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# trigonometry help

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write the equation for the graph shown in the form y = asinb(x-c)+d and in the form y = acosb(x-c)+d

I will just add the pic (Melody) Jun 3, 2021
edited by Melody  Jun 3, 2021

#1
+2

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

Jun 4, 2021

#1
+2

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

Melody Jun 4, 2021