Graphing Cosine Function: y=cos2πx

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FritoTaco

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Standard Equation: \(\displaystyle y = A sin(B(x - C))+D \)


  • A: amplitude is A
  • B: period is (2π)/|B|
  • C: phase shift is C/B
  • D: vertical shift is D

Problem: \(\displaystyle y=cos2\pi x\)



My Work:

Amplitude: \(\displaystyle 1\) (a value in front of cos)
Period: \(\displaystyle \dfrac{2\pi}{2\pi}\) = \(\displaystyle 1\)
Phase Shift: \(\displaystyle \dfrac{c}{b}\) = \(\displaystyle \dfrac{0}{2\pi}\) = \(\displaystyle 0\)
Vertical Shift: \(\displaystyle 0\) (d value)
Count: \(\displaystyle \dfrac{1}{4}\cdot\dfrac{1}{1}\) (Formula for Count: \(\displaystyle \dfrac{1}{4}\cdot period\))


\(\displaystyle x\)\(\displaystyle y=cosine2\pi x\)
\(\displaystyle 0\)\(\displaystyle cos2\pi(0) = cos0 = 1\)
\(\displaystyle 1/4\)\(\displaystyle cos2\pi(1/4)= ?\)
\(\displaystyle 1/2\)\(\displaystyle cos2\pi(1/2)= ?\)
\(\displaystyle 3/4\)\(\displaystyle cos2\pi(3/4)= ?\)
\(\displaystyle 1\)\(\displaystyle cos2\pi(1) = ?\)



How do I calculate 1/4 and the rest? The first one is easy because you multiply \(\displaystyle \dfrac{1}{2}\cdot0\) = \(\displaystyle cos0\) and on the Unit Circle that's at point (1,0) and cosine = x so the value equals to 1. If I do the next row, I have \(\displaystyle \dfrac{1}{2}\cdot\pi\) = 1.57. cos1.57 is not on the unit circle, or if it is I can't calculate it. What do you do in these situations? Thanks.
Also, if you're wondering how I get my \(\displaystyle x\) values in the table, the count is \(\displaystyle \dfrac{1}{4}\) and you usually start with the phase shift, which is \(\displaystyle 0\). So \(\displaystyle 0+\dfrac{1}{4}=\dfrac{1}{4}\), then, \(\displaystyle \dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\), then, \(\displaystyle \dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}\) and so on. So I need easly values so I can make a graph for my final step.
 
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FritoTaco said:
Standard Equation: \(\displaystyle y = A sin(B(x - C))+D \)


  • A: amplitude is A
  • B: period is (2π)/|B|
  • C: phase shift is C/B
  • D: vertical shift is D

Problem: \(\displaystyle y=cos2\pi x\)

My Work:

Amplitude: \(\displaystyle 1\) (a value in front of cos)

Period: \(\displaystyle \dfrac{2\pi}{2\pi}\)
Click to expand...
...which equals "1".

FritoTaco said:
Phase Shift: \(\displaystyle \dfrac{c}{b}\) = \(\displaystyle \dfrac{0}{2\pi}\) = \(\displaystyle 0\)

Vertical Shift: \(\displaystyle 0\) (d value)

Count: \(\displaystyle \dfrac{1}{4}\cdot\dfrac{1}{1}\) (Formula for Count: \(\displaystyle \dfrac{1}{4}\cdot |b|\)}
\(\displaystyle x\)\(\displaystyle y=cosine2\pi x\)
\(\displaystyle 0\)\(\displaystyle cos2\pi(0) = cos0 = 1\)
\(\displaystyle 1/4\)\(\displaystyle cos2\pi(1/4)= ?\)
\(\displaystyle 1/2\)\(\displaystyle cos2\pi(1/2)= ?\)
\(\displaystyle 3/4\)\(\displaystyle cos2\pi(3/4)= ?\)
\(\displaystyle 1\)\(\displaystyle cos2\pi(1) = ?\)
Click to expand...
Simplify the arguments:

\(\displaystyle y\, =\, \cos(2\pi x)\)
\(\displaystyle \cos\left(2\pi\,\left(\frac{1}{4}\right)\right)\, = \,\cos\left(\frac{\pi}{2}\right)\, =\, 1\)
\(\displaystyle \cos\left(2\pi\, \left(\frac{1}{2}\right)\right)\,=\,\cos(\pi)\, =\, ?\)
\(\displaystyle \cos\left(2\pi\,\left(\frac{3}{4}\right)\right)\,=\, \cos\left(\frac{3\pi}{2}\right)\, =\, ?\)
\(\displaystyle \cos(2\pi(1))\,=\, \cos(2\pi)\, =\, ?\)

Then complete the table by using the unit-circle values you've memorized.

FritoTaco said:
How do I calculate 1/4 and the rest? The first one is easy because you multiply \(\displaystyle \dfrac{1}{2}\cdot0\) = \(\displaystyle cos0\) and on the Unit Circle that's at point (1,0) and cosine = x so the value equals to 1. If I do the next row, I have \(\displaystyle \dfrac{1}{2}\cdot\pi\) = 1.57. cos1.57 is not on the unit circle...
Click to expand...
Yes, it is. Try using the exact values, rather than decimal approximations. ;-)
 
\(\displaystyle x\)\(\displaystyle y=cosine2\pi x\)
\(\displaystyle 0\)\(\displaystyle cos2\pi(0) = cos0 = 1\)
\(\displaystyle 1/4\)\(\displaystyle cos2\pi(1/4)= \dfrac{2\pi}{4}=cos\dfrac{\pi}{2}=0\)
\(\displaystyle 1/2\)\(\displaystyle cos2\pi(1/2)= \dfrac{2\pi}{2}=cos\pi=-1\)
\(\displaystyle 3/4\)\(\displaystyle cos2\pi(3/4)= \dfrac{6\pi}{4}=cos\dfrac{3\pi}{2}=0\)
\(\displaystyle 1\)\(\displaystyle cos2\pi(1) = cos2\pi=1\)




Oh, I forgot to simplify the period.

Also, I was trying to use the calculator to get a value for cosine, but I just needed to simplify! I see, thank you, that makes sense.

One question though, when I go another value beyond 1, it will be 5/4. So when plugging it in, I get: \(\displaystyle cos\dfrac{2\pi}{1}\cdot(\dfrac{5\pi}{4})=\dfrac{10\pi}{4}=\dfrac{5\pi}{2}\). Do I subtract \(\displaystyle \pi\) from it to get an easier value? So, \(\displaystyle \dfrac{5\pi}{2}-\dfrac{\pi}{1}=\dfrac{5\pi}{2}-\dfrac{2\pi}{2}=cos\dfrac{3\pi}{2}= 0\)? Thanks.

Another thing, I fixed the count formula, it's \(\displaystyle \dfrac{1}{4}\cdot period\). Luckily the period was the same as the b value so it didn't change anything.
 
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Standard Equation: [FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Math]s[/FONT][FONT=MathJax_Math]i[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]C[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]D[/FONT]y=Asin(B(x−C))+D



  • A: amplitude is A
  • B: period is (2π)/|B|
  • C: phase shift is C/B
  • D: vertical shift is D
Click to expand...

An important detail:

If y = A sin (B(x-C)) + D then the phase shift is C.

If y = A sin (Bx - C) + D then the phase shift is C/B.

(It won't affect your current problem because C=0 anyway, but it will in other problems.)
 
FritoTaco said:
\(\displaystyle x\)\(\displaystyle y=cosine2\pi x\)
\(\displaystyle 0\)\(\displaystyle cos2\pi(0) = cos0 = 1\)
\(\displaystyle 1/4\)\(\displaystyle cos2\pi(1/4)= \dfrac{2\pi}{4}=cos\dfrac{\pi}{2}=0\)
\(\displaystyle 1/2\)\(\displaystyle cos2\pi(1/2)= \dfrac{2\pi}{2}=cos\pi=-1\)
\(\displaystyle 3/4\)\(\displaystyle cos2\pi(3/4)= \dfrac{6\pi}{4}=cos\dfrac{3\pi}{2}=0\)
\(\displaystyle 1\)\(\displaystyle cos2\pi(1) = cos2\pi=1\)




















Oh, I forgot to simplify the period.

Also, I was trying to use the calculator to get a value for cosine, but I just needed to simplify! I see, thank you, that makes sense.

One question though, when I go another value beyond 1, it will be 5/4. So when plugging it in, I get: \(\displaystyle cos\dfrac{2\pi}{1}\cdot(\dfrac{5\pi}{4})=\dfrac{10\pi}{4}=\dfrac{5\pi}{2}\). Do I subtract \(\displaystyle \pi\) from it to get an easier value? So, \(\displaystyle \dfrac{5\pi}{2}-\dfrac{\pi}{1}=\dfrac{5\pi}{2}-\dfrac{2\pi}{2}=cos\dfrac{3\pi}{2}= 0\)? Thanks

No. You need to subtract \(\displaystyle 2\pi \) from it, not to get an "easier value" but to get a coterminal angle.

Another thing, I fixed the count formula, it's \(\displaystyle \dfrac{1}{4}\cdot period\). Luckily the period was the same as the b value so it didn't change anything.
Click to expand...
see comments in red
 
FritoTaco said:
\(\displaystyle x\)\(\displaystyle y\,=\,\cos(2\pi x)\)
\(\displaystyle 0\)\(\displaystyle \cos(2\pi(0)) \,=\, \cos(0)\, =\, 1\)
\(\displaystyle \dfrac{1}{4}\)\(\displaystyle \cos\left(2\pi\,\left(\frac{1}{4}\right)\right)\, = \,\cos\left(\frac{\pi}{2}\right)\, =\, 0\)
\(\displaystyle \dfrac{1}{2}\)\(\displaystyle \cos\left(2\pi\, \left(\frac{1}{2}\right)\right)\,=\,\cos(\pi)\, =\, -1\)
\(\displaystyle \dfrac{3}{4}\)\(\displaystyle \cos\left(2\pi\,\left(\frac{3}{4}\right)\right)\,=\, \cos\left(\frac{3\pi}{2}\right)\, =\, 0\)
\(\displaystyle 1\)\(\displaystyle \cos(2\pi(1))\,=\, \cos(2\pi)\, =\, 1\)

Oh... I was trying to use the calculator to get a value for cosine, but I just needed to simplify! I see, thank you, that makes sense.

One question though, when I go another value beyond 1, it will be 5/4. So when plugging it in, I get:

. . . . .\(\displaystyle \cos\left(\dfrac{2\pi}{1}\, \cdot\, \dfrac{5\pi}{4}\right)\,=\, \cos\left(\dfrac{10\pi}{4}\right)\, =\, \cos\left(\dfrac{5\pi}{2}\right)\)
Click to expand...
I've re-inserted the "cos" (that is, the function) in the last line above. Graders often take points off for "magic" such as disappearing trig functions, especially on tests. Be careful!

FritoTaco said:
Do I subtract \(\displaystyle \pi\) from it to get an easier value?
Click to expand...
Almost; the cycle for cosines repeats every two-pi, not every pi, units.

However, you have the right idea. This is how one uses the periodicity of the trig functions. Since the values of cosine repeat every cycle, find the first-cycle argument that corresponds to the later-cycle value you've been given.

FritoTaco said:
So, \(\displaystyle \dfrac{5\pi}{2}-\dfrac{\pi}{1}=\dfrac{5\pi}{2}-\dfrac{2\pi}{2}=cos\dfrac{3\pi}{2}= 0\)? Thanks.
Click to expand...
The value (5/2)pi is (1/2)pi more than 2pi, and is thus 1/4 of the way into the second cycle. (A cycle is 2pi.) The value (3/2)pi is 3/4 of the way through the first cycle. Why would the cosine have the same value at 1/4 of the way through the cycle as it has at 3/4 of the way through? ;)
 
stapel said:
Why would the cosine have the same value at 1/4 of the way through the cycle as it has at 3/4 of the way through? ;)
Click to expand...

I know this question may be rhetorical, but because at 1/4, it's at 90 degrees making it (0,1) and cosine equal x so it's the same as 3/4's or 270 degrees which is (0, -1).
 
\(\displaystyle x\)\(\displaystyle y=cosine2\pi x\)
\(\displaystyle 0\)\(\displaystyle cos2\pi(0) = cos0 = 1\)
\(\displaystyle 1/4\)\(\displaystyle cos2\pi(1/4)= \dfrac{2\pi}{4}=cos\dfrac{\pi}{2}=0\)
\(\displaystyle 1/2\)\(\displaystyle cos2\pi(1/2)= \dfrac{2\pi}{2}=cos\pi=-1\)
\(\displaystyle 3/4\)\(\displaystyle cos2\pi(3/4)= \dfrac{6\pi}{4}=cos\dfrac{3\pi}{2}=0\)
\(\displaystyle 1\)\(\displaystyle cos2\pi(1) = cos2\pi=1\)
\(\displaystyle \dfrac{5}{4}\)\(\displaystyle \dfrac{2\pi}{1}\cdot\dfrac{5}{4}=\dfrac{10\pi}{4}=\dfrac{5\pi}{2}=\dfrac{5\pi}{2}-\dfrac{2\pi }{1}=\dfrac{5\pi}{2}-\dfrac{4\pi}{2}=cos\dfrac{\pi}{2}=0\)
\(\displaystyle 3/2\)\(\displaystyle \dfrac{2\pi}{1}\cdot\dfrac{3}{2}=\dfrac{6\pi}{2}=cos3\pi=-1\)
\(\displaystyle 7/4\)\(\displaystyle \dfrac{2\pi}{1}\cdot\dfrac{7}{4}=\dfrac{14\pi}{4}=\dfrac{7\pi}{2}=\dfrac{7\pi}{2}-\dfrac{2\pi}{1}=\dfrac{7\pi}{2}-\dfrac{4\pi}{2}=cos\dfrac{3\pi}{2}=0\)
\(\displaystyle 2\)\(\displaystyle \dfrac{2\pi}{1}\cdot\dfrac{2\pi}{1}=cos4\pi=1\)

So would this be right if I added more points beyond 1?
 
FritoTaco said:
\(\displaystyle x\)\(\displaystyle y=cosine2\pi x\)
\(\displaystyle 0\)\(\displaystyle cos2\pi(0) = cos0 = 1\)
\(\displaystyle 1/4\)\(\displaystyle cos2\pi(1/4)= \dfrac{2\pi}{4}=cos\dfrac{\pi}{2}=0\)
\(\displaystyle 1/2\)\(\displaystyle cos2\pi(1/2)= \dfrac{2\pi}{2}=cos\pi=-1\)
\(\displaystyle 3/4\)\(\displaystyle cos2\pi(3/4)= \dfrac{6\pi}{4}=cos\dfrac{3\pi}{2}=0\)
\(\displaystyle 1\)\(\displaystyle cos2\pi(1) = cos2\pi=1\)
\(\displaystyle \dfrac{5}{4}\)\(\displaystyle \dfrac{2\pi}{1}\cdot\dfrac{5}{4}=\dfrac{10\pi}{4}=\dfrac{5\pi}{2}=\dfrac{5\pi}{2}-\dfrac{2\pi }{1}=\dfrac{5\pi}{2}-\dfrac{4\pi}{2}=cos\dfrac{\pi}{2}=0\)
\(\displaystyle 3/2\)\(\displaystyle \dfrac{2\pi}{1}\cdot\dfrac{3}{2}=\dfrac{6\pi}{2}=cos3\pi=-1\)
\(\displaystyle 7/4\)\(\displaystyle \dfrac{2\pi}{1}\cdot\dfrac{7}{4}=\dfrac{14\pi}{4}=\dfrac{7\pi}{2}=\dfrac{7\pi}{2}-\dfrac{2\pi}{1}=\dfrac{7\pi}{2}-\dfrac{4\pi}{2}=cos\dfrac{3\pi}{2}=0\)
\(\displaystyle 2\)\(\displaystyle \dfrac{2\pi}{1}\cdot\dfrac{2\pi}{1}=cos4\pi=1\)

So would this be right if I added more points beyond 1?
Click to expand...
Erm... Would what be right? I mean, you are aware that trig functions repeat forever, right? ;)
 
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