Find absolute extrma of sin(cos(x))

  • #1
John O' Meara
330
0

Homework Statement


A simple question; find the absolute extrema of f(x)=sin(cos(x)) on the interval [0,2pi].


Homework Equations


The chain rule


The Attempt at a Solution


Assuming that I am correct about the derivative being -cos(cos(x))sin(x), how do you solve
-cos(cos(x))sin(x)=0
 

Answers and Replies

  • #2
Borek
Mentor
29,096
3,707
Assuming you are right - when does x*y equal 0?
 
  • #3
John O' Meara
330
0
The absolute extrema must occur at the endpoints of the interval or at the solutions to the equation f'(x)=0. Do I change the trigonometric product into a sum using sin(A+B)+sin(A-B). When is x*y=0? when either x or y =0.
 
  • #4
Borek
Mentor
29,096
3,707
When is x*y=0? when either x or y =0.

Can't you use this information to solve your equation?
 
  • #5
John O' Meara
330
0
So you are saying that sin(x)=0 => x=0,pi,2pi and how about the cos(cos(x))=0.
 
  • #6
elibj123
240
2
for what t's cos(t)=0?
 
  • #7
John O' Meara
330
0
For what t's cos(t)=0, t= pi/2, 3pi/2
 
  • #8
Borek
Mentor
29,096
3,707
What values can cos(x) take? If so, what values can cos(cos(x)) take?
 
  • #9
John O' Meara
330
0
the cos(x) can take on all values in the interval [-1,1], depending on the value of x.
 
  • #10
Borek
Mentor
29,096
3,707
Take it a step further - can you tell anything about cos([-1,1])?

And start moving on your own, my hand hurts from spoonfeeding.
 
  • #11
John O' Meara
330
0
I know all that, but I can't see it as a solution. Obviously cos([-1,1]) is positive only. I hope your hand doesn't hurt too much. Thanks very much for the help.
 
  • #12
Borek
Mentor
29,096
3,707
x*y=0 if y is always > 0.
 

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