# Find absolute extrma of sin(cos(x))

John O' Meara

## Homework Statement

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

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

Mentor
Assuming you are right - when does x*y equal 0?

John O' Meara
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.

Mentor
When is x*y=0? when either x or y =0.

Can't you use this information to solve your equation?

John O' Meara
So you are saying that sin(x)=0 => x=0,pi,2pi and how about the cos(cos(x))=0.

elibj123
for what t's cos(t)=0?

John O' Meara
For what t's cos(t)=0, t= pi/2, 3pi/2

Mentor
What values can cos(x) take? If so, what values can cos(cos(x)) take?

John O' Meara
the cos(x) can take on all values in the interval [-1,1], depending on the value of x.

Mentor
Take it a step further - can you tell anything about cos([-1,1])?

And start moving on your own, my hand hurts from spoonfeeding.

John O' Meara
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.

Mentor
x*y=0 if y is always > 0.