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.
 A: Hint
$$f(x)=\left(1+\frac{1}{\sin ^6(x)}\right)\left(1+\frac{1}{\cos ^6(x)}\right)$$
Using the logarithmic differentiation and using multiple angles leads to
$$\frac{f'(x)}{f(x)}=\frac{1536 \cot (2 x)}{(\cos (4 x)-17) (\cos (4 x)+7)}$$ It seems to be quite simple now.
A: Since $\sin\left(\frac{\pi}{2}-x\right)=\cos(x)$, $f(x)$ is symmetric around $x=\frac{\pi}{4}$.  This implies that $x=\frac{\pi}{4}$ is a critical point.
Writing $f(x)=(1+\sin(x)^{-6})(1+\cos(x)^{-6})$, we find that
$$
f'(x)=-6\sin(x)^{-7}\cos(x)(1+\cos(x)^{-6})+6\cos(x)^{-7}\sin(x)(1+\sin(x)^{-6}).
$$
While this looks a bit messy, let's try for a second derivative.
$$
f''(x)=42\sin(x)^{-8}\cos(x)^2(1+\cos(x)^{-6})
+6\sin(x)^{-6}(1+\cos(x)^{-6})
-36\sin(x)^{-6}\cos(x)^{-6}
+42\cos(x)^{-8}\sin(x)^2(1+\sin(x)^{-6})
+6\cos(x)^{-6}(1+\sin(x)^{-6})
-36\cos(x)^{-6}\sin(x)^{-6}
$$
Again, this looks a bit messy, but we notice that almost all of the terms are positive, which gives us some home that the second derivative might only have one sign.  In fact, we can rewrite the second derivative as
$$
f''(x)=42\sin(x)^{-8}\cos(x)^2
+6\sin(x)^{-8}\cos(x)^{-4}
+6\sin(x)^{-6}(1+\cos(x)^{-6})
+42\cos(x)^{-8}\sin(x)^2
+6\cos(x)^{-8}\sin(x)^{-4}
+6\cos(x)^{-6}(1+\sin(x)^{-6}
+36\cos(x)^{-4}\sin(x)^{-4}(\sin(x)^{-2}-\cos(x)^{-2})^2.
$$
The idea in this step is to absorb the single negative term into a square. The second derivative is always positive, so the first derivative is always increasing.  Hence, the first derivative can have at most one critical point, which we already know is at $x=\frac{\pi}{4}$.
All that's left is to compute the value of $f$ at $x=\frac{\pi}{4}$ and to consider the behavior near the endpoints of $(0,\pi/2)$.
A: I took the straightforward approach of finding stationary points, and it was not complicated.
Let $s=\sin{x}$ and $c=\cos{x}$.
$f'(x)=(1+s^{-6})(6c^{-7}s)+(-6s^{-7}c)(1+c^{-6})$
$=6c^{-7}s^{-7}(s^8+s^2-c^8-c^2)$
$=6c^{-7}s^{-7}(s^8+s^2-(1-s^2)^4-(1-s^2))$
$=6c^{-7}s^{-7}(2s^6-3s^4+3s^2-1)$
$=6c^{-7}s^{-7}(2s^2-1)(s^4-s^2+1)=0$
$s^2=\frac{1}{2}\implies c^2=\frac{1}{2}\implies f(x)=81$
Assume one of these stationary points is a local maximum. The curve approaches $\infty$ near the vertical asymptotes, so there must be a minimum with $f(x)<81$, but such a minimum does not exist. So all the stationary points are minimums.$\text{ }\therefore \sqrt{m}=9$.
A: Without calculus, let $\,a = \sin^2(x)\,$, $\,b = \cos^2(x)\,$, then $\,f(x) = \left(1+\dfrac{1}{a^3}\right)\left(1+\dfrac{1}{b^3}\right)\,$, where $\,a,b \ge 0\,$ and $\,a+b=1\,$. By the generalized means inequalities:

*

*$\sqrt{ab} \le \dfrac{a+b}{2} = \dfrac{1}{2} \quad\implies\quad \dfrac{1}{ab} \ge 4 \quad\implies\quad \dfrac{1}{a^3b^3} \ge 4^3 = 64\;$;


*$\sqrt[3]{\dfrac{2}{\frac{1}{a^3}+ \frac{1}{b^3}}} \le \dfrac{a+b}{2} = \dfrac{1}{2} \quad\implies\quad \dfrac{1}{a^3}+ \dfrac{1}{b^3} \ge 16\;$.
Then the following inequality holds, with equality iff $a=b = \dfrac{1}{2}$, or $|\sin(x)| = |\cos(x)| = \dfrac{1}{\sqrt{2}}$:
$$
f(x) = \left(1+\frac{1}{a^3}\right)\left(1+\frac{1}{b^3}\right) \;=\; 1 + \frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{a^3b^3} \;\ge\; 1 + 16 + 64 = 81
$$
A: I think we don't really need to evaluate a complicated derivative.
The cosine and sine are symmetric around $\frac\pi 4$, and they are both monotone (in opposite directions).
Each of the factors varies monotonously between $2$ and $\infty$, which they achieve on the boundaries of the interval.
Consequently at the boundaries $f$ goes to $\infty$.
It follows that the minimum must be at the symmetric middle at $\frac\pi 4$.
Fill in to find:
$$f\Big(\frac \pi 4\Big)=\left(1+\frac{1}{(\frac 1{\sqrt 2})^6}\right)\left(1+\frac{1}{(\frac 1{\sqrt 2})^6}\right)=9\cdot9$$
So $m=81$ and $\sqrt m=9$.
A: 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.
A: A more efficient answer than my previous approach:
Starting in the same way:
Since $\sin\left(\frac{\pi}{2}-x\right)=\cos(x)$, $f(x)$ is symmetric around $x=\frac{\pi}{4}$.  This implies that $x=\frac{\pi}{4}$ is a critical point.
Writing $f(x)=(1+\sin(x)^{-6})(1+\cos(x)^{-6})$, we find that
$$
f'(x)=-6\sin(x)^{-7}\cos(x)(1+\cos(x)^{-6})+6\cos(x)^{-7}\sin(x)(1+\sin(x)^{-6}).
$$
This derivative can be factored as
$$
f'(x)=6\sin(x)^{-7}\cos(x)^{-7}(-\cos(x)^8-\cos(x)^2+\sin(x)^8+\sin(x)^2).
$$
Since $\cos(x)>\sin(x)$ for $0<x<\frac{\pi}{4}$ and $\cos(x)<\sin(x)$ for $\frac{\pi}{4}<x<\frac{\pi}{2}$, the cosine terms dominate for $0<x<\frac{\pi}{4}$ and the sine terms dominate for $\frac{\pi}{4}<x<\frac{\pi}{2}$.
Therefore, the derivative is negative for $0<x<\frac{\pi}{4}$ and positive for $\frac{\pi}{4}<x<\frac{\pi}{2}$.  Hence, $x=\frac{\pi}{4}$ is the global minimum.
