How do you graph sine and cosine functions when it is translated?

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How do you graph sine and cosine functions when it is translated?

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Answer 1 · 2015-03-22T14:41:12

I think you'll find a useful answer here: http://socratic.org/trigonometry/graphing-trigonometric-functions/translating-sine-and-cosine-functions

Vertical translation

Graphing $y = sin x + k$ $y=sinx+k$ Which is the same as $y = k + sin x$ $y=k+sinx$ :

In this case we start with a number (or angle) $x$ $x$ . We find the sine of $x$ $x$ , which will be a number between $- 1$ $-1$ and $1$ $1$ . The after that, we get $y$ $y$ by adding $k$ $k$ . (Remember that $k$ $k$ could be negative.)

This gives us a final $y$ $y$ value betwee $- 1 + k$ $-1+k$ and $1 + k$ $1+k$ .

This will translate the graph up if $k$ $k$ is positive ( $k > 0$ $k>0$ )
or down if $k$ $k$ is negative ( $k < 0$ $k<0$ )

Examples:

$y = sin x$ $y=sinx$
graph{y=sinx [-5.578, 5.52, -1.46, 4.09]}

$y = sin x + 2 = 2 + sin x$ $y=sinx+2 = 2+sinx$
graph{y=sinx+2 [-5.578, 5.52, -1.46, 4.09]}

$y = sin x - 4 = - 4 + sin x$ $y=sinx-4=-4+sinx$
graph{y=sinx-4 [-5.58, 5.52, -5.17, 0.38]}

The reasoning is the same for $y = cos x + k = k + cos x$ $y=cosx+k=k+cosx$ , but the starting graph looks different, so the final graph is also different:

$y = cos x$ $y=cosx$
graph{y=cosx [-5.578, 5.52, -1.46, 4.09]}

$y = cos x + 2 = 2 + cos x$ $y=cosx+2 = 2+cosx$
graph{y=cosx+2 [-5.578, 5.52, -1.46, 4.09]}

Answer 2 · 2015-03-22T15:20:12

Horizontal Translation

One way to think about horizontal translations of a function is to think about the value of $x$ $x$ that will cause us to find $f (0)$ $f(0)$ .

We know the graph of $y = f (x) = sin (x)$ $y=f(x)=sin(x)$ .

To graph $y = sin (x - 4)$ $y=sin(x-4)$ , we can think of it as graphing $y = f (x - 4)$ $y=f(x-4)$ .
Now, what value of $x$ $x$ will make me find $f (0) = sin (0)$ $f(0)=sin(0)$ ? Clearly, it is $x = 4$ $x=4$ .
So " $4$ $4$ is the new $0$ $0$ ". Everything moves $4$ $4$ to the right.

graph{y=sin(x-4) [-0.498, 7.295, -2.302, 1.596]}

To graph $y = sin (x + \frac{π}{3})$ $y=sin(x+ pi/3)$ , we ask ourselves, "What value of $x$ $x$ will cause us to find $sin (0)$ $sin(0)$ ?
That will be $x = - \frac{π}{3}$ $x=- pi/3$
So, " $- \frac{π}{3}$ $- pi/3$ is the new $0$ $0$ ". Everything moves $- \frac{π}{3}$ $- pi/3$ to the right. Wait a minute, surly it's more clear to say: Everything moves $\frac{π}{3}$ $pi/3$ to the left.

(For the graph below, remember that $\frac{π}{3}$ $pi/3$ is a little more than $1$ $1$ )

graph{y=sin(x+pi/3) [-3.02, 1.845, -1.192, 1.241]}

To start the graph of $y = sin (b x - c)$ $y=sin(bx- c)$ , we'll need to change the period and also translate.

Cosine

The reasoning for graphing $y = cos (x - h)$ $y=cos(x-h)$ is the same. The difference is that
$sin (0) = 0$ $sin(0)=0$ , so the point corresponding to the 'new $0$ $0$ ' goes on the $x$ $x$ -axis
$cos 0 = 1$ $cos0=1$ , so the point corresponding to the 'new $0$ $0$ ' goes at $y = 1$ $y=1$

$y = cos (x + \frac{π}{3})$ $y = cos (x+ pi/3)$
graph{y=cos(x+pi/3) [-3.02, 1.845, -1.192, 1.241]}

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How do you graph sine and cosine functions when it is translated?

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