Help Solving x - sinx for -pi/2 to pi/2

[mohlam12]

hey everybody
i need help solving this.. i have e function a(x) = x - sin x x belongs to -pi/2, pi/2

i need to demonstrate that for x>=0, a(x)>=0

i tried to derivate the function to get 1-cosx which is positive, therefore a(x) is positive too.. but then i'd face problems when it comes to demonstrating the same thing for x<0

any hints ?!

thanks

 

[HallsofIvy]

It's good that you "face problems when it comes to demonstrating the same thing for x< 0"! The "same thing", that a(x)>= 0, is not true for x< 0. If x< 0, then with u= -y, a(y)= -y- sin(-y)= -y+ sin(y)= -(y- sin(y)). Since y> 0, y- sin(y)> 0 so -(y- sin(y))< 0. For x<= 0, a(x)<= 0.

By the way, you can't simply say "1- cos(x) is positive, therefore a(x) is positive too". The fact that the derivative is positive tells you that x- sin(x) is increasing. You have to add the trivial fact that sin(0)= 0 in order to argue that it is then positive.