## Algebra and Trigonometry 10th Edition

If $c$ is the period of $f(t)$, then the period of $f(\frac{1}{2}t)$ is $2c$ $f(\frac{1}{2}[t+4c])=f(\frac{1}{2}t)$, because $4c$ is a multiple of $2c$, that is, $4c=2(2c)$