Integration Question

In summary, when finding the area between the curves y=4cosx and y = 4cos(2x) on the interval [0, pi], it is necessary to use a trigonometric identity to solve for the points of intersection. One useful identity is cos(2x) = cos(x), which can be rewritten in several other forms.
  • #1
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Homework Statement



Find the area between the curves y=4cosx and y = 4cos(2x) [0, pi]

Homework Equations





The Attempt at a Solution



I know I need to integrate this, but I get hung up finding the intersection of the two lines so I can split it into two different areas.

4cosx = 4 cos(2x)

0 = cos(2x) - cos(x) I think I need to use a trig identity here, but I'm not sure.
0 = sin(2x)sin(x)

x= pi?
 
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  • #2
char808 said:

Homework Statement



Find the area between the curves y=4cosx and y = 4cos(2x) [0, pi]

Homework Equations





The Attempt at a Solution



I know I need to integrate this, but I get hung up finding the intersection of the two lines so I can split it into two different areas.

4cosx = 4 cos(2x)

0 = cos(2x) - cos(x) I think I need to use a trig identity here, but I'm not sure.
Finding the points of intersection would be a very good idea, and a trig identity would be very useful.
char808 said:
0 = sin(2x)sin(x)
?? How did you go from cos(2x) - cos(x) to sin(2x)sin(x)?
char808 said:
x= pi?
That looks like a guess.
 
  • #3
Yes, I'm at a loss about where to go from here:

0 = cos(2x) - cos(x)

Possibility: (I am admittedly weak when it comes to trig, I could/can figure this out with almost any other function.)

0=cos(2x-x)
0= cos(x)

x= pi/2


?
 
  • #4
char808 said:
0 = cos(2x) - cos(x) I think I need to use a trig identity here, but I'm not sure.

Yes indeed you do. What identities do you know for cos(2x)?
 
  • #5
I don't. If you have a reference link I can go study I would appreciate that. Not covered in the book I have.
 
  • #6
char808 said:
Yes, I'm at a loss about where to go from here:

0 = cos(2x) - cos(x)

Possibility: (I am admittedly weak when it comes to trig, I could/can figure this out with almost any other function.)

0=cos(2x-x)
cos(2x) - cos(x) [itex]\neq[/itex] cos(2x - x)
char808 said:
0= cos(x)

x= pi/2


?
 
  • #7
char808 said:
I don't. If you have a reference link I can go study I would appreciate that. Not covered in the book I have.

Did you study trig at any time? If so, what did you do with your book?

khanacademy.org has a lot of lectures about a variety of math stuff. You might start there.
 
  • #8
Guessing also works. For example, cos(2x)=cos(x) means that, at one point, 2x=x (there are also other points, however). Only one point can satisfy 2x=x.
 
  • #9
Mark44 said:
Did you study trig at any time? If so, what did you do with your book?

khanacademy.org has a lot of lectures about a variety of math stuff. You might start there.

I last studied trig ~7-8 years ago. I probably sold the book.

I will check out the link.
 
  • #10
hi char808 ! :smile:

you need to learn the standard trigonometric identities …

cosA - cosB = 2 sin((A+B)/2) sin((A-B)/2) would help :wink:
 
  • #11
One that I found helpful was
cos(2x) = cos2(x) - sin2(x)

Two other forms of this are
cos(2x) = 2cos2(x) - 1
cos(2x) = 1 - 2sin2(x)

One of these can be used to write your equation cos(2x) = cos(x) as a quadratic in form.
 

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