First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

Then there exists at least one c in the open interval (a,b) such that f’(c) = 0.

Given function is:

⇒ on [ – 1,0]

We know that sine function is continuous and differentiable over R.

Let’s check the values of ‘f’ at an extremum

⇒

⇒

⇒

⇒ f( – 1) = 0

⇒

⇒

⇒ f(0) = 0 – 0

⇒ f(0) = 0

We have got f( – 1) = f(0). So, there exists a cϵ( – 1,0) such that f’(c) = 0.

Let’s find the derivative of the function ‘f’

⇒

⇒

⇒

We have f’(c) = 0

⇒

⇒

⇒

⇒

⇒

⇒

Cosine is positive between , for our convenience we take the interval to be , since the values of the cosine repeats.

We know that value is nearly equal to 1. So, the value of the c nearly equal to 0.

So, we can clearly say that cϵ( – 1,0).

∴ Rolle’s theorem is verified.