Prove the inequality sin x>x*cos x, if x is in the interval (0,pi).

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.

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