Find the minimum value of a trigonometric function

calculusderivativesinequalitymaxima-minima

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.

Best Answer

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.

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