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).
| Trig Function T-Charts | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| $ y=\sin \left( x \right)$
| $ y=\cos \left( x \right)$
| $ y=\tan \left( x \right)$
| $ y=\csc \left( x \right)$
| $ y=\sec \left( x \right)$
| $ y=\cot \left( x \right)$
| ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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.
| Trig Transformation Formula | Explanation |
| Note that typically these formulas in the form $ y=a\sin \left( {bx-c} \right)+d$, where $ \displaystyle \frac{c}{b}$ is the phase shift. I like to take out the “$ b$” and have “$ c$” as the phase shift.
$ \begin{array}{l}y=a\sin \left( {b\left( {x-c} \right)} \right)+d\\\\\\y=a\cos \left( {b\left( {x-c} \right)} \right)+d\end{array}$
Others:
$ \begin{array}{l}y=a\csc \left( {b\left( {x-c} \right)} \right)+d\\y=a\sec \left( {b\left( {x-c} \right)} \right)+d\\y=a\tan \left( {b\left( {x-c} \right)} \right)+d\\y=a\cot \left( {b\left( {x-c} \right)} \right)+d\end{array}$
| $ \boldsymbol{\left| a \right|=}$ the amplitude. This is how far up and down the graph goes from the middle (scaling factor): think “stretch” if $ a>1$ or “compression” if $ a<1$). Without a shift, $ a=1$; thus the sin and cos graphs go from $ y=-1$ to $ y=1$, with the middle at $ y=0$. Note that the “$ a$” is only called an amplitude for sin and cos graphs; otherwise it’s considered a “stretch” or “compression”. For the sin and cos graphs, the amplitude is the highest $ y$-value minus the lowest $ y$-value, divided by 2. If there is a negative sign before the $ a$, the graph is flipped vertically (it is upside down).
$ \boldsymbol{b=}$ the number of times the graph will repeat itself in the “normal” period, which is $ 2\pi$ for sin, cos, csc, and sec, and $ \pi$ for tan and cot. Note that “$ b$” is a horizontal stretch if $ 0<b<1$, and a horizontal compression if $ b>1$ (oppositive math for horizontal dilations). The period is the length, or difference of $ x$-values, of one complete cycle of the graph; it is sometimes called a frequency. It follows for the sin, cos, csc, and sec graphs that $ \displaystyle b=\frac{{2\pi }}{{\text{new period}}}$ and $ \displaystyle \text{new period}=\frac{{2\pi }}{b}$. Normally, $ b=1$, which means that the period of the graph is $ \displaystyle b=\frac{{2\pi }}{\text{1}}=2\pi $. For tan and cot, since normally the period is $ \pi$, we have: $ \displaystyle b=\frac{\pi }{{\text{new period}}}$ and $ \displaystyle \text{new period}=\frac{\pi }{b}$.
$ \boldsymbol{c=}$ the horizontal shift or phase shift of the graph. (Note that in some math books, the general equation is $ y=a\sin \left( {bx-c} \right)+d$, which means $ \displaystyle \frac{c}{b}$ is the phase shift). Since this is a horizontal shift, when $ \boldsymbol{c}$ is subtracted from $ \boldsymbol{x}$, it shifts to the right, for example. Normally, $ c=0$.
$ \boldsymbol{d=}$ the vertical shift of the graph (also called the midline) Notice that it’s on the outside of the parentheses; this shifts a graph vertically in the direction we would think it would; for example, a positive $ d$ moves the graph higher. Normally, $ d=0$. |
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).
| Trig Transformation | T-Chart | Graph | ||||||||||||
| $ \displaystyle y=\sin \left( {x+\frac{\pi }{4}} \right)$
Move graph to the left $ \displaystyle \frac{\pi }{4}$ units.
The $ y$’s stay the same; subtract $ \displaystyle \frac{\pi }{4}$ from the $ x$-values (we do the opposite math when working with the $ x$’s). |
| Period: $ 2\pi $ Amplitude: 1 $ x$-intercepts: $ \displaystyle x=-\frac{\pi }{4}+\pi k$ | ||||||||||||
| $ y=4\sin \left( {2x} \right)$
Amplitude (vertical stretch) of graph is 4.
The graph repeats itself 2 times within normal period of $ 2\pi $.
Or, to figure out the distance of a complete cycle, we can use new period = $ \displaystyle \frac{{2\pi }}{b}=\frac{{2\pi }}{2}=\pi $.
Both the $ x$-values and $ y$-values are affected. |
For the t-chart, remember that for the $ x$, we do the opposite math ($ \displaystyle \frac{1}{2}x$ instead of $ 2x$). |
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left[ {-4,4} \right]$ Period: $ \pi $ Amplitude: 4 $ x$-intercepts: $ \displaystyle x=\frac{\pi }{2}k$ | ||||||||||||
| $ y=-\cos \left( x \right)-4$
Flip graph across the $ x$-axis.
Move graph down 4 units.
These changes will affect the $ y$-values only. |
|
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left[ {-5,-3} \right]$ Period: $ 2\pi $ Amplitude: 1 Midline-intercepts: $ \displaystyle x=\frac{\pi }{2}+\pi k$ Center of Graph: $ y=-4$ | ||||||||||||
| $ \displaystyle y=-2\cos \left( {\frac{1}{3}x-\frac{\pi }{6}} \right)+1$ $ \displaystyle y=-2\cos \left( {\frac{1}{3}\left( {x-\frac{\pi }{2}} \right)} \right)+1$
Note you have to factor out the $ \displaystyle \boldsymbol {\frac{1}{3}}$ to get the equation in the correct form.
Amplitude (vertical stretch) of graph is 2. Flip graph across $ x$-axis. The new period = $ \displaystyle \frac{{2\pi }}{b}=\frac{{2\pi }}{{\frac{1}{3}}}=6\pi $. The graph is shifted $ \displaystyle \frac{\pi }{2}$ units to the right, and 1 unit up. |
For the t-chart, remember that for the $ x$, we do the opposite math ($ 3x$ instead of $ \displaystyle \frac{1}{3}x$). |
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left[ {-1,3} \right]$ Period: $ 6\pi $ Amplitude: 2 Midline-intercepts: $ x=2\pi +3\pi k$ Center of Graph: $ y=1$ |
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left[ {-1,1} \right]$
Transforming Without Using 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;\,\,\,\,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.
- 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:
| Transformed Trig Graph | Steps to Get Equation |
 | 1. Since the middle of the graph is close to (on, actually) the $ y$-axis, use the positive sin function: $ y=a\sin \left( {b\left( {x-c} \right)} \right)+d$.
2. To get the amplitude, or $ |a|$, subtract the lowest $ y$-point from the highest, and then divide by 2: $ 3-\left( {-1} \right)=4\div 2=2$. Now we have $ y=2\sin \left( {b\left( {x-c} \right)} \right)+d$.
3. To get the middle of the function, or the vertical shift ($ d$), add the amplitude to the lowest $ y$-value: $ –1+2=1$. We can also see from the graph that the middle of the graph is at $ y=1$. Now we have $ y=2\sin \left( {b\left( {x-c} \right)} \right)+1$.
4. To get $ b$, find the period of the graph by seeing how long it goes before repeating itself (subtract the two $ x$-values to get this new period). The graph repeats itself from when $ x=0$ to when $ x=\pi $, so the new period is $ \pi -0=\pi $. Then use the equation $ \displaystyle b=\frac{{2\pi }}{{\text{new period}}}=\frac{{2\pi }}{\pi }=2$. Now we have $ y=2\sin \left( {2\left( {x-c} \right)} \right)+1$.
5. Since the graph isn’t shifted to the left or right from the $ y$-axis, there is no phase shift: $ c=0$. The graph is $ y=2\sin \left( {2x} \right)+1$.
See the screens on the left to see how we can check a complete revolution of the graph in a graphing calculator – looks good! Note that in the WINDOW screen, we can use $ \pi $ when we input the Xmax for example, and the calculator converts it to 3.141592654. |
 | 1. Since the top of the graph is close to the $ y$-axis, use the positive cos function: $ y=a\cos \left( {b\left( {x-c} \right)} \right)+d$.
2. To get the amplitude, or $ |a|$, subtract the lowest $ y$-point from the highest, and then divide by 2: $ 2-\left( {-10} \right)=12\div 2=6$. Now we have $ y=6\cos \left( {b\left( {x-c} \right)} \right)+d$.
3. To get the middle of the function, or the vertical shift ($ d$), add the amplitude to the lowest $ y$-value: $ –10+6=–4$. We can also see from the graph that the middle of the graph is at $ y=-4$. Now we have $ y=6\cos \left( {b\left( {x-c} \right)} \right)-4$.
4. To get $ b$, find the period of the graph by seeing how long it goes before repeating itself (subtract the two $ x$-values to get this new period). We can see that the graph repeats half of a complete revolution from when $ x=100$ to when $ x=900$. Thus, half a revolution is $ 900-100=800$, so a complete revolution is $ 1600$. Use the equation $ \displaystyle b=\frac{{2\pi }}{{\text{new period}}}=\frac{{2\pi }}{{1600}}=\frac{\pi }{{800}}$. Now we have $ \displaystyle y=6\cos \left( {\frac{\pi }{{800}}\left( {x-c} \right)} \right)-4$.
5. To find the phase shift of the graph, see how close it is to the $ y$-axis $ (x=0)$; this value is $ c$. Since the top of the graph is closest to the $ y$-axis, and it’s a little bit to the right (100 to the right), we have a cos graph with an $ “x-c”$ situation. $ c=100$, and the graph is $ \displaystyle y=6\cos \left( {\frac{\pi }{{800}}\left( {x-100} \right)} \right)-4$.
See the screens on the left to see how we can check this half revolution of the graph in a graphing calculator – looks good! (Note that to enter $ \displaystyle \frac{\pi }{{800}}$, I used “alpha” “Y=” “n/d” in order to input this fraction in fraction mode). |
You might be asked to write a sinusoidal equation, given certain characteristics of the transformed trig graph; here is an example:
| Trig Transformation Problem | Solution |
| Write an equation of a (non-flipped) sinusoidal function $ y=\sin \left( x \right)$ with the following characteristics:
Period: $ 6\pi $; Phase Shift: right $ \displaystyle \frac{\pi }{3}$; Range: $ \left[ {4,20} \right]$ | Use the equation $ y=a\sin \left( {b\left( {x-c} \right)} \right)+d$, where $ \left| a \right|$ is the amplitude, $ \displaystyle b=\frac{{2\pi }}{{\text{period}}}$, $ c$ is the phase shift, and $ d$ is the vertical shift. Since the period is $ 6\pi $, and the phase shift is right $ \displaystyle \frac{\pi }{3}$, so far we have $ \require {cancel} \displaystyle y=a\sin \left( {\frac{{{{{\cancel{{2\pi }}}}^{1}}}}{{{{{\cancel{{6\pi }}}}^{3}}}}\left( {x-\frac{\pi }{3}} \right)} \right)+d$.
To get the amplitude, or $ |a|$, subtract the lowest $ y$-point from the highest (using the given range), and then divide by 2: $ 20-4=16\div 2=8$. Now we have $ \displaystyle y=8\sin \left( {\frac{1}{3}\left( {x-\frac{\pi }{3}} \right)} \right)+d$. To get the middle of the function, or the vertical shift ($ d$), add the amplitude to the lowest $ y$ value: $ 4+8=12.$
The transformed equation is $ \displaystyle y=8\sin \left( {\frac{1}{3}\left( {x-\frac{\pi }{3}} \right)} \right)+12$. Work backwards to make sure we get the correct characteristics; we do! |
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$ feet. To get $ b$, we have $ \displaystyle b=\frac{2\pi }{\text{new period}}=\frac{2\pi }{300}=\frac{\pi }{150}$. Now we have $ \displaystyle y=50\cos \left( {\frac{\pi }{{150}}\left( {x-c} \right)} \right)+40$.
- 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 2nd 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: (2nd 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 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).
| Trig Transformation | T-Chart | Graph | ||||||||||||
| $ \displaystyle y=-4\csc \left( {x+\frac{\pi }{4}} \right)+1$
First flip graph across the $ \boldsymbol {x}$-axis and stretch by factor of 4.
Period stays the same at $ 2\pi $.
Move graph up 1 unit.
Shift graph $ \displaystyle \frac{\pi }{4}$ to the left.
Transform asymptotes by starting with new asymptote and adding $ \displaystyle \left( {\frac{{\text{new period}}}{2}} \right)k$. |
(und = undefined)
Asymptotes: $ \displaystyle x=-\frac{\pi }{4}+\pi k,\,\,k\in \,\text{Int}$ |
Period: $ 2\pi $ Vertical Stretch: 4 Center of Graph: $ y=1$ Graph Flipped across $ x$-axis | ||||||||||||
| $ y=4\sec \left( {5x} \right)+3$
Stretch vertically by a factor of 4.
The new period = $ \displaystyle \frac{{2\pi }}{b}=\frac{{2\pi }}{5}$.
Shift graph up 3 units.
No horizontal phase shift.
Transform asymptotes by starting with new asymptote and adding $ \displaystyle \left( {\frac{{\text{new period}}}{2}} \right)k$. |
(und = undefined)
Asymptotes: $ \displaystyle x=\frac{\pi }{{10}}+\frac{\pi }{5}k,\,\,k\in \,\text{Int}$ |
Period: $ \displaystyle \frac{{2\pi }}{5}$ Vertical Stretch: 4 Center of Graph: $ y=3$ |
Domain: $ \displaystyle x\ne -\frac{\pi }{4}+\pi k$ Range: $ \left( {-\infty ,-3} \right]\cup \left[ {5,\infty } \right)$
Domain: $ \displaystyle x\ne \frac{\pi }{{10}}+\frac{\pi }{5}k$ Range: $ \left( {-\infty ,-1} \right]\cup \left[ {7,\infty } \right)$Tangent and Cotangent Transformations
The reciprocal functions tangent (tan) and cotangent (cot) are transformed the same way as the csc and sec, (with “$ 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).
| Trig Transformation | T-Chart | Graph | ||||||||||||
| $ y=-2\tan \left( {4x} \right)+1$
First flip graph across the $ x$-axis and stretch by factor of 2.
The new period = $ \displaystyle \frac{\pi }{b}=\frac{\pi }{4}$.
Then move graph up 1 unit.
No horizontal phase shift.
Transform asymptotes by starting with new asymptote and adding $ \left( {\text{new period}} \right)k$. |
(und = undefined)
Asymptotes: $ \displaystyle x=\frac{\pi }{8}+\frac{\pi }{4}k,\,\,k\in \,\text{Int}$ |
Domain: $ \displaystyle x\ne \frac{\pi }{8}+\frac{\pi }{4}k$ Range: $ \left( {-\infty ,\infty } \right)$ Period: $ \displaystyle \frac{\pi }{4}$ Vertical Stretch: 2 Graph Flipped across $ x$-axis | ||||||||||||
| $ \displaystyle y=\frac{1}{2}\cot \left( {\frac{1}{2}x-\pi } \right)$ $ \displaystyle y=\frac{1}{2}\cot \left( {\frac{1}{2}\left( {x-2\pi } \right)} \right)$
Vertical compression of $ \displaystyle \frac{1}{2}$; no vertical shift.
The new period = $ \displaystyle \frac{\pi }{b}=\frac{\pi }{{\frac{1}{2}}}=2\pi $.
Shift graph $ 2\pi $ to the right.
Transform asymptotes by starting with new asymptote and adding $ \left( {\text{new period}} \right)k$. |
(und = undefined)
Asymptotes: $ x=2\pi k$ |  Domain: $ x\ne 2\pi k$ Range: $ \left( {-\infty ,\infty } \right)$ Period: $ 2\pi $ Vertical Compression: $ \displaystyle \frac{1}{2}$ |
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:
| Trig Graph | Steps to Get Equation |
 | 1. Use sec graph, since it’s an even function and looks closest to sec (but it’s flipped). The middle of the graph is at $ y=4$.
2. The vertical stretch is 2, since we go up 2 and down 2 from $ y=4$ to start the graphs. There’s also no horizontal shift. Now we have $ y=-2\sec \left( {bx} \right)+4$ (negative because of the flip).
3. To get $ b$, find the period of the graph by seeing how long it goes before repeating itself (subtract the two $ x$-values to get this new period). The graph repeats itself from when $ \displaystyle x=-\frac{\pi }{{10}}$ to when $ \displaystyle x=\frac{{3\pi }}{{10}}$, so the new period is $ \displaystyle \frac{{3\pi }}{{10}}-\left( {-\frac{\pi }{{10}}} \right)=\frac{{4\pi }}{{10}}=\frac{{2\pi }}{5}$. Then use the equation $ \displaystyle b=\frac{{2\pi }}{{\text{new period}}}=\frac{{2\pi }}{{\left( {\frac{{2\pi }}{5}} \right)}}=\frac{{2\pi }}{1}\times \frac{5}{{2\pi }}=5$. The equation is $ y=-2\sec \left( {5x} \right)+4$. |
 | 1. We could use the cot graph, but we could also use a flipped tan graph (thus with a negative sign) – let’s try this, since the $ x$-intercept is close to the origin. There is no vertical shift.
2. The graph is shifted horizontally $ \displaystyle \frac{\pi }{{24}}$ to the left.
3. The vertical stretch is 2, since we go up 2 and down 2 from $ y=0$ for the points halfway between the middle point and the asymptotes. So far, we have $ \displaystyle y=-2\tan b\left( {x+\frac{\pi }{{24}}} \right)$.
4. To get $ b$, find the period of the graph by seeing how long it goes before repeating itself (subtract the two $ x$-values to get this new period). The graph repeats itself from when $ \displaystyle x=-\frac{\pi }{{24}}$to when $ \displaystyle x=\frac{\pi }{8}$, so the new period is $ \displaystyle \frac{\pi }{8}-\left( {-\frac{\pi }{{24}}} \right)=\frac{{3\pi }}{{24}}+\frac{\pi }{{24}}=\frac{\pi }{6}$. Use the equation $ \displaystyle b=\frac{\pi }{{\text{new period}}}=\frac{\pi }{{\left( {\frac{\pi }{6}} \right)}}=\frac{\pi }{1}\times \frac{6}{\pi }=6$. Now we have $ \displaystyle y=-2\tan 6\left( {x+\frac{\pi }{{24}}} \right)$. Try these on the graphing calculator! |
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+\pi =\frac{\pi }{2}+\pi k;\,\,\,3x=\left( {\frac{\pi }{2}-\pi } \right)+\pi k;$ $ \displaystyle 3x=-\frac{\pi }{2}+\pi k;\,\,\,x=\frac{{-\frac{\pi }{2}}}{3}+\frac{{\pi k}}{3};\,\,\,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;\,\,\,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:
| Trig Transformation | Notes | Graph |
| $ \displaystyle y=-2\cos \left( {3x+\frac{\pi }{2}} \right)+4$ $ \displaystyle y=-2\cos \left( {3\left( {x+\frac{\pi }{6}} \right)} \right)+4$ (First factor out the 3).
Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left[ {4-2,4+2} \right]=\left[ {2,6} \right]$ Period: $ \displaystyle \frac{{2\pi }}{3}$ Amplitude: 2 Phase Shift: $ \displaystyle -\frac{\pi }{6}$ Flip?: Yes | To get the range, start with the vertical shift, and add and subtract the amplitude. The graph is shifted up 4 (vertical shift or midline), and goes up and down by 2 (amplitude).
Start graph $ \displaystyle \frac{\pi }{6}$ to the left of the $ y$-axis (low point, since graph is flipped) and next cycle starts at $ \displaystyle -\frac{\pi }{6}+\frac{{2\pi }}{3}=\frac{\pi }{2}$, since $ \displaystyle \frac{{2\pi }}{3}$ is the period.
To get the high point, take the middle (average) of $ \displaystyle -\frac{\pi }{6}\text{ and }\frac{\pi }{2}$, which is $ \displaystyle \frac{\pi }{6}$.
You could also have gotten the midline-intercepts by setting the trig argument to $ \displaystyle \frac{\pi }{2}+\pi k:\,\,\,3x+\frac{\pi }{2}=\frac{\pi }{2}+\pi k;\,\,x=\frac{{\pi k}}{3}$. |  |
| $ \displaystyle y=-3\tan \left( {\frac{\pi }{2}x+\pi } \right)-3$ $ \displaystyle y=-3\tan \left( {\frac{\pi }{2}\left( {x+2} \right)} \right)-3$ (First factor out the $ \displaystyle \frac{\pi }{2}$).
Asymptotes: $ \displaystyle -1+2k$ Domain: $ \displaystyle x\ne -1+2k$ Range: $ \left( {-\infty ,\infty } \right)$ Period: $ 2$ Phase Shift: $ -2$ Flip?: Yes | To get the asymptotes, start with the regular tan asymptotes, set to the new tan argument, and solve for $ x$: $ \displaystyle \frac{\pi }{2}x+\pi =\frac{\pi }{2}+\pi k;\,\,\frac{\pi }{2}x=\left( {\frac{\pi }{2}-\pi } \right)+\pi k;\,\,\,\frac{\pi }{2}x=-\frac{\pi }{2}+\pi k;\,\,\,x=-1+2k\,(=1+2k)$. Draw the asymptotes for $ k=-1, 0, 1$, which are $ -3,-1,\,\text{and}\,\,\text{1}$. Graph will be flipped because of the negative sign. To get the period, take the regular tan period of $ \pi$ and divide by $ \displaystyle \frac{\pi }{2}$; period is 2 (distance between asymptotes).
To get the middle of each tan graph, look at vertical shift; graph will be shifted down 3 units, and the $ x$-value will be halfway between the asymptotes. To get two more points on either side of the center point, go half-way between the center point and the asymptotes; these $ x$-values will be $ -2.5,-1.5,-.5,\,\text{and}\,\text{.5}$. To get the $ y$-value for these points, apply the vertical stretch of 3 (starting with the middle points) by going up 3 and down 3. |  |
| $ y=-4\csc \left( {2x} \right)+1$
Asymptotes: $ \displaystyle \frac{\pi }{2}k$ Domain: $ \displaystyle x\ne \frac{\pi }{2}k$ Range: $ \begin{array}{l}\left( {-\infty ,1-4} \right]\cup \left[ {1+4,\infty } \right)\\=\left( {-\infty ,-3} \right]\cup \left[ {5,\infty } \right)\end{array}$ Period: $ \pi $ Phase Shift: none Flip?: Yes | To get the asymptotes, start with the regular csc asymptotes, set to the new csc argument, and solve for $ x$: $ \displaystyle 2x=\pi k;\,\,x=\frac{{\pi k}}{2}\,\,\text{or }\frac{\pi }{2}k$. Draw the asymptotes for $ k=-1,0,1$, which are $ \displaystyle -\frac{\pi }{2}$, 0, and $ \displaystyle \frac{\pi }{2}$.
The new range will be $ \displaystyle \left( {-\infty ,\text{vertical shift (d)}-\left| a \right|} \right]\cup \left[ {\text{vertical shift (d)}+\left| a \right|,\infty } \right)$, which is $ \left( {-\infty ,-3} \right]\cup \left[ {5,\infty } \right)$. Graph will be flipped from normal csc graph, starting at the $ y$-axis, since there is no phase shift.
To get the period, take the regular csc period of $ 2\pi $ and divide by 2, so period is $ \pi $ (distance between every other asymptote). |  |
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!
Comments are closed.