Melody

We answered this very recenty. Go look for it.

It has already got an answer.  If you don't like it then make a post on the original thread saying why.

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.

CPhill has continued this here:       (Thanks Chris)

For guests, you can just message via a post like this one.

One of us will probably see it.

Messaging privately is a nicer, more sure way of doing it though

Either way, don't forget that 'please' goes a long way though.

If are positive and one neg then you have   

1+1-1-1=0

If 2 are neg and one is pos you have

-1-1+1+1=0

I expect you are over thinking it.

if a+b+c=0 and none are 0 (and all are reak)  then one must be neg, one pos and the other either one.  

So when you multiply them together the answer could be positive or negative

a/|a|= plus or minus 1    etc

so maybe you are looking for all the possible combinations of

\(\pm 1\pm1\pm1\pm1\)

Or maybe you can discount some of those answers off.    Think about it a bit.

Anyway, I know someone else is also answering this right now.

It is fixed, thanks Admin  

It is happening to me too.  I have sent a message to admin. Hopefully it will be fixed soon.

Why is it that you call yourself 'thanks for the help' but you have never thanked anyone for the help?

Also, although you have only posted 4 question posts you have asked many more questions than that.

Please confine yourself to one question per post.  (like you have here)