Difference between $\frac {2π} {b}$ versus $2πb$

Question

I'm having trouble finding period involving this equation: $y = a \sin (\frac {x-h} {b}) + k$. Sometimes my teacher would use $ \frac {2π} b$ in order to solve for period and sometimes my teacher would use $2πb$ to solve for $b$. What is the difference between the two equations, and how do I know which one is the appropriate equation to use in a problem.

Answer 1

consider this reasoning $\dots$

\begin{align} 0&\le\frac{x-h}{b}\le 2\pi\\ 0&\le x-h\le 2\pi b\\ h&\le x\le 2\pi b+h\\ \end{align}

Were $\frac{x-h}{b}$ to vary by $2\pi$, how much would $x$ vary by, assuming $b\gt 0$?

Answer 2

$y = a \sin (\frac {x-h} {b}) + k=a \sin (\frac {x}{b}-\frac{h} {b}) + k$. To find the period, you use $2\pi$ to divide whatever the coefficient of $x$ is. In this case, the period is $\frac{2\pi}{1/b}=2b\pi$, assuming $b>0$.

One point to add: if the coefficient is negative, one has to divide the absoulte value of it to get the period.