Finding Wronskian of [cos(theta)]^2 and 1+cos(2theta)

  • Thread starter Thread starter physstudent1
  • Start date Start date
  • Tags Tags
    Wronskian
Click For Summary
SUMMARY

The discussion centers on calculating the Wronskian of the functions [cos(theta)]^2 and 1 + cos(2theta). The user initially sets up the determinant incorrectly, using -sin(2theta) instead of the correct derivative -2sin(2theta). After correcting this mistake and applying basic trigonometric identities, the user successfully simplifies the Wronskian to 0, confirming the functions are linearly dependent.

PREREQUISITES
  • Understanding of Wronskian determinants in differential equations
  • Knowledge of trigonometric identities, particularly for cos(2theta)
  • Familiarity with differentiation rules, including the chain rule
  • Basic algebraic manipulation skills for simplifying expressions
NEXT STEPS
  • Study the properties of Wronskians in linear algebra
  • Learn about trigonometric identities and their applications in calculus
  • Practice differentiation techniques, focusing on the chain rule
  • Explore examples of linear dependence and independence of functions
USEFUL FOR

Students studying calculus, particularly those focusing on differential equations and linear algebra, as well as educators looking for examples of Wronskian applications.

physstudent1
Messages
267
Reaction score
1

Homework Statement


take the wronskian of [cos(theta)]^2 and 1+cos(2theta)


Homework Equations





The Attempt at a Solution


so I set up the determinant

[cos(theta)]^2 1+cos(2theta) as my y1 and y2 respectively and
-2cos(theta)sin(theta) and -sin(2theta) as my y1 and y2 prime respectively

i get my determinant to be -2cos(theta)sin(theta)*[cos(theta)]^2 +2cos(theta)sin(theta)*(1+cos(2theta)) the answer in the back says it should be 0. So I'm assuming I have to do something with basic trig identities to get it to 0? I can't seem to get it to go to 0 though I've been trying for 30 minutes :(. Can anyone offer some insight?
 
Physics news on Phys.org
physstudent1 said:

Homework Statement


take the wronskian of [cos(theta)]^2 and 1+cos(2theta)


Homework Equations





The Attempt at a Solution


so I set up the determinant

[cos(theta)]^2 1+cos(2theta) as my y1 and y2 respectively and
-2cos(theta)sin(theta) and -sin(2theta) as my y1 and y2 prime respectively

i get my determinant to be -2cos(theta)sin(theta)*[cos(theta)]^2 +2cos(theta)sin(theta)*(1+cos(2theta)) the answer in the back says it should be 0. So I'm assuming I have to do something with basic trig identities to get it to 0? I can't seem to get it to go to 0 though I've been trying for 30 minutes :(. Can anyone offer some insight?

For one thing, I think your y2 prime should be -2sin(2 theta). (You briefly forgot the chain rule).
 
...wow I am an idiot! thanks I got it to simplify to 0 now haha sigh.
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
28
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K