Question

Prove that the function f given by f(x) = log sin x is strictly increasing on and strictly decreasing on

Answer 1

question is ---> Prove that the function f given by f(x) = log sin x is strictly increasing on (0,π/2) and strictly decreasing on (π/2, π).

solution:- given function f(x) = log sinx
differentiate f(x) with respect to x
f'(x) = $\bf{\frac{1}{sinx}cosx}$
= cosx/sinx = cotx
e.g., f'(x) = cotx

we know, function is strictly increasing only when f'(x) > 0 and function is strictly decreasing only when f'(x) < 0.

$\text{\underline{for strictly increasing,}}$
f'(x) = cotx > 0
it is possible when x belongs to (0,π/2) or (π,3π/2) [ when we take just 0 to 2π values ]
so, f(x) is strictly increasing on (0,π/2)

$\text{\underline{for strictly decreasing}}$
f'(x) = cotx < 0
it is possible when x belongs to (π/2, π) or (3π/2, 2π) [ taking just 0 to 2π values ]
so, f(x) is strictly decreasing on (π/2, π).

hence proved that f(x) = log sin x is strictly increasing on (0,π/2) and strictly decreasing on(π/2, π)