# Minimum value of trigonometric equation

Find the minimum value of the expression $$y=\frac{16-8\sin^{2} 2x +8\cos^{4} x}{\sin^{2} 2x} .$$ When I convert the expression completely into $2x$, cross multiply and make the discriminant of the quadratic equation greater than $0$, I get the minimum value $-\infty$. I know it is wrong, but why?

• This term is equivalent to $6\cot^{2} x + 4 \tan^{2} x$ , hoping no quick error. Done it in a hurry. Where min is 10 – Mann May 13 '15 at 17:35
• Could you please tell me how did you simplify? – user167045 May 13 '15 at 17:39
• I differentiated to get the answer as 9.798 – Hiten May 13 '15 at 17:44
• The middle term doesn't matter. Since $\sin^2 2x=4\sin^2 x\cos^2 x$ we need to minimize $(4+2t^2)/(t(1-t)$ where $t=\cos^2 x$. Routine calculation.. – André Nicolas May 13 '15 at 17:57

$$8\cos^4x = 8\left(\dfrac{1+\cos (2x)}{2}\right)^2 = 8\left(\dfrac{1+2\cos (2x)+ \cos^2(2x)}{4}\right)=2+4\cos (2x)+2\cos^2(2x) = 2+4t+2t^2, t = \cos (2x) \Rightarrow y = \dfrac{16-8(1-t^2)+2+4t+2t^2}{1-t^2} = \dfrac{10+10t^2+4t}{1-t^2}=f(t), -1 \leq t \leq 1$$. Can you take it from here?

$$y=\frac{16-8\sin^{2} 2x +8\cos^{4} x}{\sin^{2} 2x}$$

or $$y=\frac{16}{4}*\sec^2 x \csc^2 x-8 +\frac{8}{4}\frac{\cos^{4} x}{\sin^2 x* \cos^2 x}$$

or

$$y=4(\tan^2 x +1)(\cot^2 x +1)-8+2 \cot^2 x$$

$$y=4(\tan^2 x + \cot^2 x + 2)-8+2 \cot^2 x$$

Which after simplifying gives,

$$y=4\tan^2 x + 6\cot^2 x$$

Edit: as user suggested the answer in comment is not valid. But one here can easily use AM-GM inequality to reach at correct answer

• Minimum of course at $x=\pi/4$ – Mann May 13 '15 at 17:44
• @ Mann : No, it is not so. See my answer done with Maple. – user64494 May 13 '15 at 18:59
• Or well maybe close to it xD – Mann May 13 '15 at 19:11