Given:- Function f(x) = sin x
Theorem:- Let f be a differentiable real function defined on an open interval (a, b).
(i) If f’(x) > 0 for all x ∈ (a, b), then f(x) is increasing on (a, b)
(ii) If f’(x) < 0 for all x ∈ (a, b), 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,
That is 4th quadrant, where
hence, Condition for f(x) to be increasing
Thus f(x) is increasing on interval \(\big(-\frac{\pi}{2},\frac{\pi}{2} \big)\)