why $\sin(x)$ is maximized if $x = 2n\pi+ \pi/2$

Question

I was working on a homework assignment and a friend told me that $7.06\sin(30x)$ will be maximum when $30x = 2n\pi + \pi/2$, can someone explain?

Answer 1

Because for $f(x)=7.06 \sin(30x)$, the maximum is when $f'(x)=0$, where:

$$f'(x)=7.06\cdot 30 \cos(30x)\tag{1}$$

Set $(1)$ to zero, and you get: $$\cos(30x)=0\implies 30x=\frac\pi2$$ And $\cos(2n\pi+x)=\cos x$, meaning the complete solution should be:

$$30x=2n\pi+\frac\pi2\,\, \forall n \in \mathbb{Z}$$

Answer 2

The values of $\sin \theta$ are always between $-1$ and $1$, with $1$ occurring whenever $\theta$ is $90^{\circ}$ or any other angle coterminal with $90^\circ$ (for example, $450^\circ$ or $810^\circ$ or $-270^\circ$ or any other value you can get by adding or subtracting a multiple of $360^\circ$). We can summarize this by saying that the maximum value of $\sin \theta$ occurs when $$\theta = 90^\circ \pm 360n^\circ$$

If we translate this into radians, the maximum occurs when $$\theta = \frac{\pi}{2} \pm 2\pi n$$ Now you asked about the function $f(x)=7.06\sin(30x)$. This function has values that are always between $-7.06$ and $7.06$, with the maximum occurring precisely when $\sin(30x) = 1$. But this is exactly the same question just solved, with $\theta = 30x$. So we need $$30x = \frac{\pi}{2} \pm 2\pi n$$ which is what your friend told you.