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Phase shift [Solved!]

My question

Hello, I have a question about phase shift. What makes the second approach incorrect?

It makes sense that the shift is -c/b when considering it as when the expression inside the sin/cos function equals 0 (as if the function is just starting off).

However, when plugging in 0 for x, it seems like the graph would be shifted c to the left (since the value inside the paranthesis is acting as though x=c), but this isn't the case when b isn't 1

Thanks a lot!

Relevant page

3. Graphs of y = a sin(bx + c) and y = a cos(bx + c)

What I've done so far

Drawn several graphs, but couldn't conclude anything

X

Hello, I have a question about phase shift. What makes the second approach incorrect?

It makes sense that the shift is -c/b when considering it as when the expression inside the sin/cos function equals 0 (as if the function is just starting off).

However, when plugging in 0 for x, it seems like the graph would be shifted c to the left (since the value inside the paranthesis is acting as though x=c), but this isn't the case when b isn't 1

Thanks a lot!
Relevant page

<a href="/trigonometric-graphs/3-graphs-sin-cos-phase-shift.php">3. Graphs of <span class="noWrap">y = a sin(bx + c)</span> and <span class="noWrap">y = a cos(bx + c)</span></a>

What I've done so far

Drawn several graphs, but couldn't conclude anything

Re: Phase shift

Hello Andrew

I think the problem here is with the concept of "starting off". Let's use "t" as the variable and talk about when things occur.

Let's also just talk about sin to keep it simple.

If we set the expression inside the sin (that is, bt + c) to 0, we are saying "at what time does sin have value 0?", since sin 0 = 0. Then by solving, we get t = -c/b.

The effect here is to shift the graph to the left by the amount c/b. (That is, the function has value 0, or it is "starting off [from value 0]" at t = -c/b.)

Now let's think about y = a sin(bt+c) when the TIME is 0 ("starting off when we flip the switch on"). Now we have y = a sin c, which is some value on the y-axis, and that will be the starting value of y. I hope you can see that it doesn't matter what b is in such a case, since it is no longer in the expression y = a sin c.

Does that help?

I have added a new example on 3. Graphs of y = a sin(bx + c) and y = a cos(bx + c) - I hope that makes it clearer.

Regards

X

Hello Andrew

I think the problem here is with the concept of "starting off". Let's use "t" as the variable and talk about when things occur.

Let's also just talk about sin to keep it simple.

If we set the expression inside the sin (that is, bt + c) to 0, we are saying "at what time does sin have value 0?", since sin 0 = 0. Then by solving, we get t = -c/b.

The effect here is to shift the graph to the left by the amount c/b. (That is, the function has value 0, or it is "starting off [from value 0]" at t = -c/b.)

Now let's think about y = a sin(bt+c) when the TIME is 0 ("starting off when we flip the switch on"). Now we have y = a sin c, which is some value on the y-axis, and that will be the starting value of y. I hope you can see that it doesn't matter what b is in such a case, since it is no longer in the expression y = a sin c.

Does that help?

I have added a new example on <a href="/trigonometric-graphs/3-graphs-sin-cos-phase-shift.php">3. Graphs of <span class="noWrap">y = a sin(bx + c)</span> and <span class="noWrap">y = a cos(bx + c)</span></a> - I hope that makes it clearer.

Regards

Re: Phase shift

Your new example helped a lot.

Thanks.

X

Your new example helped a lot.

Thanks.

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