Composition of trigonometric functions, mean value theorem

zbot1
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Homework Statement



how to show using MVT that cos(cos x) is a contraction.

Homework Equations



| d/dx (cos(cos x)) | = | sin(cos x) sin(x) | < sin 1 < 1

The Attempt at a Solution



Using that relation, the original problem is easily solved. My question is, how do we know:

| sin(cos x) sin(x) | < sin 1 ?

-zbot1
 
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zbot1 said:

Homework Statement



how to show using MVT that cos(cos x) is a contraction.

Homework Equations



| d/dx (cos(cos x)) | = | sin(cos x) sin(x) | < sin 1 < 1

The Attempt at a Solution



Using that relation, the original problem is easily solved. My question is, how do we know:

| sin(cos x) sin(x) | < sin 1 ?

-zbot1

-1<=cos(x)<=1. What does that make the range of values for sin(cos(x))?
 
Thanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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