ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: annaoh_2003 on January 29, 2022, 07:39:14 pm
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Im having trouble figuring out how to graph a trigonometric equation? where does the "base" function start. Ive included an example of a equation I received- I understand the transformations - dilate by 2, extend period by 2 (I think) and the vertical shift of 1, but then get confused over the -pi / 3 ? plus I dont have a calculator to help me figure this out. sin, cos and tan dont make much sense to me in functions. do I need to do the transformations in some sort of order? and putting pi as the domain makes it so much harder. where do I start at all :( thanks
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Im having trouble figuring out how to graph a trigonometric equation? where does the "base" function start. Ive included an example of a equation I received- I understand the transformations - dilate by 2, extend period by 2 (I think) and the vertical shift of 1, but then get confused over the -pi / 3 ? plus I dont have a calculator to help me figure this out. sin, cos and tan dont make much sense to me in functions. do I need to do the transformations in some sort of order? and putting pi as the domain makes it so much harder. where do I start at all :( thanks
The base function refers to sin, cos or tan. You should learn the key features of each. The cosine graph for its 1st positive cycle has a maximum at (0,1), an x-intercept at (pi/2, 0), a minimum at (pi, -1), another x-intercept at (3pi/2, 0) and finally a maximum at (2pi, 1). One method of sketching trig graphs is to use these as the pre-image to transformation matrices, which will give you an image i.e. transformed Cartesian coordinates
The -pi/3 is supposed to indicate a horizontal translation, however, you can only determine the horizontal translation if the coefficient in front of the x is 1. In this case, there's a 2x, so you must take out the 2 and make it 2(x - pi/6), so the horizontal translation is pi/6 units in the +ve direction of the x-axis (note that the translation is opposite in sign i.e. if it says cos(x+1), it's actually -1 unit from the positive direction of the x-axis).
Transformations are always done in the order D(ilation) R(eflection) T(ranslation). Both D and R have priority over T, but D and R aren't prioritised over each other. The easiest way to sketch trig functions is to solve for f(x) = 0 for all x-intercepts, find f(0) which will be your y-intercept, find the end-points i.e. y-values at the boundaries of the restricted domain (in this case, f(-pi) and f(2pi)). For the turning points, derive the function and let f'(x) = 0 to find the x-values of the turning points, then sub these values into the original function. Use a sign diagram to predict the nature of the turning points (i.e. local maximum/local minimum)
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Hi Annaoh!
where does the "base" function start ... I understand the transformations - dilate by 2, extend period by 2 (I think) and the vertical shift of 1, but then get confused over the -pi / 3 ?
When the question says "standard" cosine function it is referring to \(y=\cos \left(x\right)\) and it is asking - what transformations need to be applied to \(y=\cos \left(x\right)\) to get \(f\left(x\right)=2\cos \left(2x-\frac{\pi }{3}\right)+1\).
It might be easier to look at \(f\left(x\right)\) as \(y=2\cos \left(2\left(x-\frac{\pi }{6}\right)\right)+1\) and determine the transformations by inspection.
If you struggle to see the transformation like that you can alternatively rearrange and solve the following to figure out what the transformation is as \(\left(x,\:y\right)\rightarrow \left(x',\:y'\right)\) where \(y=\cos \left(x\right)\) and \(y'=2\cos \left(2\left(x'-\frac{\pi }{6}\right)\right)+1\)
Rearranging gets us \(y'=2\cos \left(2\left(x'-\frac{\pi }{6}\right)\right)+1\Rightarrow \:\frac{y'-1}{2}=\cos \left(2\left(x'-\frac{\pi }{6}\right)\right)\)
Now we solve \(\frac{y'-1}{2}=y\) and \(2\left(x'-\frac{\pi }{6}\right)=x\) for \(y'\) and \(x'\). Solving those equations will get us \(y'=2y+1\) and \(x'=\frac{x}{2}+\frac{\pi }{6}\)
Therefore the mapping of the transformation: \(\left(x,\:y\right)\rightarrow \left(x',\:y'\right)\) (from here we just sub in \(x'\) and \(y'\)) is \(\left(x,\:y\right)\rightarrow \left(\frac{x}{2}+\frac{\pi \:}{6},\:2y+1\right)\)
This represents the transformations that were applied to \(y=\cos \left(x\right)\) to get \(f(x)\), which are:
- Dilation of factor 2 from the \(x\) axis
- Dilation of factor \(\frac{1}{2}\) from the \(y\) axis
- Translation of \(\frac{\pi }{6}\) units in the positive direction of the \(x\) axis
- Translation of 1 unit in the positive direction of the \(y\) axis - (which you correctly said)
do I need to do the transformations in some sort of order?
As mentioned by Billuminati, the order of transformations does matter - if you for instance translated and the translation was followed by a dilation you would need to also dilate the translation if you know what I mean. So as a general rule of thumb dilations first when determining the sequence of transformations.
plus I dont have a calculator to help me figure this out.
I think this question assumes you have memorised the important values of the unit circle below
(https://matterofmath.com/wp-content/uploads/2021/07/Unit-Circle-Table-1024x511.jpg)
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- Dilation of factor \(\frac{1}{2}\) from the \(y\) axis
(https://matterofmath.com/wp-content/uploads/2021/07/Unit-Circle-Table-1024x511.jpg)
ahhh thank you guys, could you explain why you get 1/2 and not 2? do I need divide my y values by 1/2 to get the new "dilated" graph ?
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ahhh thank you guys, could you explain why you get 1/2 and not 2? do I need divide my y values by 1/2 to get the new "dilated" graph ?
Dilations or reflections in a given axis don't affect that same axis' value, they affect the value of the other axis ie dilations from y-axis only affect x-values and dilations from x-axis only affect y-values.