Given:- Function f(x) = cos² x

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = cos² x

⇒

⇒ f’(x) = 3cosx(–sinx)

⇒ f’(x) = –2sin(x)cos(x)

⇒ f’(x) = –sin2x ; as sin2A = 2sinA cosA

Now, as given

⇒ 2x ∈ (0,π)

⇒ Sin(2x)> 0

⇒ –Sin(2x)< 0

⇒ f’(x) < 0

hence, Condition for f(x) to be decreasing

Thus f(x) is decreasing on interval

Hence proved