Prove the inequality sin x>x*cos x, if x is in the interval (0,pi).
First, we'll create the function f(x) = sin x - x*cos x and we'll have to prove that f(x)>0.
To study the behavior of the function, namely if it is an increasing or a decreasing function, we'll have to do the first derivative test.
A function is strictly increasing if it's first derivative is positive and it is decreasing if it's first derivative is negative.
We'll re-write the function, based on the fact that the sine function is odd:
f(x) = sin x - x*cos x
We'll calculate the first derivative:
f'(x)= (sin x - x*cos x)'
f'(x) = (sin x)' - (x*cos x)'
We notice that the 2nd term is a product, so we'll apply the product rule:
f'(x) = cos x - cos x - x*sin x
We'll eliminate like terms:
f'(x)= -x*sin x
Since the sine function is positive over the interval (0 ; pi), the values of x are positive in this range and the product is negative, the first derivative is negative.
The function is decreasing over the range (0, pi), therefore the inequality x*cos x < sin x is verified.