trigonometry help

It is really good to practice these.  

 

I'm getting in a mess trying to explain it, maybe my method is a bit 'muffled'

 

I will think about the cos curve first    The parent cos curve can be presented as   

y= amplitude * cos[(2pi/wavlength)(x-phaseshift)+vertical shift  = 1cos [(2pi/2pi)(x+0)+0 = cosx

 

But your curve has amplitude = 3 and it is upside down so that is -3

the wavelength is pi

there is no phaseshift (not if you put the - in front of the 3)

And it has a vertical shift of -1

so

\(y=-3cos[(\frac{2\pi}{\pi})(x-0)]-1\\ y=-3cos(2x)-1\\\)

or

\(y=3cos[\frac{2\pi}{\pi}(x-\frac{\pi}{2})]-1\\ y=3cos(2x-\pi)-1\\ \)

 

For the sin curve the wavelength and vertical shift stays the same.  The amplitude also stays the same I will look at only the +3 version

To find the phase shift I want the middle of the wave to sit on the y axis s, at present the closest (to the y axis) middle of the graph with a positive gradient, sits at (pi/4,-1)  I need to move the graph back pi/4

The phase shift is pi/4

So

\(y=3sin[\frac{2\pi}{\pi}(x-\frac{\pi}{4})]-1\\ y=3sin(2x-\frac{\pi}{2})-1\\\)

 

I have an interactive graph so you can see that all 3 of these are in fact the same wave.

 

https://www.desmos.com/calculator/7uupa1klvx