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

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In summary, the wronskian of [cos(theta)]^2 and 1+cos(2theta) is equal to 0. This can be simplified by setting up the determinant with y1=[cos(theta)]^2 and y2=1+cos(2theta), and y1 prime=-2cos(theta)sin(theta) and y2 prime=-2sin(2theta). By applying the chain rule correctly, the determinant simplifies to 0.
  • #1
physstudent1
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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?
 
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  • #2
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).
 
  • #3
...wow I am an idiot! thanks I got it to simplify to 0 now haha sigh.
 

What is the Wronskian of [cos(theta)]^2 and 1+cos(2theta)?

The Wronskian is a mathematical tool used to determine if a set of functions are linearly independent. In this case, the Wronskian of [cos(theta)]^2 and 1+cos(2theta) is equal to -sin(2theta).

How is the Wronskian calculated?

The Wronskian is calculated by taking the determinant of a matrix formed by the derivatives of the given functions. In this case, the matrix would be [2cos(theta)(-sin(theta)), -2sin(2theta)] and the determinant would be -2sin(2theta).

What does a negative Wronskian value indicate?

A negative Wronskian value indicates that the given functions are linearly dependent. In other words, one function can be expressed as a constant multiple of the other function.

Can the Wronskian be used to determine if two functions are orthogonal?

No, the Wronskian is only used to determine linear independence. To determine orthogonality, you would need to use the inner product of the two functions.

Why is the Wronskian important in mathematics?

The Wronskian is important because it allows us to determine if a set of functions are linearly independent, which is a fundamental concept in linear algebra. It also has many applications in differential equations, where it can be used to determine if a set of solutions is complete.

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