Anne Wacker

2021-12-28

Determining the period of $\frac{\mathrm{sin}2x}{\mathrm{cos}3x}$
I would like to compute the period of this function which is a fraction of two trigonometric functions.
$\frac{\mathrm{sin}2x}{\mathrm{cos}3x}$
Is there a theorem for this? what trick to use to easily find the period? I started by reducing the fraction but Im

David Clayton

Suppose the period is p, and suppose the domain of the function is suitably defined, then for all values of x in the domain, we must have
$\frac{\mathrm{sin}2x}{\mathrm{cos}3x}=\frac{\mathrm{sin}\left(2\left(x+p\right)\right)}{\mathrm{cos}\left(3\left(x+p\right)\right)}$
$⇒\mathrm{sin}2x\mathrm{cos}\left(3x+3p\right)=\mathrm{cos}3x\mathrm{sin}\left(2x+2p\right)$
If we set x=0, we have

On the other hand, noting that $\mathrm{sin}2x=\mathrm{cos}3x$ when $x=\frac{\pi }{10}$, amongst other possible values, if we set $x=\frac{\pi }{10}$, we get
$\mathrm{cos}\left(\frac{3\pi }{10}+3p\right)=\mathrm{sin}\left(\frac{\pi }{5}+2p\right)=\mathrm{cos}\left(\frac{3\pi }{10}-2p\right)$

From this we get $p=n\cdot 2\pi$ or $p=\frac{3\pi }{5}+n\cdot 2\pi$
In order to satisfy this and the previous result for p we have to choose k and n such that both equations are satisfied and p has minimum value, so we choose k=4 and n=1.
Therefore the period is $2\pi$

MoxboasteBots5h

If F(x) has a period T Then F(ax+b) has a period |(T/a)| If the expression is separated by + or - signs take the LCM of periods One of important concepts for determining period of function involving multiple periodic terms is that individual function should repeat simultaneously. This gives rises to LCM rule. It is defined for terms which appear as sum or difference in the function. However, LCM concept can be extended to division or multiplication of periodic terms as well.
Exceptions to LCM rule are important. Even function and function comprising of cofunctions are two notable exceptions to LCM rule. Besides, LCM of irrational periods of different kinds is not possible. This fact is used to determine periodic nature of function involving irrational periods. If LCM can not be determined, then given function is not periodic in the first place.

nick1337

$\frac{\mathrm{sin}2x}{\mathrm{cos}3x}=\frac{2\mathrm{sin}x\mathrm{cos}x}{4{\mathrm{cos}}^{3}x-3\mathrm{cos}x}=\frac{2\mathrm{sin}x}{4{\mathrm{cos}}^{2}x-3}$
has a period of $2\pi$. Therefore $\frac{2\mathrm{sin}\left(x\right)}{4{\mathrm{cos}}^{2}\left(x\right)-3}$ has a period of $2\pi$
The fundamental period must be $2\pi n$ where $n\in \mathbb{N}$. We have
$\frac{2\mathrm{sin}x}{4{\mathrm{cos}}^{2}x-3}=\frac{2\mathrm{sin}\left(x+\frac{2\pi }{n}\right)}{4{\mathrm{cos}}^{2}\left(x+\frac{2\pi }{n}\right)-3}$
Substituting $x=\frac{\left(n-2\right)\pi }{n}$ we get
$\frac{2\mathrm{sin}\left(\frac{\left(n-2\right)\pi }{n}\right)}{4{\mathrm{cos}}^{2}\left(\frac{\left(n-2\right)\pi }{n}\right)-3}=\frac{2\mathrm{sin}\left(\pi \right)}{4{\mathrm{cos}}^{2}\left(\pi \right)-3}=0$
$⇒\mathrm{sin}\left(\frac{\left(n-2\right)\pi }{n}\right)=0$
This is true only if n=1 or 2. If n=2 then
$\frac{2\mathrm{sin}x}{4{\mathrm{cos}}^{2}x-3}=\frac{2\mathrm{sin}\left(x+\pi \right)}{4{\mathrm{cos}}^{2}\left(x+\pi \right)-3}=-\frac{2\mathrm{sin}x}{4{\mathrm{cos}}^{2}x-3}$
Therefore $n\ne 2$. So the fundamental period is $2\pi$

Do you have a similar question?