Dilating sinusoidal period at certain abscissa $x$?

Question

Lets take a look at this function :

$$f(x) = \sin\left(\frac{\pi x }{2}\right)$$

when $x$ tends to $1$ this functions get closer to $1$ by bigger values

now look at this one :

$$\sin\left(\frac{\pi(x+2)}{2\cdot3}\right)$$

this function is again nearer to $1$ at same abscissa $x$ !!

is there any relation which by we can retrieve more effective functions that brings us to 1 again with closer ordinates ??

Answer 1

i thought of deleting my entire post but thinking of the ability of learning something from this i v decided to answer my own question since i found a convenient answer by myself ...

$$sin({\Pi a}+b ) = 1 \Rightarrow \Pi a + b= \frac{2k+1}{2}\Pi \Rightarrow \frac{2k+1}{2}\Pi - a\Pi= b \Rightarrow f(x)=sin(( a (x-1)+ \frac{2k+1}{2}) \Pi) $$

the value of k which ensure higher summit is even

the values of a that ensure us larger periods are lower ....... so f becomes

$$ f(x)=sin(( \frac{x-1}{a}+ \frac{1}{2}) \Pi) \ \ \ with\ \ a\in \mathbb{R} \\ infinitely\ bigger $$

you can verify this relation by tracing curves according to bigger unsigned values of a

this matlab script illustates the preceding :

for i=1 : 1000


y=sin((1/i*(x-1)+1/2)*pi);

a=plot(x, y(:));

set(gca, 'XLim',[-1 3], 'YLim',[-1 1]);

grid;

pause(0.1);

delete(a);

end

thanks for your attention !