## Trigonometry (11th Edition) Clone

The least possible value of $t$ is $0.0014$ seconds.
$$i=I_{max}\sin(2\pi ft)$$ For $i=\frac{1}{2}I_{max},f=60$, we have $$\frac{1}{2}I_{max}=I_{max}\sin(2\pi\times60t)$$ (since $i=\frac{1}{2}I_{max}$, we can replace $i$ with $\frac{1}{2}I_{max}$) $$\frac{1}{2}I_{max}=I_{max}\sin(120\pi t)$$ Because $I_{max}$ is the maximum current, $I_{max}\ne0$. And as $i=I_{max}$, $i\ne0$. So now we can divide both sides of the equation by $I_{max}$: $$\sin(120\pi t)=\frac{1}{2}$$ $t$ refers to time, so as a rule, $t\in[0,\infty)$ Also $t$ is minimum when $120\pi t$ is minimum, as $120\pi$ is a constant. Therefore, we would be able to find the least of $t$ as we find the first value of $120\pi t$ that would have $\sin(120\pi t)=\frac{1}{2}$ over the interval $[0,2\pi)$. Such a value of $120\pi t$ can be found using the inverse function for sine: $$120\pi t=\sin^{-1}\frac{1}{2}$$ $$120\pi t=\frac{\pi}{6}$$ $$t=\frac{\pi}{6\times120\pi}$$ $$t=\frac{1}{720}\approx0.0014(seconds)$$ Therefore, the least possible value of $t$ satisfying the given data is $0.0014$ seconds.