i wonder how this curve will look:
y1:y=sin(x) that i know how it look
y2:y=(3sin(4x/5))/2 that i have a problem with. i know that the curve will go up to 1.5.
$\displaystyle \displaystyle \[f(x) = \frac{{3\sin \left( {\frac{{4x}}{5}} \right)}}{2}\]$
I believe period is what you're basically looking for
$\displaystyle \displaystyle \[T = \frac{{2\pi }}{{\left( {\frac{4}{5}} \right)}} = \frac{{5\pi }}{2}\]$
You have to analyze your function within one period, because after that it's the same.
The difference will be like
P.S. Another way of imagining that would be to realize that point $\displaystyle \displaystyle \[O\left( {0;0} \right)\]$ stays in its place whilst modified function is $\displaystyle \displaystyle \[\frac{3}{2}\]$ times stretched along Y-axis and $\displaystyle \displaystyle \[\frac{5}{4}\]$ times stretched along X-axis...
Well, basically, if you have:
$\displaystyle y = a\sin(bx - c) + d$
a is the amplitude, or how high can the trig function go up to and how low it can get to. A negative amplitude is a trig function reflected through the line y = d (where d is explained later)
b is the period of the function, where there are b cycles in 360 degrees or 2pi rad.
c is the path difference, and indicates how much more to the left or to the right the function has been translated. (-ve value of c is translation to the left)
d is the 'middle line', where the trig function 'rests'. When d = 0, the 'middle line' is y = 0, then d = 1, the 'middle line' is y = 1, etc.
I hope it helps you a little with understand trig ratios. Same goes for cos and tan