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# Translating Sine and Cosine Functions

## Graph shifted trig functions

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Translating Sine and Cosine Functions

Your best friend asks you to describe how the graph of \begin{align*}y= \sin (x-\pi)-2\end{align*} is different from the graph of \begin{align*}y= \sin x\end{align*}. What do you tell your friend?

### Translating Sine and Cosine Functions

Just like other functions, sine and cosine curves can be moved to the left, right, up and down. The general equation for a sine and cosine curve is \begin{align*}y=a \sin (x-h)+k\end{align*} and \begin{align*}y=a \cos(x-h)+k\end{align*}, respectively. Also, just like in other functions, \begin{align*}h\end{align*} is the horizontal shift, also called a phase shift, and \begin{align*}k\end{align*} is the vertical shift. Notice, that because it is \begin{align*}x - h\end{align*} in the equation, \begin{align*}h\end{align*} will always shift in the opposite direction of what is in the equation.

Let's graph \begin{align*}y=\cos \left(x - \frac{\pi}{4}\right)\end{align*}.

This function will be shifted \begin{align*}\frac{\pi}{4}\end{align*} units to the right. The easiest way to sketch the curve, is to start with the parent graph and then move it to the right the correct number of units.

Now, let's graph \begin{align*}y=\sin (x+2)+3\end{align*}.

Because -2 is not written in terms of \begin{align*}\pi\end{align*} (like the \begin{align*}x\end{align*}-axis), we need to estimate where it would be on the axis. \begin{align*}- \frac{3 \pi}{4}=2.35 \ldots\end{align*} So, -2 will be shifted not quite to the \begin{align*}- \frac{3 \pi}{4}\end{align*} tic mark. Then, the entire function will be shifted up 3 units. The red graph is the final answer.

Finally, let's find the equation of the sine curve below.

First, we know the amplitude is 1 because the average between 2 and 0 (the maximum and minimum) is 1. Next, we can find the vertical shift. Recall that the maximum is usually 1, in this equation it is 2. That means that the function is shifted up 1 unit \begin{align*}(2-1)\end{align*}. The horizontal shift is the hardest to find. Because sine curves are periodic, the horizontal shift can either be positive or negative.

Because \begin{align*}\pi\end{align*} is \begin{align*}3.14 \ldots\end{align*}, we can say that “moves back almost \begin{align*}\pi\end{align*} units” is -3 units. So, the equation is \begin{align*}y=\sin (x+3)+1\end{align*}. If we did the positive horizontal shift, we could say that the equation would be \begin{align*}y=\sin (x-3.28)+1\end{align*}.

To determine the value of the horizontal shift, you might have to estimate. For example, we estimated that the negative shift was -3 because the maximum value of the parent graph is at \begin{align*}x=\frac{\pi}{2}\end{align*} and the maximum to the left of it didn’t quite make it to \begin{align*}x=- \frac{\pi}{2}\end{align*} (the distance between \begin{align*}\frac{\pi}{2}\end{align*} and \begin{align*}-\frac{\pi}{2}\end{align*} is \begin{align*}\pi\end{align*}). Then, to determine the positive shift equation, recall that a period is \begin{align*}2 \pi\end{align*}, which is \begin{align*}6.28 \ldots\end{align*} So, the positive shift would be \begin{align*}2 \pi - 3\end{align*} or \begin{align*}6.28 - 3 = 3.28.\end{align*}.

### Examples

#### Example 1

Earlier, you were asked to tell your friend how the graph of \begin{align*}y= \sin (x-\pi)-2\end{align*} is different from the graph of \begin{align*}y = \sin x\end{align*}

If you compare \begin{align*}y= \sin (x-\pi)-2\end{align*} to the general equation \begin{align*}y=a \sin(x-h)+k\end{align*}, you see that \begin{align*}h=\pi\end{align*} and \begin{align*}k=-2\end{align*}.

\begin{align*}h\end{align*} is the horizontal shift, so the function is \begin{align*}\pi\end{align*} units to the right of \begin{align*}y= \sin x\end{align*}.

\begin{align*}k\end{align*} is the vertical shift, so the function is 2 units down from \begin{align*}y= \sin x\end{align*}.

Therefore, the graph of \begin{align*}y= \sin (x-\pi)-2\end{align*} is shifted \begin{align*}\pi\end{align*} units to the right and two units down from the graph of \begin{align*}y= \sin x\end{align*}.

In Examples 2 & 3, graph the function from \begin{align*}[\pi, 3 \pi]\end{align*}.

#### Example 2

\begin{align*}y=-1+\sin x\end{align*}

Shift the parent graph down one unit.

#### Example 3

\begin{align*}y=\cos\left(x+\frac{\pi}{3}\right) - 2\end{align*}

Shift the parent graph to the left \begin{align*}\frac{\pi}{3}\end{align*} units and down 2 units.

#### Example 4

Find the equation of the cosine curve below.

The parent graph is in green. It moves up 3 units and to the right \begin{align*}\frac{3 \pi}{4}\end{align*} units. Therefore, the equation is \begin{align*}y=\cos \left(x-\frac{3 \pi}{4}\right)+3\end{align*}.

If you moved the cosine curve backward, then the equation would be \begin{align*}y=\cos \left(x+\frac{5 \pi}{4}\right)+3\end{align*}.

### Review

For questions 1-4, match the equation with its graph.

1. \begin{align*}y=\sin \left(x-\frac{\pi}{2}\right)\end{align*}
2. \begin{align*}y=\cos \left(x-\frac{\pi}{4}\right)+3\end{align*}
3. \begin{align*}y=\cos \left(x+\frac{\pi}{4}\right)-2\end{align*}
4. \begin{align*}y=\sin \left(x-\frac{\pi}{4}\right)+2\end{align*}

Which graph above also represents these equations?

1. \begin{align*}y=\cos(x-\pi)\end{align*}
2. \begin{align*}y=\sin \left(x+\frac{3 \pi}{4}\right)-2\end{align*}
3. Write another sine equation for graph A.
4. Writing How many sine (or cosine) equations can be generated for one curve? Why?
5. Fill in the blanks below.
1. \begin{align*}\sin x=\cos(x-\underline{\;\;\;\;\;})\end{align*}
2. \begin{align*}\cos x=\sin(x-\underline{\;\;\;\;\;})\end{align*}

For questions 10-15, graph the following equations from \begin{align*}[-2\pi, 2\pi]\end{align*}.

1. \begin{align*}y=\sin \left(x+\frac{\pi}{4}\right)\end{align*}
2. \begin{align*}y=1+\cos x\end{align*}
3. \begin{align*}y=\cos(x+\pi)-2\end{align*}
4. \begin{align*}y=\sin(x+3)-4\end{align*}
5. \begin{align*}y=\sin \left(x-\frac{\pi}{6}\right)\end{align*}
6. \begin{align*}y=\cos(x-1)-3\end{align*}
7. Critical Thinking Is there a difference between \begin{align*}y=\sin x +1\end{align*} and \begin{align*}y=\sin(x+1)\end{align*}? Explain your answer.

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 14.2.

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