# Critial points

1. Oct 18, 2011

### athrun200

1. The problem statement, all variables and given/known data
Q4(b)
http://www.physics.hku.hk/~phys1315/doc/MPI_HW3.pdf" [Broken]

2. Relevant equations

3. The attempt at a solution
After I set fx and fy =0
I get cosx=cosy.
How to solve it?

Last edited by a moderator: May 5, 2017
2. Oct 19, 2011

### lanedance

so I'm assuming f is a function of 2 variables... well what do you know about the cos function?

3. Oct 19, 2011

### athrun200

Sorry for my stupid mistake, I have forgotten to upload the question.

I know about the cos function.
After solving, it should be x=2*Pi*n+y.
However, the question only need to find critial pt within 0<y,x<Pi

4. Oct 19, 2011

### lanedance

ok so what do you get for fx and fy?

5. Oct 19, 2011

### athrun200

fx=cos(x+y)+cos(x)
fy=cos(x+y)+cos(y)

6. Oct 19, 2011

### HallsofIvy

Ys, since both derivatives are 0, you can set them equal to each other and reduce, that is cos(x+y)+ cos(x)= cos(x+y)+ cos(y) reduces to cos(x)= cos(y), but then you still need the equations themselves.

It might help you to know that cos(x+y)= cos(x)cos(y)- sin(x)sin(y). If cos(x)= cos(y) then $sin(x)= \pm sin(y)$

7. Oct 19, 2011

### athrun200

I don't understand this statement...

8. Oct 19, 2011

### HallsofIvy

Do you know that $cos^2(x)+ sin^2(x)= 1$? From that $cos(x)= \pm\sqrt{1- sin^2(x)}$. That means that cos(x)= cos(y) gives $\sqrt{1- sin^2(x)}= \pm\sqrt{1- sin^2(y)}$. Squaring both sides, $1- sin^2(x)= 1- sin^2(y)$ so that $sin^2(x)= sin^2(y)$ and, taking the square root again- $sin(x)= \pm sin(y)$.

My purpose in saying that was after using the identity I gave you to reduce cos(x+y)+ cos(x)= 0 to $cos(x)cos(y)- sin(x)sin(y)+ cos(x)= 0$ you can use the fact that cos(x)= cos(y) and either sin(x)= sin(y) or sin(x)= -sin(y) to reduce to an equation in x only:
if sin(x)= sin(y), then $cos^2(x)- sin^2(x)+ cos(x)= 0$ or
if sin(x)= -sin(y), then $cos^2(x)+ sin^2(x)+ cos(x)= 0$

Now use $sin^2(x)= 1- cos^2(x)$ to reduce those to equations in cos(x) only.

9. Oct 19, 2011

### athrun200

Oh no. How complicated the answer is!
I know these trigonometric identities but I never think of I should use them like this.

I would like to ask what is the problem if I solve cos(x)=cos(y) directly?

10. Oct 19, 2011

### HallsofIvy

You can't. That is a single equation in two variables- there are an infinite number of x, y pairs that satisfy it. y= x= any number is a solution.

11. Oct 19, 2011

### lanedance

i think we may be over complicating it a bit, so rather than equating fx and fy, how about noticing the following

fx=cos(x+y)+cos(x)=0
gives
cos(x)=-cos(x+y)

similarly for fy
fy=cos(x+y)+cos(y)=0
gives
cos(y)=-cos(x+y)

so it is true that for the above to be satisfied requires cos(x)=cos(y), though that only means fx=fy, not fx=fy=0, which is the overall goal.

however it will be instructive to first consider cos(x)=cos(y), to solve this, consider a graph of cos(x), clearly it repeats on every 2.pi, so if y is s solution, so is y+2.pi

Now confine x to (-pi,pi), there will be 2 solutions to cos(x)=cos(y), clearly y=x and using the fact cos is symmetric cos(y)=cos(-y), gives y=-x.

Thus all the solutions are of the form:
y = x+2.pi
y = -x+2.pi

Now going back to the main requirement for fx=fy=0:
cos(x)=-cos(x+y)
cos(y)=-cos(x+y)

can you do a similar process to find allowable (x,y) pairs?

12. Oct 25, 2011