<s>[INST] Find the minimum value of a trigonometric function If the minimum value of $f\left(x\right)=\left(1+\frac{1}{\sin ^6\left(x\right)}\right)\left(1+\frac{1}{\cos ^6\left(x\right)}\right),\:x\:∈\:\left(0,\:\frac{\pi }{2}\right)$ is $m$, find $\sqrt m$.
How do I differentiate this function without making the problem unnecessarily complicated? If there are any other methods to finding the minimum value I am open to those too.
 [/INST] Let $s\equiv\sin x$ and $c\equiv\cos x$ then
$$
\begin{align}
f(x)
&=
\left(1+{1\over s^6}\right)
\left(1+{1 \over c^6}\right)
\\
&=
{s^6+1\over s^6}{c^6+1 \over c^6}
\\
&=
{(sc)^6+s^6+c^6+1\over (sc)^6}
\\
\end{align}
$$
Now $s^6+c^6=(s^2+c^2)^3-3(sc)^2(s^2+c^2)=1-3(sc)^2$
so
$$
f(x)=1+{2\over (sc)^6}-{3\over (sc)^4}\\
=1+{2\times2^6\over \sin^6 2x}
-{3\times 2^4\over\sin^42x}
$$
and
$$
f'(x)=\cos 2x\times(96/\sin^5 2x-384/\sin^72x)
$$
Since $-1\leq\sin \leq 1$ there are no solutions to $96\sin^2 2x-384=0$, the stationary point must be at $\cos 2x=0$, hence $x=\pi/4$. We can easily see it's a minimum because $f(x)$ consists of positive powers of $1/\sin 2x$,
$$
f(x)=1+{32-24\sin^2 2x\over\sin^7 2x}
$$
and
$\sin^7\leq\sin^2$, and $24\sin^2<32$, and $\sin 2x$ is a maximum at this point.
</s>