We learned how to transform Basic Parent Functions in the **Parent Functions and Transformations** section. Now we will transform the **six Trigonometric Functions**. Later we’ll be transforming the **Inverse Trig Functions** here. Some prefer to do all the transformations **with t**

**-charts**like we did earlier, and some prefer it

**without**

**t****-charts**(see

**here**for the sine and cosine transformations specifically, and

**here**for help with all the trig functions).

**Word of caution: In the trig transformation formulas below, I like to use formulas like **$ y=a\sin \left( {b\left( {x-c} \right)} \right)+d$, **with a phase shift of** $ c$**, instead of **$ y=a\sin \left( {bx-c} \right)+d$**, with a phase shift of **$ \displaystyle \frac{c}{b}$**. Sorry for the confusion, but the results will be the same!**

## T-Charts for the Six Trigonometric Functions

Here are the **trig parent function t-charts** I like to use; starting and stopping points may be changed, as long as they cover a cycle. Note that each covers

**one period**(one

**complete cycle**of the graph before it starts repeating itself) for each function. Also note that “undef” means the function is

**undefined**for that $ x$-value; there is a

**vertical asymptote**there. Note also that when the original functions (like

**sin**,

**cos**, and

**tan**) have

**0**’s as $ y$-values, their respective reciprocal functions are undefined at those points (because of division of

**0**).

## Sin and Cos Transformations

Below are the general formulas in order to **transform **a** sin **or **cos function**, as well as the remaining four trig functions. Note that sometimes you’ll see the formula arranged differently; for example, with “$ a$” being the vertical shift at the beginning.

### Drawing Transformed Graphs for Sin and Cos

Here are some examples of drawing transformed trig graphs, first with the **sin** function, and then the **cos** (the rest of the trig functions will be addressed later).** **You will probably be asked to **sketch one complete cycle** for each graph, label significant points, and list the **Domain**, **Range**, **Period** and **Amplitude** for each graph.

Notice that for the ** t**-charts, the $ x$-points turn into $ \displaystyle \frac{1}{b}x+c$ (opposite math, starting with multiplication/division), and the $ y$-points turn into $ ay+d$ (regular math).

**Transforming Without Using ***t*-charts (steps for all trig functions are here)

*t*-charts (steps for all trig functions are here)

Many teachers teach trig transformations without using *t*-charts; here is how you might do that for **sin** and **cosine**:

Since we can get the **new period** of the graph (how long it goes before repeating itself), by using $ \displaystyle \frac{2\pi }{b}$, and we know the phase shift, we can graph key points, and then draw the curve based on whether it is sin or cosine, and positive or negative. We can use **5 key points** for a whole period of a graph.

Using the example $ \displaystyle y=-2\cos \left( {\frac{1}{3}\left( {x-\frac{\pi }{2}} \right)} \right)+1$ again, graph without using a $ t$-chart:

- The graph is centered at $ y=1$, because of the vertical shift. From here we go
**up 2**and**down 2**(the amplitude) so the $ y$-part of the graph goes from $ -1$ to $ 3$. - To get new
**midline-intercepts**: parent function midline intercepts ($ x$-intercepts) are at $ \pi k$ for**sin**and $ \displaystyle \frac{\pi }{2}+\pi k$ for**cos**. Set the transformed trig argument to the parent function $ x$-intercepts, and solve for $ x$. In our example, since the trig function is**cos**, $ \displaystyle \frac{1}{3}\left( {x-\frac{\pi }{2}} \right)=\frac{\pi }{2}+\pi k;\,\,\,x-\frac{\pi }{2}=\frac{{3\pi }}{2}+3\pi k;$ $ \displaystyle \,\,x=2\pi +3\pi k$. Draw these midline-intercepts and the points in between by taking averages of the $ x$’s, using the new “$ y$” from the amplitude and vertical shift. If the trig function is**negative**, flip the graph vertically (make it upside down). - Other way without using midline-intercepts:
- The
**first (leftmost) point**is $ \displaystyle \frac{\pi }{2}$ to the**right**, and it is “upside down” for cosine (because of the**negative sign**before the**2**). This point is**2**units down from the middle of the graph. which is $ y=1$, so the point is $ \displaystyle \left( {\frac{\pi }{2},-1} \right)$. (Without the negative, the point would be $ \displaystyle \left( {\frac{\pi }{2},3} \right)$). - Since the
**period**is $ 6\pi $, the**middle point**is $ 3\pi $ to the right of $ \displaystyle x=\frac{\pi }{2}$, so the next (highest) point is at $ \displaystyle \left( {\frac{\pi }{2}+3\pi ,3} \right)=\left( {\frac{{7\pi }}{2},3} \right)$. The**last (rightmost) point**is again at the lowest $ y$-point at $ \displaystyle \left( \frac{7\pi }{2}+3\pi ,-1 \right)=\left( \frac{13\pi }{2},-1 \right)$. - The intermediate points are halfway in between each
**2**of the**3**points that we just found. To get the point between $ \displaystyle \left( \frac{\pi }{2},-1 \right)$ and $ \displaystyle \left( {\frac{{7\pi }}{2},3} \right)$, take the average of the $ x$-values and $ y$-values to get $ \left( 2\pi ,1 \right)$. We do the same for the point between $ \displaystyle \left( {\frac{{7\pi }}{2},3} \right)$ and $ \displaystyle \left( {\frac{{13\pi }}{2},-1} \right)$ to get $ \left( 5\pi ,1 \right)$. Now we have the**5**points and we can draw the graph.

- The
- Note that the new
**domain**is $ \left( {-\infty ,\infty } \right)$, the new**range**is $ \left[ {-1,3} \right]$.

You may also be asked to perform **absolute value transformations**. If the absolute value is on the outside, like $ y=\left| {\sin x} \right|$, reflect all the $ y$-values across the $ x$-axis, and for $ y=\sin \left| x \right|$, “erase” all the negative $ x$-values and reflect the positive $ y$-values across the $ y$-axis:

## Writing Equations from Transformed Graphs for Sin and Cos

You may be asked to write trig function equations, **given transformed graphs**. (Writing equations from trig functions other than **sin** and **cos** may be found **here**). Here are the steps to do this; examples will follow.

**Note that there are typically many “correct” answers for these problems!**

- Write out the
**generic transformed equation**for**sin**or**cos**: $ y=a\sin \left( {b\left( {x-c} \right)} \right)+d$ or $ y=a\cos \left( {b\left( {x-c} \right)} \right)+d$. To determine if it is a**sin**graph or a**cos**graph, see if the graph has a peak near the $ y$-axis (**cos**graph), or a middle (vertical shift) near the $ y$-axis (**sin**graph). Note that the graph may appear flipped across the $ x$-axis, but by shifting the graph to the left or right, you may not have to write the equation as a flipped function (with a negative). - Find the amplitude ($ |a|$) of the transformed function by subtracting the bottom $ y$-value from the top $ y$-value, and then dividing by
**2**. (Remember that for the**csc**,**sec**,**tan**, and**cot**graphs, this is just called a “stretch”, not an amplitude.) - To get $ d$, or the vertical shift of the function, add the amplitude to the bottom $ y$-value. This is the $ y$-value for the
**middle of the function**(the “average” of the bottom $ y$ and the top $ y$); we also can just observe this from the graph. - To get $ b$, first find the period of the graph: see how long it goes before repeating itself. Subtract the two $ x$-values to get this new period. For example, if the graph “starts” at $ \pi $ and then starts repeating itself again at $ 5\pi $ (a complete revolution), its period is $ 4\pi $. Once you get this period, $ \displaystyle b=\frac{2\pi }{\text{new period}}$. (Remember that for the
**tan**and**cot**graphs, $ \displaystyle b=\frac{\pi }{\text{new period}}$). - Find the
**phase shift**of the graph by seeing how close it is to the $ y$-axis ($ x=0$); this value is $ c$. You can write the graph as a**cos**graph or a**sin**graph; both will be correct, depending on what $ c$ is, but you typically want $ c$ to be the smallest it can be. Look at the $ x$-value of the $ y$ highest point for**cos**(since**cos**typically starts at $ y=1$), or**middle**point of graph for**sin**(since**sin**typically starts at $ y=0$), and compare to $ x=0$. Since the generic formula has $ “x-c”$ in it, if the graph is to the**right**, $ c$ is**positive**; if it is to the**left**, $ c$ is**negative**. - If the graph happens to be
**flipped**from what it should be, add a**negative sign**before $ a$. - Check your equation in the
**graphing calculator**by setting the $ x$-**window**exactly to the**leftmost point**of the graph (Xmin) and to the**rightmost point**of the graph (Xmax). Similarly, set the $ y$-**window**to the**lowest point**on the graph (Ymin) and to the**highest point**on the graph (Ymax). When you graph, you should see the exact graph for that problem.

Here are some examples; note that answers may vary:

## Sinusoidal Applications

Uh oh – more word problems! These aren’t too bad, once you get the hang of them. And now that you know how to **transform** **sin** and **cos** functions, that’s really all we’re doing here.

A **sinusoidal function**, or** sinusoid **is a fancy name for the **sin** (or **cos**) waves that we’ve been working with. Sinusoids are quite useful in **many scientific fields**; sine waves are everywhere! With sinusoidal applications, you’ll typically have to decide between using a **sin graph** or a **cos graph**. Sometimes it helps to remember that the **sin** graphs start in the middle of the graph, and the **cos** graphs start at the top of the graph. Also, sometimes, the graphs will be “upside down” which means you might need to reflect the **sin** or **cos** (using a negative coefficient).

Let’s just start with an example, and see the steps:

**Roller Coaster Problem:**

A part of the track of a roller coaster has the shape of a **sinusoidal function**. The **highest** and **lowest** points on the roller coaster are **150 feet apart horizontally** and **100 feet apart vertically**. The lowest point of the roller coaster was actually built **10 feet below the ground**.

Let $ y=$ the height of the track (with $ y=0$ as the ground), and $ x=$ the number of feet horizontally, with $ x=0$ at the highest point of the track.

**(a)** Write the **sinusoidal equation** of this part of the track of the roller coaster.

**(b)** How high is the highest point of the roller coaster? How high is the coaster at $ x=$ **15** feet? **100** feet?

**(c)** How long (horizontally) is the roller coaster when the track is **75** feet above the ground? **15** feet above the ground?

**Solution:**

**(a)** Get the equation with these steps:

- Graph the function’s
**high points**and**low points**, and figure out the coordinates of these points, given what we know. Since the track is**100**feet tall, and the lowest point is at $ y=-10$, we know the highest point is at $ (0,100-10)=(0,90)$ (since the highest point is when $ x=0$). Since the horizontal distance between the highest point and lowest point is**150**, the lowest point is at $ (150,-10)$. Here’s what we can graph: - Use a
**cos graph**, since the highest point of the graph is on the $ y$-axis: $ y=a\cos \left( {b\left( {x-c} \right)} \right)+d$. - To get the amplitude, or $ |a|$, subtract the lowest $ y$-point from the highest, and then divide by
**2**. Thus, $ a=90-\left( -10 \right)=100\div 2=50$. Now we have $ y=50\cos \left( {b\left( {x-c} \right)} \right)+d$. - To get the middle of the function, or the vertical shift ($ d$), add the amplitude to the lowest $ y$-value: $ -10+50=40$. (We could also have just taken the average of
**90**and**–10**). Now we have $ y=50\cos \left( {b\left( {x-c} \right)} \right)+40$. - To get the period of the graph, we know that the
**horizontal distance**between the highest point and the lowest point is**one half of a period**. Thus, the period of the graph is $ \left( 2 \right)\left( {150} \right)=300$ - Since the highest point is on the $ y$-axis ($ x=0$), there is no horizontal phase shift. Thus, the sinusoidal function is $ \displaystyle y=50\cos \left( {\frac{\pi }{{150}}x} \right)+40$. We could put it in a graphing calculator to check it.

**(b)** At the highest point, the roller coaster is **90 feet** above the ground ($ \displaystyle y=50\cos \left( \frac{\pi }{150}*0 \right)+40=90$). At $ x=15$ feet, the roller coaster is $ \displaystyle y=50\cos \left( \frac{\pi }{150}*15 \right)+40$, or about **87.55 feet** tall. At $ x=100$ feet, the roller coaster is $ \displaystyle y=50\cos \left( \frac{\pi }{150}*100 \right)+40$, or **15 feet** tall. It’s easiest to put the function $ \displaystyle y=50\cos \left( {\frac{\pi }{{150}}x} \right)+40$ in the graphing calculator, and use the **2 ^{nd} TRACE** (

**CALC) value**function to get these values (see the

**WINDOW**you can use below).

**(c)** When the track is **75 feet** above the ground, $ y=75$. The easiest way to get the $ x$ at that point is to use the **Intersect** feature in the graphing calculator: (**2 ^{nd} Trace** (Calc),

**5**,

**ENTER**,

**ENTER**,

**ENTER**). Thus, when the track is

**75**

**feet**above the ground, the roller coaster is about

**37.98 feet**from the highest point at $ x=0$. Note the window I used to match the graph of the roller coaster.

You can use the same steps to see that when the roller coaster track is **15 feet** above the ground, the roller coaster is **100 feet** from the beginning point.

**Bouncing Spring Problem:**

The weight on a long spring bounces up and down sinusoidally with time. You are looking at a second hand on a clock and notice that when the clock reads **.2 seconds**, the weight first reaches a high point that is **50 centimeters** (cm) above the ground. The next low point is at **30 cm** above the ground, and this occurs at **1.5 seconds**.

**(a)** Draw the graph that represents this situation, and write the sinusoidal equation that expresses the distance from the ground in terms of the number of seconds that has passed.

**(b)** What is the approximate distance from the ground when the clock reads **18 seconds**?

**(c)** What is the approximate distance from the ground when the clock was at $ t=$ **0** seconds?

**(d)** What is the first positive value for the time when the weight is **45 cm** above the ground?

**Solution:** **(a)** Get the equation with these steps:

- Graph the function’s
**high points**and**low points**, and figure out the coordinates of these points, given what we know. Since the high point is at**50 cm**at**.2 sec**, we know the high point is at $ (.2,50)$. We also know that the low point is at $ (1.5,30)$. Here’s what we can graph:Use a**cos graph**again, since the highest point of the graph is on the $ y$-axis: $ y=a\cos \left( {b\left( {x-c} \right)} \right)+d$. - To get the amplitude, or $ |a|$, subtract the lowest $ y$-point from the highest, and then divide by
**2**. Thus, $ a=50-30=20\div 2=10$. Now we have $ y=10\cos \left( {b\left( {x-c} \right)} \right)+d$. - To get the middle of the function, or the vertical shift ($ d$), add the amplitude to the lowest $ y$-value: $ 30+10=40$. (We could also have taken the average of
**30**and**50**). Now we have $ y=10\cos \left( {b\left( {x-c} \right)} \right)+40$. - To get the period of the graph, we know that the
**horizontal distance**between the highest point and the lowest point is**one half of a period**. Thus, the period of the graph is $ (1.5-2)(2)=2.6$**cm**. To get $ b$, we have $ \displaystyle b=\frac{2\pi }{\text{new period}}=\frac{2\pi }{2.6}=\frac{10\pi }{13}$. Now we have $ \displaystyle y=10\cos \left( {\frac{{10\pi }}{{13}}\left( {x-c} \right)} \right)+40$. - Since the highest point is at $ x=.2$, the horizontal phase shift is
**.2 to the right**. Thus, $ c=.2$. The sinusoidal function is then $ \displaystyle y=10\cos \left( {\frac{{10\pi }}{{13}}\left( {x-.2} \right)} \right)+40$. We could put it in a graphing calculator to check it.

**(b)** When the clock reads **18 seconds**, we can plug in **18** for $ x$ to get $ y$ (the distance from the ground): $ \displaystyle y=10\cos \left( {\frac{{10\pi }}{{13}}\left( {18-.2} \right)} \right)+40$, which is about **45.68 cm**. This is after many bounces, as you could see if you graphed the function and made the window of $ x$ larger.

**(c)** The approximate distance from the ground when the clock is at $ t=0$ seconds is $ \displaystyle y=10\cos \left( {\frac{{10\pi }}{{13}}\left( {0-.2} \right)} \right)+40$, which is about **48.85**** cm**.

**(d)** Use the graphing calculator to find the first positive value when the weight is **45 cm** above the ground. After you hit **GRAPH**, you may have to use the **TRACE** button to get the cursor closer to the first point of intersection before you use **intersect**. This value is approximately **.63 seconds: **

**Tidal Problem:**

Let’s do one more, where we’ll use a **sin** function:

A tsunami or tidal wave is an ocean wave caused by an earthquake. The water first goes down from its normal level and then rises an equal distance above its normal level, and so on. Let’s say the **amplitude** for this particular tsunami is **12 meters**, it’s **period** is about **20 minutes**, and it’s **normal depth** is **10 meters**. Assuming the depth of the water varies sinusoidally with time, find the sinusoidal function for this tsunami.

**Solution:** First draw a graph, assuming that the water is at its **regular depth** when $ t=0$, and then goes down to its lowest point, and then up to its highest point. (Notice that the water technically goes below the surface of the ocean; we won’t worry about the scientific consequences of this.)

We know the lowest point is at **5 minutes**, and the period is **20 minutes**, we can figure out that the highest point is at half the distance of the period (**10** minutes) from that lowest point. We can plot the following points and draw the graph:

For this graph, use the **sin function** since the middle of the function goes through the $ y$-axis ($ x=0$). But notice how the graph is **flipped**, so we will use $ -\sin$.

We are already given the amplitude (**12 meters**), vertical shift (normal depth is at **10 meters**), and period (**20 minutes**), so $ \displaystyle b=\frac{2\pi }{\text{new period}}=\frac{2\pi }{20}=\frac{\pi }{10}$. There is no horizontal phase shift, so the sinusoidal function is $ \displaystyle y=-12\sin \left( {\frac{\pi }{{10}}x} \right)+10$.

## Secant and Cosecant Transformations

The **reciprocal **functions secant (**sec**) and cosecant (**csc**) are transformed the same way as the sin and cos, yet the “$ a$” part of the transformation is **not called an amplitude**, but a **stretch**, as we are used to with non-trig transformed functions.

Remember, again, like the sin and cos transformations, the $ \displaystyle \text{new period}=\frac{{2\pi }}{b}$. If asked for **asymptotes** of transformed functions, perform the same transformations on them as you would the $ x$-values of the graph.

You can look at one of the new** asymptotes** of the transformed graph, and then **add** $ \displaystyle \left( {\frac{{\text{new period}}}{2}} \right)k$, since there are two asymptotes per period for the **csc** and **sec** graphs. You can also just set the arguments of the trig functions to the asymptotes ($ \pi k$ for **csc** and $ \displaystyle \frac{\pi }{2}+\pi k$ for **sec**) and solve for $ x$ to get the new asymptotes.

We have the parent graph ** t**-charts from

**above**; let’s go right to examples. Notice that for the

**-charts, the $ x$-points turn into $ \displaystyle \frac{1}{b}x+c$ (opposite math, starting with multiplication/division), and the $ y$-points turn into $ ay+d$ (regular math).**

*t*## Tangent and Cotangent Transformations

The **reciprocal **functions tangent (**tan**) and cotangent (**cot**) are transformed the same way as the **csc** and **sec**, (with “$ a$” part of the transformation a **stretch** and not an amplitude). The difference however is that, since the **period** of the **tan** and **cot** functions (how long the graph goes before repeating itself) is $ \pi$ instead of $ 2\pi$, we have $ \displaystyle \text{new period}=\frac{\pi }{b}$.

If asked for **asymptotes** of transformed functions, perform the same transformations on them as you would the $ x$-values of the graph. You can look at one of the new** asymptotes** of the transformed graph, and then **add** $ (\text{new period})k$, since there is one asymptote per period for the **tan** and **cot** graphs. You can also just set the arguments of the trig functions to the asymptotes ($ \displaystyle \frac{\pi }{2}+\pi k$ for **tan** and $ \pi k$ for **cot**) and solve for $ x$ to get the new asymptotes.

We have the parent graph *t*-charts from **above**; let’s go right to examples. Notice that for the ** t**-charts, the $ x$-points turn into $ \displaystyle \frac{1}{b}x+c$ (opposite math, starting with multiplication/division), and the $ y$-points turn into $ ay+d$ (regular math).

## Writing Equations from Transformed Graphs for Sec, Csc, Tan, and Cot

Here are a few examples where we get the equations of trig functions other than **sin** and **cos** from graphs. Note that there may be varying answers for these equations:

## Transformation of all Trig Functions without T-Charts

Let’s go over once again how to transform trig functions without *t*-charts. **Note that in order to perform the transformations accurately and quickly, you must know your 6 trig functions graphs inside out! **

**Word of caution: In the trig transformation formulas below, I like to use formulas like $ y=a\sin \left( {b\left( {x-c} \right)} \right)+d$, with a phase shift of $ c$, instead of $ y=a\sin \left( {bx-c} \right)+d$, with a phase shift of $ \displaystyle \frac{c}{b}$. Sorry for the confusion, but the results will be the same!**

Here are the steps for **sin** and **cos** graphs; note that there’s an example of this **here**.

- Put the trig function in the $ y=a\sin \left( {b\left( {x-c} \right)} \right)+d\,\,\,\text{or}\,\,\,y=a\cos \left( {b\left( {x-c} \right)} \right)+d$ format.
- The new
**domain**will still be $ \left( {-\infty ,\infty } \right)$. - The new
**range**will be $ \left[ {\text{vertical shift (}d\text{)}-\left| a \right|\text{,}\,\text{vertical shift (}d\text{)}+\left| a \right|} \right]$. ($ \left| a \right|$ is amplitude) - New
**period**will be $ \displaystyle \frac{{2\pi }}{b}$. **Phase shift**will be $ c$; start graph at this point, and if $ a$ is negative, make graph upside down.- First draw line where
**vertical shift**($ d$) is (the**midline**), and lines where lower and upper ranges are. Graph will be centered vertically at the vertical shift, go up to the top line, and down to the bottom line. - Then start graph (first point) where phase shift starts and count over (to right) an amount that is the
**new period**(make complete**revolution**or**cycle**of the sin or cos). Draw the last point here. Draw middle point halfway between; this will be at the halfway mark of the complete cycle. We’ll have a total of five points. - If preferred, instead of the step above, draw the
**midline-intercepts**to graph. To get new midline-intercepts: parent function midline intercepts ($ x$-intercepts) are at $ \pi k$ for**sin**and $ \displaystyle \frac{\pi }{2}+\pi k$ for**cos**. Set the transformed trig argument to the parent function $ x$-intercepts, and solve for $ x$.

Here are the steps for **tan** and **cot** graphs:

- Put the trig function in the $ y=a\tan \left( {b\left( {x-c} \right)} \right)+d\,\,\,\text{or}\,\,\,y=a\cot \left( {b\left( {x-c} \right)} \right)+d$ format.
- Remember that
**asymptotes**for**tan**are at $ \displaystyle \frac{\pi }{2}+\pi k$ and for**cot**are $ \pi k$ (the $ c$’s (**cot**and**csc**) have the “easier” ones). To get the new asymptotes, set the trig argument to the asymptotes, and solve for $ x$. For example, if we have $ \displaystyle \tan \left( {3x+\pi } \right)\,\,(\text{which would be}\,\tan \left( {3\left( {x+\frac{\pi }{3}} \right)} \right))$, we would solve: $ \displaystyle 3x=-\frac{\pi }{2}+\pi k;\,\,\,x=\frac{{-\frac{\pi }{2}}}{3}+\frac{{\pi k}}{3};$ $ \displaystyle x=-\frac{\pi }{6}+\frac{{\pi k}}{3}\,\,(=\frac{\pi }{6}+\frac{{\pi k}}{3})$. It’s best “math grammar” to start the asymptote at the smallest positive value; thus, I added $ \displaystyle \frac{{\pi \cdot 1}}{3}=\frac{{2\pi }}{6}$ to $ \displaystyle -\frac{\pi }{6}$ to get $ \displaystyle \frac{\pi }{6}$ above. - The new
**range**will still be $ \left( {-\infty ,\,\infty } \right)$. **Draw asymptotes first**; phase shift will take care of itself. (To get an asymptote starting point, you can set $ k=0$ in your new asymptote equation, $ k=-1$ for one to the left, $ k=1$ for one to the right). Right in between the asymptotes (you can take average of the $ x$’s), draw the middle of the graph (but make sure it is shifted up or down according to the**vertical shift**, or $ d$).**New period**will be $ \displaystyle \frac{\pi }{b}$ (since tan and cot have periods of $ \pi$). This should be the distance between asymptotes (the coefficient of “$ k$” in the asymptote equations).- If trig function is
**negative**, make sure you flip the graph. - New
**domain**will be all $ x$-values except for the asymptotes. - To get
**two more points**on either side of the center point, take the center or average again of the middle point of the asymptotes for the $ x$’s. To get the $ y$-value for these points (since these are normally at $ y=1$ and $ y=-1$), shift the $ y$ according according to the vertical stretch, or “slope” (for example, if the function is $ 3\tan \left( {x+\pi } \right)-4$, the $ y$’s would be $ 3(1)-4=-1$ for one of the points, and $ 3(1)-4=-7$ for the other one.

Here are the steps for **csc** and **sec** graphs:

- Put the trig function in the $ y=a\csc \left( {b\left( {x-c} \right)} \right)+d\,\,\,\text{or}\,\,\,y=a\sec \left( {b\left( {x-c} \right)} \right)+d$ format.
- Remember that
**asymptotes**for**sec**are at $ \displaystyle \frac{\pi }{2}+\pi k$ and asymptotes for**csc**are $ \pi k$ (the $ c$’s (**cot**and**csc**) have the “easier” ones). To get the new asymptotes, set the trig argument to the asymptotes, and solve for $ x$. For example, if we have $ \displaystyle \csc \left( {3x-\frac{\pi }{6}} \right)$, we would solve: $ \displaystyle 3x-\frac{\pi }{6}=\pi k;\,\,\,3x=\frac{\pi }{6}+\pi k;$ $ \displaystyle x=\frac{\pi }{{18}}+\frac{{\pi k}}{3}=\frac{\pi }{{18}}+\frac{{6\pi k}}{{18}}$. - The new
**range**will be $ \left( {-\infty ,\,\,\text{vertical shift (d)}-\left| a \right|} \right]\cup \left[ {\text{vertical shift (d)}+\left| a \right|,\infty } \right)$. ($ \left| a \right|$ is amplitude). **Draw asymptotes**first; the “cups” will fall into these areas. (To get an asymptote starting point, you can set $ k=0$ in your new asymptote equation, $ k=-1$ for one to the left, $ k=1$ for one to the right). The center of the “cups” will fall halfway between the asymptotes.**New period**will be $ \displaystyle \frac{{2\pi }}{b}$ (since**csc**and**sec**have periods of $ 2\pi $). (This should be the distance between every other asymptote).- To get where graphs start (“cup up” or “cup down”), shift the graph according to the phase shift ($ c$) (using the $ y$-axis as the starting point), and draw the graph according to the original trig function (
**csc**or**sec**) and whether or not it’s**negative**(if negative, you flip it vertically). - New
**domain**will be all $ x$-values except for the asymptotes.

It’s a good idea to graph your answers on a graphing calculator (radians) with window of one period (with the Xmin and Xmax) and range (with Ymin and Ymax) to check your graphs.

Here are some examples:

**Understand these problems, and practice, practice, practice!**

**For Practice**: Use the **Mathway** widget below to try a **Trig** **Transformation** problem. Click on **Submit** (the blue arrow to the right of the problem) and click on **Graph** to see the answer.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on **Tap to view steps**, or **Click Here**, you can register at **Mathway** for a **free trial**, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

**On to The Inverse Trigonometric Functions – you’re ready! **

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