# Finding the minimum point looks easy with a graph but hard with a formula

My research has lead me to the following function:

$$\frac{\sin(x) [\sin^2(x)\cdot F+ \cos^2(x)/F ]} { 1 - \cos(x) }$$

$F$ is a parameter, and I would like to find the minimum value of this function in the range $x \in [0,\pi/4]$ as a function of $F$.

The graph of the function looks pretty simple - when $F\geq 3$ there is a single minimum point which changes smoothly as $F$ changes (the minimum value seems to converge to 4 when $F$ grows).

However, I couldn't find an analytic expression to the minimum point. I tried to find the derivative of the function using Wolfram Alpha's calculator, but the derivative looks so complicated, I have no idea how to find its zero point.

Is there any trick with which I can find at least an approximate expression to the minimum point as a function of $F$?

• Mutiple top and bottom by $$1+ \cos x$$ will simply the algebra me thinks – Chinny84 Dec 13 '14 at 22:07

## 2 Answers

for the numerator of the first derivative i have got after simplification i have got $$2\,{F}^{2} \left( \cos \left( x \right) \right) ^{4}-3\,{F}^{2} \left( \cos \left( x \right) \right) ^{3}-{F}^{2} \left( \cos \left( x \right) \right) ^{2}-2\, \left( \cos \left( x \right) \right) ^{4}+3\,{F}^{2}\cos \left( x \right) +3\, \left( \cos \left( x \right) \right) ^{3}-{F}^{2}+ \left( \cos \left( x \right) \right) ^{2}-2\,\cos \left( x \right)$$ it is a polynomial in $\cos(x)$

Multiplying top and bottom leads to the following

$$(1+\cos x)\sin x\left[F^2 +\cot ^2 x\right]\frac{1}{F}$$ The derivative with respect to x of the above is $$-\sin^2 x \left[F^2 +\cot ^2 x\right]\frac{1}{F} +(\cos x+\cos^2x)\left[F^2 +\cot ^2 x\right]\frac{1}{F} + (1+\cos x)\sin x\left[2\cot x \left(-\mathrm{csc}^2 x\right)\right]\frac{1}{F}$$ This leads to $$\frac{1}{F\sin^2 x}\left[\left(2\cos^2 x + \cos x -1\right)\left[F^2 +\left(1-F^2\right)\cos^2 x\right]-2(\cos x+\cos^2 x)\right]$$