Question
Prove that the function f given by f(x) = log sinx is increasing on $\left( {0,\frac{\pi }{2}} \right)$ and decreasing on $\left( {\frac{\pi }{2},\pi } \right)$.

Answer

Given: f(x) = log sin x
$\Rightarrow f'\left( x \right) = \frac{1}{{\sin x}}\frac{d}{{dx}}(\sin x) = \frac{1}{{\sin x}}\cos x = \cot x$
On the interval $\left( {0,\frac{\pi }{2}} \right)$ i.e., in first quadrant,
f'(x) = cotx > 0
Therefore, f(x) is strictly increasing on $\left( {0,\frac{\pi }{2}} \right)$.
On the interval $\left( {\frac{\pi }{2},\pi } \right)$ i.e., in second quadrant,
f'(x) = cot x < 0
Therefore, f(x) is strictly decreasing on $\left( {\frac{\pi }{2},\pi } \right)$.

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