- #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: