# Show that f(x) = sin x is an increasing function on (–π/2, π/2).

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 , 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) = sin x

f’(x) = cosx

Now, as given

cosx> 0

f’(x) > 0

hence, Condition for f(x) to be increasing

Thus f(x) is increasing on interval

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Rapid Quiz challenge - Cell Structure & Cell Division52 mins
Different kind of mappings58 mins
Functions - 0152 mins
Functions - 1047 mins
Functions - 0558 mins
Functions - 0748 mins
Functions - 0253 mins
Functions - 0361 mins
Functions - 0840 mins
Functions - 1252 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses