{6})&-1\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
period\qquad \cfrac{2\pi }{B}\iff\cfrac{2\pi }{3}
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(tan²(<em>θ</em>) cos²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>))

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so cos²(<em>θ</em>) cancels with the cos²(<em>θ</em>) in the tan²(<em>θ</em>) term:

(sin²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>))

Recall the double angle identity for cosine,

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Recall the Pythagorean identity,

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sin²(<em>θ</em>) - 1 = -cos²(<em>θ</em>):

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Cancel the cos²(<em>θ</em>) terms to end up with

(tan²(<em>θ</em>) cos²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>)) = -1/2

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