- #1
karush
Gold Member
MHB
- 3,269
- 5
$\textsf{got ? on the Greek notation. if Period = T}$
\begin{align}
\displaystyle
Y_{tan}&=A\tan\left[\omega\left(x-\frac{\phi}{\omega} \right) \right]+B
\implies A\tan\left(\omega x-\phi \right)+B \\
T&=\left(\frac{\phi}{\omega}\right) \\
PS&=\phi
\end{align}
$\textsf{so on:}$
\begin{align}
\displaystyle
Y_{49}&=1+\frac{1}{2}\tan\left({2x-\frac{\pi}{4}}\right) \\
T&=\frac{\pi}{2} \\
PS&=\frac{\pi}{4}
\end{align} $\textsf{not sure on this one} $
\begin{align}
\displaystyle
Y_{tan}&=A\tan\left[\omega\left(x-\frac{\phi}{\omega} \right) \right]+B
\implies A\tan\left(\omega x-\phi \right)+B \\
T&=\left(\frac{\phi}{\omega}\right) \\
PS&=\phi
\end{align}
$\textsf{so on:}$
\begin{align}
\displaystyle
Y_{49}&=1+\frac{1}{2}\tan\left({2x-\frac{\pi}{4}}\right) \\
T&=\frac{\pi}{2} \\
PS&=\frac{\pi}{4}
\end{align} $\textsf{not sure on this one} $
Last edited: