Lissajous curve - highest velocity

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We have the following parametric representation:

$$x(t) = 2\cos t - \cos(2t)$$

$$y(t) = 2\sin t - \sin (2t) $$

A point P is described by the above equations. Voor $t=0$ and $t=2\pi$ P is in (1,0). We want the highest velocity of point P.

So we can find the velocity with $$ v(t) = \sqrt{ (x'(t))^2 + (y'(t))^2} $$

In the correction sheet, they find the derivatives and put them in the root, and eventually, after some algebra, end up with $$ v = \sqrt{8 - 8 \cos t}$$

Instead of taking its derivative and setting it equal to 0, they fill in $t=-1$ and get $v=4$. In my book it also says that you can use mathematical reasoning (so without calculations) to fill in -1 or 1 for $\sin t$ or $\cos t$.

But why can we assume the maximum speed is always attained at $t = \arcsin (1 / -1 )$ or $t = \arccos (1 / -1)$?

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since $v'=\dfrac{asin(t)}{...}=0$ means that $sin(t)=0$ which means $cos(t)=1 \ or -1$