Science:Math Exam Resources/Courses/MATH102/December 2017/Question 15 (a)/Solution 1

By the Hint, using the periodicity of the sin function, we have

$$\sin \left({\dfrac {5}{3}}t+2\pi \right)=\sin \left({\dfrac {5}{3}}t\right).$$

Factoring ${\textstyle {\dfrac {5}{3}}}$, we have

$$\sin \left({\dfrac {5}{3}}\left(t+{\dfrac {6\pi }{5}}\right)\right)=\sin \left({\dfrac {5}{3}}t\right).$$

This implies that the period of $\sin \left({\dfrac {5}{3}}t\right)$ is ${\dfrac {6\pi }{5}}$. Since a constant multiple of a function (a vertical dilation of a graph) and a translation don't change the period of a function, the period of $h$ is same with that of $\sin \left({\dfrac {5}{3}}t\right)$.

More precisely, let us multiply by ${\textstyle 3}$ and add ${\textstyle 3}$ to both sides:

$$3+3\sin \left({\dfrac {5}{3}}\left(t+{\dfrac {6\pi }{5}}\right)\right)=3+3\sin \left({\dfrac {5}{3}}t\right).$$

This means that

$$h\left(t+{\dfrac {6\pi }{5}}\right)=h(t).$$

So the period of ${\textstyle h}$ is ${\textstyle {\dfrac {6\pi }{5}}}$.

Answer: the period is ${\textstyle {\color {blue}{\dfrac {6\pi }{5}}}}$.