Question

Let $g\left(x\right)=f\left(\sin x\right)+f\left(\cos x\right),f{}^{\prime }\left(\sin x\right)>0,\forall xϵ(0,\pi /2)$.Discuss the monotonicity of $g\left(x\right)$ in $(0,\pi /2)$

Answer 1

$g\left(x\right)=f\left(sin\right(x\left)\right)+f\left(cos\right(x\left)\right)$
$g^{\prime }\left(x\right)=f^{\prime }\left(sin\right(x\left)\right).cos\left(x\right)-f^{\prime }\left(cos\right(x\left)\right).sin\left(x\right)$
Now
$g^{\prime }\left(x\right)>0$ implies
$f^{\prime }\left(sin\right(x\left)\right).cos\left(x\right)-f^{\prime }\left(cos\right(x\left)\right).sin\left(x\right)>0$
$\frac{f^{\prime }\left(sin\right(x\left)\right)}{f^{\prime }\left(cos\right(x\left)\right)}>tan\left(x\right)$.
Now $tan\left(x\right)>0$ in $(0,\frac{\pi }{2})$.
Therefore
$\frac{f^{\prime }\left(sin\right(x\left)\right)}{f^{\prime }\left(cos\right(x\left)\right)}>0$ for $(0,\frac{\pi }{2})$.
If
$f^{\prime }\left(sinx\right)>0$ in $(0,\frac{\pi }{2})$.
Then $f^{\prime }\left(cos\right(x\left)\right)<0$ in the above domain.
Hence
$g\left(x\right)=h\left(x\right)+m\left(x\right)$
Where
$h\left(x\right)=f\left(sin\right(x\left)\right)$ and $m\left(x\right)=f\left(cos\right(x\left)\right)$.
Considering that $h\left(x\right)$ is increasing at the same rate as $m\left(x\right)$ is decreasing in $(0,\frac{\pi }{2})$, we get $g^{\prime }\left(x\right)$ more or less constant.