Why is the phase shift for trig functions C/B instead of just C?

[mewmew]

Ok, this is going to sound really dumb but in my school they never went over why things are the way they are but basically just had us memorize stuff, so I have a super basic question on phase shift for trig functions. I don't really logicaly understand why in the equation y= Sin(Bx+C), for example, the phase shift is C/B. To me it seems as though the phase shift should just be C, as you are adding C to every value of x, regardless of what happens to x. I understand that you can get it from taking Bx+c=0 to find where it "starts", which would give you x= -C/B but still can't really convince myself that it makes logical sence, I am sure someone can easily help me understand this more though.

Also, does anyone know any good trig. books or websites that go over "why" instead of just stating what everything is? Thanks alot

 

[Hurkyl]

To me it seems as though the phase shift should just be C, as you are adding C to every value of x, regardless of what happens to x.

Well, you're not adding C to every value of x; you're adding C to every value of Bx.

 

[JonF]

Think about how sine works in the unit circle

ok let us start with just looking at sin(x+c)

If we take a look at sin from 0 to pi at every pi/4 we get
Sin(0) = 0
sin(pi/4) = 7.07
sin(pi/2) = 1
sin(3pi/4) = . 707
sin(pi) = 0

now look what happens if we add +pi/4 to all of those
Sin(0+pi/4) = .707
sin(pi/4+pi/4) = 1
sin(pi/2+pi/4) = .707
sin(3pi/4+pi/4) = 1
sin(pi+pi/4) = -.707


Ok now let’s just look at sin(bx)
Sin(0) = 0
sin(pi/4) = 7.07
sin(pi/2) = 1
sin(3pi/4) = . 707
sin(pi) = 0

now look what happens if we, let’s say times each of these by 1/3
sin(0*1/3) = 0
sin(pi/4*1/3) = .2588
sin(pi/2*1/3) = .5
sin(3pi/4*1/3) = . 707
sin(pi*1/3) = .866

Notice what happens. The sin values increased much slower but they will still eventually reach all the same values. This basically stretches the sin curve in Cartesian coordinates to the left and right. If our b>1 then it would be stretched up and down instead. Also notice these values will repeat every 2pi/b. So in our case 6pi. This is known as the period.


So what does this have to do with phase shifts? This is how I think of the curve. At what smallest value of x do we does our curve = 0. In other words when does sin(bx+c) = 0 for the first x.

Ok now for your standard easy sin wave [sin(x)] that place is at sin(0). But we have moved the sin curve to the left or right and stretched it. So when you solve for your shift you have to take into account how squished your curve is because that determines the period. If it has the period isn’t 2pi then first place sin(x+c)=0 where c does not equal 0 won’t be 0 any more. Picture the waves we talked about previously. Sin(3x) was stretched to about 3 times the size of sin(x). So just shifting sin back +c won’t take into account how stretched that curve is!

 

[robphy]

Ok, this is going to sound really dumb but in my school they never went over why things are the way they are but basically just had us memorize stuff, so I have a super basic question on phase shift for trig functions. I don't really logicaly understand why in the equation y= Sin(Bx+C), for example, the phase shift is C/B. To me it seems as though the phase shift should just be C, as you are adding C to every value of x, regardless of what happens to x. I understand that you can get it from taking Bx+c=0 to find where it "starts", which would give you x= -C/B but still can't really convince myself that it makes logical sence, I am sure someone can easily help me understand this more though.

Also, does anyone know any good trig. books or websites that go over "why" instead of just stating what everything is? Thanks alot

The "phase shift" is, as you say, C. It has the units of phase: radians.
The B has units of radians/meter (assuming x is measured in meters).
C/B has units of meters (as you can verify from what you wrote) and can be interpreted as a shift in position (a translation).