# Using the mean value theorem on trig functions

1. Jul 27, 2011

### magik heepo

1. The problem statement, all variables and given/known data

let g be a function mapping x to xcosx-sinx.
use the mean value theorem to prove that g(x) < 0 for x in (0,pi]

2. Relevant equations

well the function is both continuous and differentiable on the interval so that's a start...

3. The attempt at a solution

basically i thought i'd separate the interval into smaller sections coinciding with the roots of xcosx and sinx, ie:

i ticked off the interval [pi/2,pi] as xcosx is negative and sinx is still positive so g(x) <0;

currently considering xcosx for x in (0,1) and trying to find a turning point but i'm not seeing anywhere i can actually apply the MVT at the moment... halp!

Last edited: Jul 27, 2011
2. Jul 27, 2011

### micromass

Staff Emeritus
Hi magik heepo!

The trick is to show that the derivative of g is always negative on [0,pi]. You should apply the mean value theorem to get that g is decreasing.

3. Jul 27, 2011

### magik heepo

So now I write out the mvt as

((acos(a)-sin(a))-(bcos(b)-sin(b)))/a-b = -csin(c) and state that

since sin(c)>0 for c in [0,pi] and c>0 so -c<0. Also we've set a>b so a-b is positive, leaving

(acos(a)-sin(a))-(bcos(b)-sin(b)) = (b-a)csin(c) < 0

so g(x) <0...

well i'm convinced, thanks again :)