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

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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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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.
 
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